Complex Number -

complex number 4 marks(SAQ)

Note: This section is of mathematical interest and students should be encouraged to read it. In particular, it is helpful for them to understand why the complex numbers are not really any more mathematically abstract than the reals. However, knowledge of this section is not required by the current intermediate syllabus and is not necessary for an understanding of how to use complex numbers to solve equations.

The set of complex numbers is a number system, just like the set of reals or the set of integers. Like the reals or even the integers, they were developed to close a mathematical ‘gap’ in another number system. So what is this mathematical gap?

Consider the linear equation $2x-3=0$ . This has one solution in the real numbers: $x=\frac{3}{2}$ .
Now consider the equation $x^{2}+1=0$ . Algebraic manipulation shows that this is equivalent to solving $x=\sqrt{-1}$ . Since squares of real numbers are non-negative we would say that this has ‘no real solutions’.
Consider finally a general quadratic equation $ax^{2}+bx+c=0$ . To solve this, we use the quadratic formula, which gives us $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ where $\Delta =b^{2}-4ac$ is the discriminant. If the discriminant is negative, we would again say that the equation has ‘no real solutions’.

What has happened here is that squares of real numbers are always non-negative. We cannot square a real number and get a negative number. To close this gap, we extend the reals to a number system where squares can also be negative. The easiest way to achieve this is to introduce some number whose square is $-1$ . This number $i$ , is defined by $i^{2}=-1$ . Then if we want to solve $x^{2}=-c$ where $c$ is some positive real, we get $x^{2}=i^{2}c$ , or $x=\pm i\sqrt{c}$ . If we put all numbers of the form $ib$ (where $b$ is real) in our new number system, we can now solve any quadratic equation with real coefficients. Numbers of the form $ib$ ( $b$ real) are known as ‘imaginary numbers’.

In order to turn our set of numbers into a proper number system, we want to introduce some operations so that we can do things with these numbers. The two most fundamental operations of any set (or field) of numbers are addition ( $+$ ) and multiplication ( $\times$ ). We will define these operations properly later. For now, all you need to know is that if you take two real numbers and add them using complex addition, the result is the same as if you added them using real addition. Similarly, if you take two real numbers and multiply them using complex multiplication, the result is the same as if you multiplied them using real multiplication.

Since the real numbers are closed under addition ( $+$ ) and multiplication ( $\times$ ), we want this to hold true for our new number system too. What this means is that if we take any two numbers in the number system (e.g. $a$ and $ib$ ), we want their sum $(a+ib)$ to be in the number system (closure under addition).We also want their product ( $a \times ib$ ) to be in the number system (closure under multiplication). If we take the closure of the real and imaginary numbers, we get all numbers of the form $a+ib$ ( $a$ , $b$ real) must be in the new number system. A little bit of complex number arithmetic shows that this is enough to guarantee closure under addition and multiplication.

So now we have a new set of numbers, the complex numbers $C$ , where each complex number $z$ can be written in the form $a+ib$ (where $a$ , $b$ are real and $i^{2}=-1$ ). The set of complex numbers is closed under addition and multiplication. Furthermore, each real number $a$ is in the set of complex numbers, $a=a+0i$ , so that the real numbers are a subset of the complex numbers (see Figure 1). Finally, any quadratic equation with real coefficients, or even any polynomial with real coefficients, has solutions that can be represented as complex numbers.
Side note: Indeed what makes the complex numbers so powerful is that all degree n polynomials with complex coefficients have n (not necessarily distinct) complex roots. We say that $C$ is ‘algebraically closed’.

Figure 1:

This figure shows that the naturals are a subset of the integers, which are a subset of the rationals, which are a subset of the reals, which are a subset of the complex numbers. The set of numbers in the reals which are not rationals are known as the irrationals.
Students should note that whilst the real numbers and the imaginary numbers are both subsets of the complex numbers, there are complex numbers (such as $1+i$ ) which are neither real nor imaginary.

Arithmetic of Complex Numbers

Recall that every complex number is the sum of a real number and imaginary number. We say that $z$ has a real part, $Re(z)$ , and an imaginary part, $lm(z)$ . If $z=a+ib$ , these are given by:
$Re(z)=a, Im(z)=b$

$z=Re(z)+iIm(z)$
In the above notation, notice how much a complex number looks like a surd (e.g. compare $1+2\sqrt{2}$ and $1+2i=1+2\sqrt{-1}$ ). The only difference is that the number under the square root sign is negative.In fact, when it comes to arithmetic, complex numbers can be treated like surds. This concept is useful when remembering how to add, subtract, multiply or divide complex numbers. Pretend that your complex number is a surd, perform the same operations and everything should work out.

Addition and Subtraction

In order to add two complex numbers, their real and imaginary parts should be added separately. In order to subtract two complex numbers, their real and imaginary parts should be subtracted separately.
Note: Adding $-x$ is the same as subtracting $x$ , and subtracting $-x$ \ is the same as adding $x$ .

Addition: $z_{1}+z_{2}=(Re(z_{1})+Re(z_{2}))+i(Im(z_{1})+Im(z_{2}))$

Or

$(a+ib)+(c+id)=(a+c)+i(b+d)$

Subtraction:

$z_{1}-z_{2}=(Re(z_{1})-Re(z_{2}))+i(Im(z_{1})-Im(z_{2}))$

Or

$(a+ib)-(c+id)=(a-c)+i(b-d)$

Example 1

If the complex numbers $z_{1}$ and $z_{2}$ are given by $z_{1}=1+2i$ and $z_{2}=3-5i$ , determine:

a) $z_{1}+z_{2}$

b) $z_{1}+z_{2}$

Solution 1

a) $z_{1}+z_{2}=(1+2i)+(3-5i)$

$=(1+3)+(2-5)i$ $=4-3i$ b) $z_{1}-z_{2}=(1+2i)-(3-5i)$

$=(1-3)+(2-(-5))i$ $=-2+7i$

Notice that this is just like adding or subtracting surds:

$(1+2\sqrt{2})+(3-5\sqrt{2})=(1+3)\sqrt{2}=4-3\sqrt{2}$ $(1+2\sqrt{2})-(3-5\sqrt{2})=(1-3)+(2-(-5))\sqrt{2}=-2+7\sqrt{2}$

Multiplication of Complex Numbers

In order to multiply two complex numbers, each number should be treated like a surd and the FOIL method applied as appropriate. Students should note that just as the square of the irrational part of a surd gives a rational number, the square of an imaginary number gives a real number.

$(a+ib)\times (c+id)=ac+iad+ibc+i^{2}bd$
$=(ac-bd)+i(ad+bc)$

Example 2

If the complex numbers $z_{1}$ and $z_{2}$ are given by $z_{1}=1+\frac{1}{2}i$ and $z_{2}=3-5i$ , determine $z_{1}z_{2}$ .

Solution 2

$z_{1}z_{2}=(1+\frac{1}{2}i)\times (3-5i)$ $=3-5i+\frac{3}{2}i-\frac{5}{2}i^{2}$ (applying FOIL)
$=(3-(-\frac{5}{2}))+(5+\frac{3}{2})i$ (collecting ‘like’ terms: real and imaginary parts)

$=\frac{11}{2}+\frac{13}{2}i$

Complex Conjugates and their Properties
What are Complex Numbers and why do we need them?

For a given complex number $z=x+iy$ , its complex conjugate is denoted by $\bar{z}$ and is given by:

$\bar{z}=x-iy$

Students are required to know how to determine the complex conjugate of a given complex number. The following example illustrates how this is done.

Example 3

(HSC 1994 Question 2aii)
Let $z=a+ib$ , where $a$ and $b$ are real.Find $\bar{(3iz)}$ in the form $x+iy$ , where $x$ and $y$ are real.

Solution 3

First, express the complex number under the bar as the sum of its real and imaginary components:
$\bar{(3iz)}=\bar{(3i(a+ib))}$
$=\bar{(3ai-3b)}$
$=\bar{(-3b+3ai)}$
To find the conjugate, replace the imaginary component with its negative counterpart
$=-3b-3ai$

Complex conjugates have a number of nice properties:

$\bar{z_{1}+z_{2}}=\bar{z_{1}}+\bar{z_{2}}$
$\bar{z_{1}-z_{2}}=\bar{z_{1}}-\bar{z_{2}}$
$\bar{z_{1}z_{2}}=\bar{z_{1}}\times \bar{z_{2}}$
$\bar{(\frac{z_{1}}{z_{2}})}=\frac{\bar{z_{1}}}{\bar{z_{2}}}$
$\bar{cz}=c\bar{z} for c real$
$\bar{z^{n}}=(\bar{z})^{n}$

Students will eventually be required to prove all these properties (note that proving the 4^th and 6^th properties requires reading ahead to the next few section). The next example illustrates how one would go about proving such results.

Example 4

For any complex numbers $z_{1}$ and $z_{2}$ , prove that $\bar{z_{1}+z_{2}}=\bar{z_{1}}+\bar{z_{2}}$ .

Solution 4

Let $z_{1}=x_{1}+iy_{1}$ , $z_{2}=x_{2}+iy_{2}$

$\bar{z_{1}+z_{2}}= \bar{((x_{1}+x_{2})+i(y_{1}+y_{2}))}$
$=(x_{1}+x_{2})-i(y_{1}+y_{2})$
$=(x_{1}-iy_{1})+(x_{2}-iy_{2})$
$=\bar{z_{1}}+\bar{z_{2}}$

Division of Complex Numbers

One nice thing about complex conjugates is that if a complex number is multiplied with its complex conjugate, the result is a real number: $z\bar{z}=(x+iy)(x-iy)$ $=x^{2}-i^{2}y^{2}$ $=x^{2}+y^{2}$

$z\bar{z}=x^{2}+y^{2}$

Notice how this is analogous to multiplying a surd with its conjugate surd, where the result is a rational number.

This result becomes very useful when taking the reciprocals of complex numbers. Just as multiplying the top and bottom of a surd fraction by the conjugate surd of the denominator rationalises the denominator, multiplying top and bottom of a complex number fraction by the complex conjugate of the denominator makes the denominator real. $\frac{1}{z}=\frac{\bar{z}}{z\bar{z}}=\frac{\bar{z}}{x^{2}+y^{2}}=\frac{x}{x^{2}+y^{2}}-i\frac{y}{x^{2}+y^{2}}$

Similarly, when dividing by a complex number $z$ , simply multiply top and bottom by the complex conjugate of $z$ and then apply complex multiplication to the numerator.

Example 5

If the complex numbers $z_{1}$ and $z_{2}$ are given by $z_{1}=1+\sqrt{2i}$ and $z_{2}=\frac{1}{3}-i$ , determine $\frac{z_{1}}{z_{2}}$ .

Solution 5

$\frac{z_{1}}{z_{2}}=\frac{(1+\sqrt{2i})}{(\frac{1}{3}-i)}$ $=\frac{(1+\sqrt{2i})}{(\frac{1}{3}-i)}\times \frac{(\frac{1}{3}+i)}{(\frac{1}{3}+i)}$ $=\frac{3(1-3\sqrt{2})}{10}+\frac{3(1+\sqrt{2})}{10}i$

Equality of Complex Numbers

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

$z_{1}=z_{2}$

if and only if

$Re(z_{1})=Re(z_{2})$ AND $Im(z_{1})=Im(z_{2})$

Finding the Square Roots of Real and Complex Numbers

To determine the square root of a real number $x$ :

If $=c,$ $c>0$ , its square roots are the real numbers $\sqrt{c}$ and $-\sqrt{c}$ .
If $x=-c, c>0$ , $\sqrt{c}$ and $\sqrt{c}$ are real numbers. Therefore $i\sqrt{c}$ and $i\sqrt{c}$ are imaginary numbers and $(i+\sqrt{c})^{2}=i^{2}(\sqrt{c})^{2}=-c$ , $(-i\sqrt{c})^{2}=i^{2}(-\sqrt{c})^{2}=-c$ . Therefore the square roots of $x$ are $\pm i\sqrt{c}=\pm i\sqrt{-x}$ .

Determining the square root of a complex number $z=x+iy$ is a slightly more complicated process. One way to think about it is that it is equivalent to determining which complex numbers can be squared to give $z$ . That is, find all $a,b$ real such that $(a+ib)^{2}=z=x+iy$ . The general method is as follows:

Use complex multiplication to obtain the real and imaginary parts of the LHS.
$LHS=(a+ib)^{2}=a^{2}+2abi-b^{2}=(a^{2}-b^{2})+(2ab)i$
Since two complex numbers are equal if and only if their real and imaginary parts are equal, equate the real and imaginary parts of the LHS and RHS to obtain two simultaneous equations:
- Comparing real parts gives $a^{2}-b^{2}=x$
- Comparing imaginary parts gives $2ab=y$
Solve for $a$ and $b$ , either by inspection or by substitution.

The following example demonstrates how both variants of the above method can be used to determine the square roots of a complex number.

Example 6

Find all complex solutions to the equation $z^{2}=15-8i$

Solution 6

Let the solutions be of the form $z=x+iy$ .

Then $15-8i=(x+iy)^{2}=x^{2}+2xyi-y^{2}$

so that $x^{2}-y^{2}=15$ and $2xy=-8\to xy=-4$ .

Variant 1:

By inspection, $x=\pm 4$ , $y=ml$ satisfies these simultaneous equations

Checking:

$x^{2}-y^{2}=4^{2}-(-1)^{2}=16-1=15$ and $xy=(4)(-1)=-4$ so that $4-i$ is a solution

$x^{2}-y^{2}=(-4)^{2}-1^{2}=16-1=15$ and $xy=(-4)(1)=-4$ so that $-4+i$ is a solution

Variant 2:

From the second equation, $xy=-4$ so that $y=\frac{-4}{x}$ .

Substituting into the first equation yields $x^{2}-y^{2}=x^{2}-(\frac{-4}{x})^{2}=x^{2}-\frac{16}{x^{2}}=15$

so that $x^{4}-16=15x^{2}$ $\to (x^{2})^{2}-15(x^{2})-16=0\to (x^{2}-16)(x^{2}+1)=0$ and $x^{2}=16$ or $x^{2}=-1$ .

Since the second equation has no real solutions, $x^{2}=16$ and $x=\pm 4$ .

Corresponding y can be determined using $y=\frac{-4}{x}$ so that $y=m1$ .

Therefore $z=4-i$ and $z=-4+i$ are the solutions to the given equation.

Note that the first method, solve by inspection, is a lot shorter and tidier than the second method. Thus if students are able to determine the solutions to the two simultaneous equations by inspection, they are encouraged to use this method. Just to safe, however, they should always include a few lines of checking.

The second method has the advantage of providing a series of steps which is guaranteed to yield an answer. If a student cannot easily determine the answer by inspection they should proceed to solve the equations algebraically.

Students should note that since two simultaneous equations in two variables ( $a$ and $b$ ) yield exactly two (not necessarily distinct) solutions, each complex number has two, not necessarily distinct,complex square roots. This can be seen in Figure 2. Furthermore, as the above example illustrates, if $w=a+ib$ is a square root then $-w=-a-ib$ is also one. Therefore if the two complex square roots are not distinct, $a=-a$ and $b=-b$ so that $w=-w=0$ .

The conclusion is that each non-zero complex number has exactly two distinct complex square roots, and $0$ has exactly one square root.

Using Complex Numbers to Solve Equations

Solving Quadratic Equations with Real Coefficients

Consider a general quadratic equation $ax^{2}+bx+c=0$ .This can be solved by factorising the equation, completing the square or using the quadratic formula. The first method is usually done by inspection and so extending it to encompass complex roots would requires a degree of intuition about the behaviour of complex numbers that is beyond that which is expected from students. The second and third, however, involve algebraic manipulations that can easily be extended by using complex numbers and performing standard arithmetic calculations on complex numbers. The following example demonstrates how the roots of a quadratic equation with real coefficients can be determined using both the extended versions both of completing the square and using the quadratic formula.

Example 7

Find all complex solutions to the equation $x^{2}+2x+3=0$ .

Solution 7

Method 1- Completing the Square:

Starting with the equation $x^{2}+2x+3=0$ , completing the square yields

$(x^{2}+2x+1)+2=0$ $\to (x+1)^{2}=-2$

$x+1=\pm \sqrt{2i}$ $x=-1\pm \sqrt{2i}$ Method 2- Using the Quadratic Formula:

$x^{2}+2x+3=0$ $x=\frac{-2\pm \sqrt{2^{2}-4(1)(3)}}{2}=\frac{-2\pm \sqrt{-8}}{2}=\frac{-2}{2}\pm \frac{2\sqrt{2i}}{2}=-1\pm \sqrt{2i}$ $x=-1\pm \sqrt{2i}$

Solving Quadratic Equations with Complex Coefficients

Consider again a general quadratic equation $az^{2}+bz+c=0$ . This time, $a$ , $b$ and $c$ are complex numbers. However, this doesn’t change the method that we use to determine the possible value of $z$ . Solving by inspection may be a little harder, however completing the square and using the quadratic formula are still equally viable methods.

The following example illustrates how to find the roots of a quadratic equation with complex coefficients.

Example 8

Find all complex solutions to the equation $z^{2}-4(1+i)z+10i=0$ .

Solution 8

Method 1- Completing the Square:

Completing the square on the left hand side of the equation yields

$z^{2}-4(1+i)z+10i$
$=z^{2}-4(1+i)z+(2+2i)^{2}-(2+2i)^{2}+10i$
$=(z-(2+2i))^{2}-(4+8i-4)+10i$
$=(z-2-2i)^{2}+2i$

so that $(z-2-2i)^{2}+2i=0$ and $(z-2-2i)^{2}=-2i$

Let $z-2-2i=x+iy$ , $x$ , $y$ real.

Notice how in this step, we don’t automatically jump to ‘let $z=x+iy$ ’. This is because we know that the quantities that we’re most familiar with dealing with the real and imaginary parts of the number being squared. Of course, it is also fine to let $z=x+iy$ , but the working will probably be a little more tedious and a little less familiar.

Then $(x+iy)^{2}=-2i$ so that $x^{2}-y^{2}=0$ and $2xy=-2$ .

The first equation gives $x^{2}-y^{2}=0\to y=\pm x$

The second equation gives $y=\frac{-1}{x}=m\frac{1}{y}\to y^{2}=m1$ so that $\vert y\vert =1$ and $y=-x=\pm 1$ .

$z-2-2i=1-i$ or $z-2-2i=-1+I$

$z=3+i$ or $z=1+3i$

Method 2- Using the Quadratic Formula:

$z^{2}-4(1+i)z+10i=0$
$z=\frac{4(1+i)\pm \sqrt{(-4-4i)^{2}-4(1)(10i)}}{2}=\frac{4(1+i)\pm \sqrt{(16+32i-16)-40i}}{2}$
$=\frac{4(1+i)}{2}\pm \frac{\sqrt{-8i}}{2}$
$=2(1+i)\pm \sqrt{-2i}$

Here the square roots of $-2i$ need to be obtained. This can be done using the same method as the one used above, where all pairs of $x$ and $y$ satisfying $(x+iy)^{2}=-2i$ are determined. This leads to solving the simultaneous equations $x^{2}-y^{2}=0$ and $2xy=-2$ .

The solutions to the simultaneous equations can be attained by inspection or through algebraic manipulation and yield $y=-x=\pm 1$ .

$z=2(1+i)+(1-i)$ or $z=2(1+i)+(-1+i)$

$z=3+i$ or $z=1+3i$

The Argand Diagram

For every real $a$ and $b$ there exists a complex number given by $a+ib$ . From before, if the real parts and the imaginary parts of two complex numbers are equal, then they are the same number. This means that for every real $a$ and $b$ there exists a unique complex number given by $a+ib$ . In other words, there is a 1-1 correspondence between ordered pairs of reals $(a,b)$ and the complex numbers. Extending the idea a little further, there is a 1-1 correspondence between the complex numbers and the points in the x-y plane.

$z=a+ib\leftrightarrow (a, b)$

This correspondence immediately suggests a way to diagrammatically represent a complex number $z$ : on the x-y plane, with x and y coordinates corresponding to $Re(z)$ and $Im(z)$ respectively. Such a diagram is known as an ‘Argand diagram’ and the plane on which the points are plotted is sometimes referred to as the ‘Argand plane’ or the ‘complex plane’. The x-axis, along which the value of $Re(z)$ can be determined, is referred to as the ‘real axis’ and the y-axis is referred to as the ‘imaginary axis’.

The following example demonstrates how to represent complex numbers of the form $z=a+ib$ on an Argand diagram.

Example 9

$a=rcos\theta$ and $b=rsin\theta$

If $\theta$ is restricted by $-\pi <\theta \le \pi$ , the value of $\theta$ is unique. If not, there is an infinite family of solutions given by adding $2n\pi$ ( $n$ an integer) to the value of $\theta$ between $-\pi$ and $\pi$ .
Note that if $a=b=0$ (i.e. $z=0$ ), $r=0$ and all values of $\theta$ are possible.

For a compvvWhat are Complex Numbers and why do we need them?

Note: This section is of mathematical interest and students should be encouraged to read it. In particular, it is helpful for them to understand why the complex numbers are not really any more mathematically abstract than the reals. However, knowledge of this section is not required by the current HSC syllabus and is not necessary for an understanding of how to use complex numbers to solve equations.

The set of complex numbers is a number system, just like the set of reals or the set of integers. Like the reals or even the integers, they were developed to close a mathematical ‘gap’ in another number system. So what is this mathematical gap?

Consider the linear equation $2x-3=0$ . This has one solution in the real numbers: $x=\frac{3}{2}$ .
Now consider the equation $x^{2}+1=0$ . Algebraic manipulation shows that this is equivalent to solving $x=\sqrt{-1}$ . Since squares of real numbers are non-negative we would say that this has ‘no real solutions’.
Consider finally a general quadratic equation $ax^{2}+bx+c=0$ . To solve this, we use the quadratic formula, which gives us $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ where $\Delta =b^{2}-4ac$ is the discriminant. If the discriminant is negative, we would again say that the equation has ‘no real solutions’.

What has happened here is that squares of real numbers are always non-negative. We cannot square a real number and get a negative number. To close this gap, we extend the reals to a number system where squares can also be negative. The easiest way to achieve this is to introduce some number whose square is $-1$ . This number $i$ , is defined by $i^{2}=-1$ . Then if we want to solve $x^{2}=-c$ where $c$ is some positive real, we get $x^{2}=i^{2}c$ , or $x=\pm i\sqrt{c}$ . If we put all numbers of the form $ib$ (where $b$ is real) in our new number system, we can now solve any quadratic equation with real coefficients. Numbers of the form $ib$ ( $b$ real) are known as ‘imaginary numbers’.

In order to turn our set of numbers into a proper number system, we want to introduce some operations so that we can do things with these numbers. The two most fundamental operations of any set (or field) of numbers are addition ( $+$ ) and multiplication ( $\times$ ). We will define these operations properly later. For now, all you need to know is that if you take two real numbers and add them using complex addition, the result is the same as if you added them using real addition. Similarly, if you take two real numbers and multiply them using complex multiplication, the result is the same as if you multiplied them using real multiplication.

Since the real numbers are closed under addition ( $+$ ) and multiplication ( $\times$ ), we want this to hold true for our new number system too. What this means is that if we take any two numbers in the number system (e.g. $a$ and $ib$ ), we want their sum $(a+ib)$ to be in the number system (closure under addition).We also want their product ( $a \times ib$ ) to be in the number system (closure under multiplication). If we take the closure of the real and imaginary numbers, we get all numbers of the form $a+ib$ ( $a$ , $b$ real) must be in the new number system. A little bit of complex number arithmetic shows that this is enough to guarantee closure under addition and multiplication.

So now we have a new set of numbers, the complex numbers $C$ , where each complex number $z$ can be written in the form $a+ib$ (where $a$ , $b$ are real and $i^{2}=-1$ ). The set of complex numbers is closed under addition and multiplication. Furthermore, each real number $a$ is in the set of complex numbers, $a=a+0i$ , so that the real numbers are a subset of the complex numbers (see Figure 1). Finally, any quadratic equation with real coefficients, or even any polynomial with real coefficients, has solutions that can be represented as complex numbers.
Side note: Indeed what makes the complex numbers so powerful is that all degree n polynomials with complex coefficients have n (not necessarily distinct) complex roots. We say that $C$ is ‘algebraically closed’.

Figure 1:

This figure shows that the naturals are a subset of the integers, which are a subset of the rationals, which are a subset of the reals, which are a subset of the complex numbers. The set of numbers in the reals which are not rationals are known as the irrationals.
Students should note that whilst the real numbers and the imaginary numbers are both subsets of the complex numbers, there are complex numbers (such as $1+i$ ) which are neither real nor imaginary.

Arithmetic of Complex Numbers

Recall that every complex number is the sum of a real number and imaginary number. We say that $z$ has a real part, $Re(z)$ , and an imaginary part, $lm(z)$ . If $z=a+ib$ , these are given by:

$Re(z)=a, Im(z)=b$
$z=Re(z)+iIm(z)$

In the above notation, notice how much a complex number looks like a surd (e.g. compare $1+2\sqrt{2}$ and $1+2i=1+2\sqrt{-1}$ ). The only difference is that the number under the square root sign is negative.In fact, when it comes to arithmetic, complex numbers can be treated like surds. This concept is useful when remembering how to add, subtract, multiply or divide complex numbers. Pretend that your complex number is a surd, perform the same operations and everything should work out.

Addition and Subtraction

In order to add two complex numbers, their real and imaginary parts should be added separately. In order to subtract two complex numbers, their real and imaginary parts should be subtracted separately.
Note: Adding $-x$ is the same as subtracting $x$ , and subtracting $-x$ \ is the same as adding $x$ .

Addition: $z_{1}+z_{2}=(Re(z_{1})+Re(z_{2}))+i(Im(z_{1})+Im(z_{2}))$

$(a+ib)+(c+id)=(a+c)+i(b+d)$

Subtraction:

$z_{1}-z_{2}=(Re(z_{1})-Re(z_{2}))+i(Im(z_{1})-Im(z_{2}))$

$(a+ib)-(c+id)=(a-c)+i(b-d)$

Example 1

If the complex numbers $z_{1}$ and $z_{2}$ are given by $z_{1}=1+2i$ and $z_{2}=3-5i$ , determine:

a) $z_{1}+z_{2}$

b) $z_{1}+z_{2}$

Solution 1

a) $z_{1}+z_{2}=(1+2i)+(3-5i)$

$=(1+3)+(2-5)i$ $=4-3i$ b) $z_{1}-z_{2}=(1+2i)-(3-5i)$

$=(1-3)+(2-(-5))i$ $=-2+7i$

Notice that this is just like adding or subtracting surds:

$(1+2\sqrt{2})+(3-5\sqrt{2})=(1+3)\sqrt{2}=4-3\sqrt{2}$ $(1+2\sqrt{2})-(3-5\sqrt{2})=(1-3)+(2-(-5))\sqrt{2}=-2+7\sqrt{2}$

Multiplication of Complex Numbers

In order to multiply two complex numbers, each number should be treated like a surd and the FOIL method applied as appropriate. Students should note that just as the square of the irrational part of a surd gives a rational number, the square of an imaginary number gives a real number.

$(a+ib)\times (c+id)=ac+iad+ibc+i^{2}bd$
$=(ac-bd)+i(ad+bc)$

Example 2

If the complex numbers $z_{1}$ and $z_{2}$ are given by $z_{1}=1+\frac{1}{2}i$ and $z_{2}=3-5i$ , determine $z_{1}z_{2}$ .

Solution 2

$z_{1}z_{2}=(1+\frac{1}{2}i)\times (3-5i)$ $=3-5i+\frac{3}{2}i-\frac{5}{2}i^{2}$ (applying FOIL)
$=(3-(-\frac{5}{2}))+(5+\frac{3}{2})i$ (collecting ‘like’ terms: real and imaginary parts)

$=\frac{11}{2}+\frac{13}{2}i$

Complex Conjugates and their Properties

For a given complex number $z=x+iy$ , its complex conjugate is denoted by $\bar{z}$ and is given by:

$\bar{z}=x-iy$

Students are required to know how to determine the complex conjugate of a given complex number. The following example illustrates how this is done.

Example 3

(HSC 1994 Question 2aii)
Let $z=a+ib$ , where $a$ and $b$ are real.Find $\bar{(3iz)}$ in the form $x+iy$ , where $x$ and $y$ are real.

Solution 3

Complex conjugates have a number of nice properties:

Example 4

For any complex numbers $z_{1}$ and $z_{2}$ , prove that $\bar{z_{1}+z_{2}}=\bar{z_{1}}+\bar{z_{2}}$ .

Solution 4

Let $z_{1}=x_{1}+iy_{1}$ , $z_{2}=x_{2}+iy_{2}$

$\bar{z_{1}+z_{2}}= \bar{((x_{1}+x_{2})+i(y_{1}+y_{2}))}$
$=(x_{1}+x_{2})-i(y_{1}+y_{2})$
$=(x_{1}-iy_{1})+(x_{2}-iy_{2})$
$=\bar{z_{1}}+\bar{z_{2}}$

Division of Complex Numbers

One nice thing about complex conjugates is that if a complex number is multiplied with its complex conjugate, the result is a real number: $z\bar{z}=(x+iy)(x-iy)$ $=x^{2}-i^{2}y^{2}$ $=x^{2}+y^{2}$

$z\bar{z}=x^{2}+y^{2}$

Notice how this is analogous to multiplying a surd with its conjugate surd, where the result is a rational number.

Similarly, when dividing by a complex number $z$ , simply multiply top and bottom by the complex conjugate of $z$ and then apply complex multiplication to the numerator.

Example 5

If the complex numbers $z_{1}$ and $z_{2}$ are given by $z_{1}=1+\sqrt{2i}$ and $z_{2}=\frac{1}{3}-i$ , determine $\frac{z_{1}}{z_{2}}$ .

Solution 5

Equality of Complex Numbers

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

$z_{1}=z_{2}$

if and only if

$Re(z_{1})=Re(z_{2})$ AND $Im(z_{1})=Im(z_{2})$

Finding the Square Roots of Real and Complex Numbers

To determine the square root of a real number $x$ :

If $=c,$ $c>0$ , its square roots are the real numbers $\sqrt{c}$ and $-\sqrt{c}$ .
If $x=-c, c>0$ , $\sqrt{c}$ and $\sqrt{c}$ are real numbers. Therefore $i\sqrt{c}$ and $i\sqrt{c}$ are imaginary numbers and $(i+\sqrt{c})^{2}=i^{2}(\sqrt{c})^{2}=-c$ , $(-i\sqrt{c})^{2}=i^{2}(-\sqrt{c})^{2}=-c$ . Therefore the square roots of $x$ are $\pm i\sqrt{c}=\pm i\sqrt{-x}$ .

Use complex multiplication to obtain the real and imaginary parts of the LHS.
$LHS=(a+ib)^{2}=a^{2}+2abi-b^{2}=(a^{2}-b^{2})+(2ab)i$
Since two complex numbers are equal if and only if their real and imaginary parts are equal, equate the real and imaginary parts of the LHS and RHS to obtain two simultaneous equations:
- Comparing real parts gives $a^{2}-b^{2}=x$
- Comparing imaginary parts gives $2ab=y$
Solve for $a$ and $b$ , either by inspection or by substitution.

The following example demonstrates how both variants of the above method can be used to determine the square roots of a complex number.

Example 6

Find all complex solutions to the equation $z^{2}=15-8i$

Solution 6

Let the solutions be of the form $z=x+iy$ .

Then $15-8i=(x+iy)^{2}=x^{2}+2xyi-y^{2}$

so that $x^{2}-y^{2}=15$ and $2xy=-8\to xy=-4$ .

Variant 1:

By inspection, $x=\pm 4$ , $y=ml$ satisfies these simultaneous equations

Checking:

$x^{2}-y^{2}=4^{2}-(-1)^{2}=16-1=15$ and $xy=(4)(-1)=-4$ so that $4-i$ is a solution

$x^{2}-y^{2}=(-4)^{2}-1^{2}=16-1=15$ and $xy=(-4)(1)=-4$ so that $-4+i$ is a solution

Variant 2:

From the second equation, $xy=-4$ so that $y=\frac{-4}{x}$ .

Substituting into the first equation yields $x^{2}-y^{2}=x^{2}-(\frac{-4}{x})^{2}=x^{2}-\frac{16}{x^{2}}=15$

so that $x^{4}-16=15x^{2}$ $\to (x^{2})^{2}-15(x^{2})-16=0\to (x^{2}-16)(x^{2}+1)=0$ and $x^{2}=16$ or $x^{2}=-1$ .

Since the second equation has no real solutions, $x^{2}=16$ and $x=\pm 4$ .

Corresponding y can be determined using $y=\frac{-4}{x}$ so that $y=m1$ .

Therefore $z=4-i$ and $z=-4+i$ are the solutions to the given equation.

The conclusion is that each non-zero complex number has exactly two distinct complex square roots, and $0$ has exactly one square root.

Using Complex Numbers to Solve Equations

Solving Quadratic Equations with Real Coefficients

Example 7

Find all complex solutions to the equation $x^{2}+2x+3=0$ .

Solution 7

Method 1- Completing the Square:

Starting with the equation $x^{2}+2x+3=0$ , completing the square yields

$(x^{2}+2x+1)+2=0$ $\to (x+1)^{2}=-2$

$x+1=\pm \sqrt{2i}$ $x=-1\pm \sqrt{2i}$ Method 2- Using the Quadratic Formula:

$x^{2}+2x+3=0$ $x=\frac{-2\pm \sqrt{2^{2}-4(1)(3)}}{2}=\frac{-2\pm \sqrt{-8}}{2}=\frac{-2}{2}\pm \frac{2\sqrt{2i}}{2}=-1\pm \sqrt{2i}$ $x=-1\pm \sqrt{2i}$

Solving Quadratic Equations with Complex Coefficients

The following example illustrates how to find the roots of a quadratic equation with complex coefficients.

Example 8

Find all complex solutions to the equation $z^{2}-4(1+i)z+10i=0$ .

Solution 8

Method 1- Completing the Square:

Completing the square on the left hand side of the equation yields

$z^{2}-4(1+i)z+10i$
$=z^{2}-4(1+i)z+(2+2i)^{2}-(2+2i)^{2}+10i$
$=(z-(2+2i))^{2}-(4+8i-4)+10i$
$=(z-2-2i)^{2}+2i$

so that $(z-2-2i)^{2}+2i=0$ and $(z-2-2i)^{2}=-2i$

Let $z-2-2i=x+iy$ , $x$ , $y$ real.

Then $(x+iy)^{2}=-2i$ so that $x^{2}-y^{2}=0$ and $2xy=-2$ .

The first equation gives $x^{2}-y^{2}=0\to y=\pm x$

The second equation gives $y=\frac{-1}{x}=m\frac{1}{y}\to y^{2}=m1$ so that $\vert y\vert =1$ and $y=-x=\pm 1$ .

$z-2-2i=1-i$ or $z-2-2i=-1+I$

$z=3+i$ or $z=1+3i$

Method 2- Using the Quadratic Formula:

$z^{2}-4(1+i)z+10i=0$
$z=\frac{4(1+i)\pm \sqrt{(-4-4i)^{2}-4(1)(10i)}}{2}=\frac{4(1+i)\pm \sqrt{(16+32i-16)-40i}}{2}$
$=\frac{4(1+i)}{2}\pm \frac{\sqrt{-8i}}{2}$
$=2(1+i)\pm \sqrt{-2i}$

The solutions to the simultaneous equations can be attained by inspection or through algebraic manipulation and yield $y=-x=\pm 1$ .

$z=2(1+i)+(1-i)$ or $z=2(1+i)+(-1+i)$

$z=3+i$ or $z=1+3i$

The Argand Diagram

$z=a+ib\leftrightarrow (a, b)$

The following example demonstrates how to represent complex numbers of the form $z=a+ib$ on an Argand diagram.

Example 9

lex number $\theta$ , the corresponding values of $r$ and $\theta$ are given special names.

The quantity $r=\sqrt{a^{2}+b^{2}}$ is called the modulus of $r=\sqrt{a^{2}+b^{2}}$ and is denoted by |z|.
All possible values of $\theta$ are called arguments of $z$ and are denoted by $arg(z)$ . The unique value of $\theta$ that is between $-\pi$ and $\pi$ is known as the principal argument of $z$ and is denoted by $Arg(z)$ .
Note: The capitalised A is very important: it differentiates the principal argument from other arguments.
- If $\theta$ is not the principal argument then it is incorrect to write $\theta =Arg(z)$ .
- If $\theta =Arg(z)$ it is still correct to write $\theta =arg(z)$ , but some information is lost in doing so.
Note also that the complex number 0 does not have a defined principal argument. By convention, 0 does not have an argument.

Modulus: $\vert z\vert =\sqrt{Re(z)^{2}+Im(z)^{2}}=\sqrt{z\bar{z}}$

Arguments:

$arg(z)=\theta$

where $Re(z)=\vert z\vert \cos \theta$ and $Im(z)=\vert z\vert \sin \theta$

Principal Argument:

$Arg(z)=\theta$

where $Re(z)=\vert z\vert \cos \theta , Im(z)=\vert z\vert \sin \theta$ and $-\pi <\theta \le \pi$

Just as two complex numbers are equal if and only if their real and imaginary parts are both equal, two complex numbers are equal if and only if their moduli and principal arguments are equal. (The obvious exception is the complex number 0, which does not have a defined principal argument.)

The modulus of a complex number of the form $z=a+ib$ is easily determined. Students tend to struggle more with determining a correct value for the argument. One method is to find the principal argument using a diagram and some trigonometry. The following example illustrates how this can be done.

Example 10

For the complex number $z=1-\sqrt{3i}$ , find $|z|$ and $arg(z)$ .

Solution 10

$\vert z\vert =\sqrt{(1)^{2}+(-\sqrt{3})^{2}}=\sqrt{1+3}=\sqrt{4}=2$ solution ten of modulus and argument

Let $Arg(z)=\theta$ . From the diagram, $-\frac{\pi }{2}<\theta <0$ and $tan\theta =-\sqrt{3}$ .

Most students would solve the trigonometric equation by plugging $\tan ^{-1}(-\sqrt{3})$ into their calculator. This gives $\theta =-\frac{\pi }{3}+2k\pi$ , not necessarily $\theta =-\frac{\pi }{3}$ . In this case, we see from checking possible values of $\theta$ that $\theta =-\frac{\pi }{3}$ is indeed the value of the principal argument.

$arg(z)=-\frac{\pi }{3}$

From the relationships between $|z|$ , $\theta$ , $Re(z)$ and $Im(z)$ , it can be determined that a complex number $z$ can also be written in the form $z=|z|(cos\theta +isin\theta )$ or $z=|z|cis\theta$ , where $cis\theta$ is standard shorthand notation for $cos\theta +isin\theta$ . This is known as the ‘modulus-argument form’, or ‘mod-arg form’ of a complex number.

Mod – Arg Form: $z=\vert z\vert cis \theta$

The next example demonstrates the process in a typical conversion between Cartesian form and mod-arg form, and illustrates how to decide on a form to use.

Example 11

Let $z=a+ib$ , where $a$ and $b$ are real numbers and $a\ne 0$

i. Express $|z|$ and $tan(arg(z))$ in terms of $a$ and $b$ .

ii. Express $\frac{z}{3+5i}$ in the form $x+iy$ , where $x$ and $y$ are real.

Solution 11

i. $\vert z\vert =\sqrt{a^{2}+b^{2}}$
Let $arg(z)=\theta$ . Then $\vert z\vert cos\theta =a$ , $\vert z\vert sin\theta =b$ so that dividing the second equation by the first gives $tan\theta =\frac{|z|sin\theta }{|z|cos\theta }=\frac{b}{a}$

$\tan (arg(z))=\frac{b}{a}$ ii. This part involves division of complex numbers, which is usually simpler to do using mod-arg form. However, it presents everything in Cartesian form, so it makes sense to do the question in that form.
$\frac{z}{3+5i}=\frac{a+ib}{3+5i}\times \frac{3-5i}{3-5i}$ (multiplying top and bottom by the complex conjugate of the denominator)
$=\frac{3a-5ai+3bi+5b}{3^{2}+5^{2}}$ (using multiplication of complex numbers)

$=\frac{(3a+5b)+i(3b-5a)}{34}$

$=\frac{3a+5b}{34}+i\frac{3b-5a}{34}$

Multiplication and Division of Complex Numbers and Properties of the Modulus and Argument

Students should ensure that they are familiar with how to transform between the Cartesian form and the mod-arg form of a complex number. This is because questions involving complex numbers are often much simpler to solve using one form than the other form. In particular, when complex numbers are multiplied or divided, their moduli and arguments have much simpler and cleaner relationships than their real and imaginary components. This means that in questions involving multiplication and division of complex numbers, it is often recommended to convert to modulus-argument form and complete the calculation before converting back (if conversion is necessary).

The relationships that students are required to know how to derive are:

Properties of the Modulus: $\vert z_{1}z_{2}\vert =\vert z_{1}\vert \times \vert z_{2}\vert$

$\vert \frac{z_{1}}{z_{2}}\vert =\frac{\vert z_{1}\vert }{\vert z_{2}\vert }$

$\vert z^{n}\vert =\vert z\vert ^{n}$

Properties of the Argument: $arg(z_{1}z_{2})=arg(z_{1})+arg(z_{2})$

$arg(\frac{z_{1}}{z_{2}})=arg(z_{1})-arg(z_{2})$

$arg(z^{n})=narg(z)$

Students should be able to prove all of these relations. The following example illustrates how to prove the first property in each box.

Example 12

Let $z_{1}=r_{1}cis\theta _{1}$ and $z_{2}=r_{2}cis\theta _{2}$ .

a) Express $z_{1}z_{2}$ in modulus-argument form

b) Hence show that $\vert z_{1}z_{2}\vert =|z_{1}|\times |z_{2}|$ and $arg(z_{1}z_{2})=arg(z_{1})+arg?(z_{2})$

Solution 12

a) $z_{1}z_{2}=\lbrack r_{1}(\cos \theta _{1}+isin\theta _{1})\rbrack [r_{1}(cos\theta _{2}+isin\theta _{2})]$

$=r_{1}r_{2}(cos\theta _{1}\cos \theta _{2}+icos\theta _{1}\sin \theta _{2}+isin\theta _{1}\cos \theta _{2}-sin\theta _{1}\sin \theta _{2}$ $=r_{1}r_{2}((\cos \theta _{1}\cos \theta _{2}-sin\theta _{1}\sin \theta _{2})+i(cos\theta _{1}\sin \theta _{2}+sin\theta _{1}\cos \theta� _{2}))$ $=r_{1}r_{2}(\cos (\theta _{1}+\theta _{2})+isin(\theta _{1}+\theta _{2}))$ $=r_{1}r_{2}cis(\theta _{1}+\theta _{2})$ b) From (a), $\vert z_{1}z_{2}\vert =r_{1}r_{2}=\vert z_{1}\vert \times |z_{2}|$ as required.
Similarly from (a), $arg(z_{1}z_{2})=\theta _{1}+\theta _{2}=arg(z_{1})+arg(z_{2})$ as required.

The proof for the second property is similar to that of the first. For positive integers $n>0$ the third property can be derived using the first property and mathematical induction. For negative integers $n<0$ the third property can be derived using $-n$ and the second property. These proofs are left as an exercise to the reader.

The following worked examples illustrate how the above properties can be used in calculations in mod-arg form.

Example 13

Given $z=\frac{\sqrt{3}+i}{1+i}$ ,

i. Find the argument and modulus of $z$

ii. Find the smallest positive integer $n$ such that $z^{n}$ is real

Solution 13

$\vert z\vert =\frac{\vert \sqrt{3}+i\vert }{\vert 1+i\vert }$ (using the property that ) $\vert \frac{z_{1}}{z_{2}}\vert =\vert \frac{z_{1}}{z_{2}}\vert$

$=\frac{\sqrt{ (\sqrt{3})^{2}+1^{2}}}{\sqrt{1^{2}+1^{2}}}$

$=\frac{\sqrt{4}}{\sqrt{2}}$

$=\sqrt{2}$
$arg(z)=arg(\sqrt{3}+i)-arg(1+i)$ (using the property that $arg(\frac{z_{1}}{z_{2}})=arg(z_{1})-arg(z_{2})$ )
$=\tan ^{-1}(\sqrt{3})-\tan ^{-1}(1)$ (where using inverse tan is ok because both arguments are in the first quadrant)

$=\frac{\pi }{3}-\frac{\pi }{4}$ $=\frac{\pi }{12}$ ii.

If $z^{n}$ is real then $arg(z^{n})=2k\pi$ for some integer $k$ .
Since $arg(z^{n})=narg(z)=\frac{n\pi }{12}$ (using the property that $arg(z^{n})=narg(z)$ )we want the smallest integer n such that $\frac{n}{12}$ is a multiple of 2. This gives $n=24$ .
The smallest positive integer $n$ such that $z^{n}$ is real is $n=24$ .

Example 14 (HSC 1997 Q2d)

Let $w=\frac{3+4i}{5}$ and $z=\frac{5+12i}{13}$ , so that $\vert w\vert =\vert z\vert =1$ .

i. Find $wz$ and $wz$ in the form $x+iy$

ii. Hence find two distinct ways of writing $65^{2}$ as the sum $a^{2}+b^{2}$ , where $a$ and $b$ are integers and $0<a<b$ .

Solution 14

$wz=\frac{3+4i}{5}\times \frac{5+12i}{13}=\frac{15+36i+20i-48}{65}=\frac{-33}{65}+\frac{56}{65}i$ $wz=\frac{3+4i}{5}\times \frac{5-12i}{13}=\frac{15-36i+20i+48}{65}=\frac{63}{65}-\frac{16}{65}i$ ii.

We note two interesting things given in the question: we want to find $65^{2}$ , where 65 is in the denominator of all the terms in $wz$ and $wz$ , and $\vert w\vert =\vert z\vert =1$ . This immediately makes us think to use the moduli of $wz$ and $wz$ .

By the properties of the modulus, $\vert wz\vert =\vert w\vert \vert z\vert =1$ and $\vert w\overline{z}\vert =\vert w\vert \vert \overline{z}\vert =\vert w\vert \vert z\vert =1$
$(\frac{-33}{65})^{2}+ (\frac{56}{65})^{2}=1$ so that $33^{3}+56^{2}=65^{2}$ .

Similarly $63^{2}+16^{2}=65^{2}$ .

Example 15

i. If $z=cos\theta +isin\theta$ , show that $1+z=2cos\frac{\theta }{2}(\cos \frac{\theta }{2}+isin\frac{\theta }{2})$

ii. $z_{1}$ , $z_{2}$ are complex numbers such that $\vert z_{1}\vert =\vert z_{2}\vert =1$ .
If $z_{1}$ , $z_{2}[l/atex]� have arguments [latex]a,\beta$ respectively, where $-\pi <a,\beta \le \pi$ , show that $\frac{z_{1}+z_{1}z_{2}}{z_{1}+1}$ has modulus $\frac{\cos \frac{\beta }{2}}{\cos \frac{a}{2}}$ .

Solution 15

i. We wish to prove that $cos\theta +isin\theta +1=2cos\frac{\theta }{2}(\cos \frac{\theta }{2}+isin\frac{\theta }{2})$ .
Since we know double angle formulae better than half angle formulae, let us start with the RHS and show that it is equal to the LHS.

$RHS=2cos\frac{\theta }{2}(\cos \frac{\theta }{2}+isin\frac{\theta }{2})$ $=2 (\cos \frac{\theta }{2})^{2}+i(2cos\frac{\theta }{2}\sin \frac{\theta }{2})$ These groups of terms nearly look like the double angle formulas for $cos\theta$ and $sin\theta$ respectively so we made the necessary adjustments

$=(2 (\cos \frac{\theta }{2})^{2}-1)+1+i(2cos\frac{\theta }{2}\sin \frac{\theta }{2})$ $=cos\theta +1+isin\theta$ $=LHS$ ii. The given expression is ugly, so we want to simplify it as much as possible before plugging in any expressions. We note first that the expression on top factorises:

$\frac{z_{1}+z_{1}z_{2}}{z_{1}+1}=\frac{z_{1}(1+z_{2})}{z_{1}+1}$

Since it doesn’t look like we can simplify this any more, we use the properties of the modulus to obtain

$\vert \frac{z_{1}+z_{1}z_{2}}{z_{1+1}}\vert =\frac{\vert z_{1}\vert |1+z_{2}|}{|z_{1}+1|}$

But we know the first term is 1, and we know what $1+z_{1}$ and $1+z_{2}$ are from part (i).

$=\frac{1\times |2cos\frac{\beta }{2}cis\frac{\beta }{2}|}{|2cos\frac{a}{2}cis\frac{a}{2}|}$ $=\frac{2cos\frac{\beta }{2}}{2cos\frac{a}{2}}$ $=\frac{\cos \frac{\beta }{2}}{\cos \frac{a}{2}}$ $LHS=RHS$ as required.

Geometric Relationships between points on an Argand Diagram

There are nice geometrical relationships between points representing certain complex numbers. In particular, students are required to recognise the relationship between the point representing a complex number $z$ and the points representing the complex numbers $\overline{z}$ , $cz$ ( $c$ real) and $iz$ . These can be summarised as follows:

$z\to \bar{z} : reflect over the x axis$ $z\to cz : dilate about the origin by a factor of c$ $z\to iz : rotate anticlockwise about the origin by 90deg$

Students are not required to know the proofs of these properties. However, the proofs have been included as they help to illustrate some basic principles used in graphing complex numbers and relationships between complex numbers.

Relationship between Points Representing $z$ and $\overline{z}$

If $z=a+ib$ is represented by the point $P(a,b)$ then $\overline{z}=a-ib$ is represented by the point $Q(a,-b)$ . This is the reflection of $P$ over the x-axis.

Similarly, if $z=rcis\theta$ is represented by the point $P(r,\theta )$ then

$\overline{z}=r(cos\theta +isin\theta )=r(cos\theta -isin\theta )=r(\cos (-\theta )+isin(-\theta ))=rcis(-\theta )$
and $\overline{z}$ is represented by the point $Q(r,-\theta )$ , which is again obviously the reflection of $P$ over the x-axis.

relationship between points one of geometric relationship

Relationship between Points Representing $z$ and $cz$ ( $c$ is real)

If $z=a+ib$ is represented by the point $P(a,b)$ then $cz=ca+icb$ is represented by the point $Q(ca,cb)$ .

Similarly, if $z=rcis\theta$ is represented by the point $P(r,\theta )$ then $z=crcis\theta$ is represented by the point $Q(cr,\theta )$ .

This means that the vector $\overrightarrow{OQ}$ is the dilation of the vector $\overrightarrow{OP}$ about the origin with factor $c$ .

That is:

If $c>0$ , stretch out $\overrightarrow{OP}$ by a factor of $c$
If $c<0$ , stretch out $\overrightarrow{OP}$ by a factor of $-c$ and rotate it by $180deg$ about the origin
If $c=0$ then the point representing $cz$ is the point $(0,0)$

relationship between points two of geometric relationship

Relationship between Points Representing $z$ and $iz$

If $z=rcis\theta$ is represented by the point $P(r,\theta )$ then $iz=(cis\frac{\pi }{2})(rcis\theta )=rcis(0+\frac{\pi }{2})$ is represented by the point $Q(r,\theta +\frac{\pi }{2})$ . This means that the vector $\overrightarrow{OQ}$ is the vector $\overrightarrow{OP}$ rotated anticlockwise $\frac{\pi }{2}$ radians, or $90deg$ , about the origin, since its magnitude is the same but its angle of rotation is $\frac{\pi }{2}$ radians larger.

relationship between points three of geometric relationship

Note that in this case, since the calculations involve multiplying complex numbers, it is a lot simpler to use mode-arg form. If $z$ were instead represented in Cartesian form, then $z=a+ib$ represented by the point $P(a,b)$ and $iz=ia-b=-b+ia$ is represented by the point $Q(-b,a)$ .The relationship between these points is not as immediately obvious.

The following example demonstrates how the relationships between the points representing $z$ , $\overline{z}$ , $cz$ ( $c$ real) and $iz$ can be used to determine the complex numbers represented by certain points in terms of the complex numbers represented by other points.

Example 16

(i) Let $OABC$ be a square on an Argand diagram where $O$ is the origin. The points $A$ and $C$ represent the complex numbers $z$ and $iz$ respectively. Find the complex number represented by $B$ .

(ii) The square is now rotated about $O$ through $45deg$ in an anticlockwise direction to $OA^{‘}B^{‘}C'$ . Find the complex numbers represented by the points $A'$ , $B'$ and $C'$ .

Solution 16

a) A good approach to an Argand diagram question always begins with a diagram. Diagrams help you to see properties or make links that you otherwise would have missed.
solution sixteen of geometric relationship

Here, we see that $OABC$ is just a special case of a parallelogram, so by the parallelogram rule $B$ represents the complex number $z+iz$ . B represents the number $z+iz$ .

b) Again, start with a diagram

solution sixteen b of geometric relationship

We note that rotation by $\frac{\pi }{4}$ is the same as multiplying by $cis\frac{\pi }{4}$ . We can motivate this move from the previously studied relationships: rotation by $\frac{\pi }{2}$ is the same as multiplying by $cis\frac{\pi }{2}=i$

$A'$ represents $\frac{z(1+i)}{\sqrt{2}}$ , $B^{‘}$ represents $\frac{z(1+i)(1+i)}{\sqrt{2}}=\sqrt{2iz}$ and $C'$ represents $\frac{z(1+i)}{\sqrt{2}}=\frac{z(-1+i)}{\sqrt{2}}$

Representing Complex Numbers as Vectors on an Argand Diagram

We showed before that there is a 1-1 correspondence between the complex numbers and the points in the x-y plane, so that each complex number can be represented by a unique point $P$ in the complex plane. This means that there is also a 1-1 correspondence between the complex numbers and the vectors $\overrightarrow{OP}$ , where $O$ is the origin and $P$ is a point in the complex plane.

We represent each complex number $z=a+ib=rcis\theta$ by the vector $\overrightarrow{OP}$ , where $P=(a,b)$ . From the definition of the modulus and argument, we see that this means that:

The length of the vector $\overrightarrow{OP}$ is the distance from $P$ to the origin, which is the modulus
i.e. $\vert z\vert =\vert \overrightarrow{OP}\vert$
The direction of the vector $\overrightarrow{OP}$ is defined by the angle from the positive x axis to the ray $\overrightarrow{OP}$ , which is the argument. This means that the direction of the vector $\overrightarrow{OP}$ is uniquely defined by $Arg(z)$ .

So we see that this vector representation is similar to the Argand diagram representation. What this representation adds to the point representation is that vectors can be moved about, whereas points are fixed in the plane. Figure 3 illustrates this concept. This means that a lot of geometric properties (and hence algebraic properties) can be determined by rearranging the vectors.

figure three of representing complex numbers

Figure 3

Although vectors $\overrightarrow{a}$ , $\overrightarrow{b}$ and $\overrightarrow{c}$ all represent the complex number $1+i$ and the vectors $\overrightarrow{d}$ and $\overrightarrow{e}$ both represent the complex number $-1-2i$ , they are placed in different positions on the complex plane. In particular, vectors $\overrightarrow{c}$ and $\overrightarrow{e}$ are placed tail to tail, whilst vectors $\overrightarrow{e}$ and $\overrightarrow{b}$ are placed tip to tail. These vectors are therefore easier to relate than say, $\overrightarrow{a}$ and $\overrightarrow{e}$ .

Constructing Vector Representations of Complex Numbers

Given the vector representations of two complex numbers $z_{1}$ and $z_{2}$ , students are required to construct the vectors $z_{1}+z_{2}$ , $z_{1}-z_{2}$ , $z_{2}-z_{1}$ and $z_{1}z_{2}$ . This can be done by considering the geometrical properties of the vectors.

Two vectors, $\overrightarrow{a}$ and $\overrightarrow{b}$ , can be added either by using the triangle method or by using the parallelogram method. Each method is equally viable and in different situations one method will be more useful than the other.

The triangle method, or the tip-to-tail method, involves placing the tip of the vector $\overrightarrow{a}$ at the tail of the vector $\overrightarrow{b}$ . The vector pointing from the remaining tail to the remaining tip, i.e. from the tail of $\overrightarrow{a}$ to the tip of $\overrightarrow{b}$ , gives the vector $\overrightarrow{a}+ \overrightarrow{b}$ .
The parallelogram method involves placing the tails of each vector at the same spot. If the parallelogram with $\overrightarrow{a}$ and $\overrightarrow{b}$ as two of its adjacent sides is constructed, the vector along the diagonal from the tails to the opposite vertex gives the vector $\overrightarrow{a}+ \overrightarrow{b}$ .
constructing vector representation of complex numbers

Note that the vector $-\overrightarrow{a}$ is the vector with the same magnitude as $\overrightarrow{a}$ but in the opposite direction.

Note also that subtracting a vector $\overrightarrow{b}$ is the same as adding the vector $-\overrightarrow{b}$ .

Constructing the Point Representing $z_{1}+z_{2}$ from the Points Representing $z_{1}$ and $z_{2}$

Given the points representing $z_{1}$ and $z_{2}$ , the vectors representing $z_{1}$ and $z_{2}$ with tail at the origin can be obtained by joining the points with the origin. Since the vectors representing $z_{1}$ and $z_{2}$ are placed tail to tail, the parallelogram method can then be used to construct the vector representing $z_{1}+z_{2}$ with tail at the origin. The tip of this vector gives the point representing $z_{1}+z_{2}$ .

Students must include:·

Vectors representing $z_{1}$ and $z_{2}$ with tail at the origin and corresponding labels
Lines indicating construction of parallelogram
Vector representing $z_{1}+z_{2}$ with label

constructing the point of constructing vector of relationship

Constructing Vectors Representing $z_{1}-z_{2}$ and $z_{1}-z_{2}$ from Vectors Representing $z_{1}$ and $z_{2}$

Given the vectors $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ representing $z_{1}$ and $z_{2}$ respectively, vectors representing $-z_{1}$ and $-z_{2}$ with tip at the origin are given by $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ respectively. Since $z_{1}$ and $-z_{2}$ are placed tip to tail, the triangle method can then be used to construct a vector representing $z_{1}-z_{2}$ by adding $z_{1}$ and $-z_{2}$ . The same method holds for constructing a vector representing $z_{2}-z_{1}$ .

Students must include:·

Vectors representing $z_1$ and $z_2$ or $z_1$ and $z_2$ with corresponding labels.
Clear tip-to-tail vector representation of $z_1-z_2$ with label.

constructing vertors representing one of constructing vector representation

constructing vectors representing two of constructing representation

Constructing the Point Representing $z_{1}z_{2}$ from the Points Representing $z_{1}$ and $z_{2}$

Suppose the points representing $z_{1}$ and $z_{2}$ are given by $P$ and $Q$ . Denote by $I$ the point $(0,1)$ . Use equal angles to construct the similar triangles $\Delta OIP$ and $\Delta OQR$ . Then $R$ is the point representing $z_{1}z_{2}$ .

Students must include:

Indication of two pairs of equal angles.
Coordinates of point $I$ .
Labels of $z_{1}, z_{2}$ or $z_{1}z_{2$ for each respective vector.

constructing the point representing three of constructing vector representations

To see why this is true, consider $|OR|$ and the angle from the positive x axis to the ray $\overrightarrow{OR}$ , $ROI$ .

By similar triangles, $\frac{|OR|}{|OP|}=\frac{|OQ|}{|OI|}$ so that $\vert OR\vert =\frac{|OQ||OP|}{|OI|}=\frac{|z_{1}||z_{2}|}{1}=|z_{1}||z_{2}|$ .

Again by similar triangles, $ROQ=POI$
so that $ROI=?ROQ+?QOI=?POI+?QOI=arg(z_{1})+arg(z_{21})=arg?(z_{1}z_{2})$ .

Together, these tell us that $R$ is the point representing $z_{1}z_{2}$ .

Triangle Inequality

Students are expected to be able to prove the triangle inequality for the moduli of complex numbers:

$\vert z_{1}+z_{2}\vert \le \vert z_{1}\vert +\vert z_{2}\vert$

The geometrical proof of this given in the following worked example and demonstrates one application of vector representations.

Example 17

Given two arbitrary complex numbers $z_{1}$ and $z_{2}$ , prove that $\vert z_{1}+z_{2}\vert \le \vert z_{1}\vert +|z_{2}|$ (where $|z|$ is the modulus of $z$ ).

Solution 17

Let the two vectors $\overrightarrow{OP}$ and $\overrightarrow{PQ}$ represent $z_{1}$ and $z_{2}$ respectively. By the triangle method, the vector $\overrightarrow{OQ}$ represents the complex number $z_{1}+z_{2}$ . $\vert z_{1}+z_{2}\vert =| \overrightarrow{OQ}|$ $\le \vert \overrightarrow{OP}\vert +\vert \overrightarrow{PQ}\vert$ by the geometric triangle inequality $=\vert z_{1}\vert +|z_{2}|$

solution seventeen of constructing vector representations

Worked Examples involving Vector Representations of Complex Numbers

HSC questions do not simply ask students to draw a memorised diagram. Students are often required to synthesise the information embedded in the above techniques, such as by identifying the third side of the triangle $z_{1}$ , $z_{2}$ as the vector $z_{1}-z_{2}$ , or even using algebra to determine other geometric properties. The following examples illustrate a few applications of the techniques presented in this section.

Example 18

Let $u$ and $v$ be two complex numbers, where $u=-2+i$ and $v$ is defined by $\vert v\vert =3$ and $argv=\frac{\pi }{3}$

(i) On an Argand diagram plot the points $A$ and $B$ representing the complex numbers $u$ and $v$ respectively.

(ii) Plot the points $C$ and $D$ represented by the complex numbers $u-v$ and $iu$ respectively. Indicate any geometric relationships between the four points $A$ , $B$ , $C$ and $D$ .

Solution 18

Graphs for (i) and (ii):

solution eighteen one of constructing vector representation solution eighteen two of contructing vector representation

(i) Since $u$ and $v$ are given in Cartesian and mod-arg form respectively, the points were plotted using those respective forms. This is fine, as long as for the Cartesian form both the x and y coordinates are specified, and for the mod-arg form both the modulus and argument are specified.

(ii) By the triangle rule for vector subtraction the vector $\overrightarrow{AB}$ is equal to $u-v$ . Therefore the point $C$ is such that $\overrightarrow{OC}=\overrightarrow{AB}$ , that is, $OCAB$ is a parallelogram.
To obtain $iu$ rotate point $A$ $90deg$ about the origin. This means that $OA=OD$ and $?AOD=90deg$ . The diagram looks like so:
Geometric relationships:
$OCAB$ is a parallelogram
$OA=OD$ and $?AOD=90deg$

Example 19

example nineteen of constructing vector representations

The diagram shows a complex plane with origin $O$ . The points $P$ and $Q$ represent arbitrary non-zero complex numbers $z$ and $w$ respectively. Thus the length of $PQ$ is $|z-w|$ .(i) Use the diagram to show that $\vert z-w\vert \le \vert z\vert +|w|$ .(ii) Construct the point $R$ representing $Z+W$ . What can be said about quadrilateral $OPRQ$ ?(iii) If $\vert z-w\vert \vert z\vert +|w|$ , what can be said about the complex number $\frac{w}{z}$ ?

Solution 19

(i) The proof follows the same lines as the proof for $\vert z-w\vert \le \vert z\vert +|w|$ .
From the diagram, $|z|=OP$ , $\vert w\vert =OQ$ and $\vert z-w\vert =PQ$ are the sides of a triangle. Therefore by the geometric triangle inequality, $\vert z-w\vert =PQ\le OP+OQ=\vert z\vert +|w|$ .

(ii) The point $R$ can be obtained using the parallelogram rule for addition:

solution nineteen of constructing vector representations

The quadrilateral $OPQR$ is a parallelogram.

We guess that the previous parts of the question should help us to answer this question.
From the diagram, $|z-w|$ and $|z+w|$ are the diagonals of $OPQR$ , which is a parallelogram.
This means that if $\vert z-w\vert =|z+w|$ then $OPQR$ will be a rectangle.
This tells us that the angle from $OP$ to $OQ$ is a right angle.
We know from our construction of $iz$ from $z$ that right angles involve multiplying by $i$ , so we think that $\frac{w}{z}$ will be some multiple of $i$ .
Checking this using args (since we have information about args): $arg(\frac{w}{z})=arg(w)arg(z)=\frac{\pi }{2}$ so that $\frac{w}{z}$ is purely imaginary, of the form $ci$ where $c$ is positive.

Example 20

In an Argand Diagram, $B$ , $C$ and $D$ represent the complex numbers $a$ , $\beta$ , $\gamma$ and $\delta$ respectively.

(i) Describe the point which represents $\frac{1}{2}(a+\gamma )$ .

(ii) Deduce that if $a+\gamma =\beta +\delta$ , then $ABCD$ is a parallelogram.

Solution 20

(i) First, we draw a diagram. Since we know how to find the point that represents $a+\gamma$ we find that, call it $P$ and then we know that the midpoint of $OP$ is the point that we want.

solution twenty of contructing vector representation
However, since we obtained $P$ using the parallelogram rule, we know that $OAPC$ is a parallelogram so that $AC$ and $OP$ , its diagonals, bisect each other. Therefore if $M$ is the midpoint of $OP$ then $M$ is also the midpoint of $AC$ and the point that we want is the midpoint of $AC$ .
The point that represents $\frac{1}{2}(\alpha +\gamma )$ is the midpoint of $AC$ .

(ii) We want to use part (i) to give us some information about $a+\gamma$ and $\beta +\delta$ . From part (i) the point that represents $\frac{1}{2}(\alpha +\gamma )$ is the midpoint of $AC$ . Since we want to relate $a+\gamma$ with $\beta +\delta$ we note that similarly the point that represents $\frac{1}{2}(\beta +\delta )$ is the midpoint of $BD$ .
From part (i) we see that if $a+\gamma =\beta +\delta$ , the midpoints of $AC$ and $BD$ are the same point. This means that if $AC$ and $BD$ are the diagonals of $ABCD$ , then the diagonals of this quadrilateral bisect each other so that $ABCD$ is a parallelogram.

De Moivre’s Theorem

De Moivre’s Theorem states that for all integers $n$ :

$(cis \theta )^{n}=cis (n\theta )$

This result is equivalent to $(cos\theta +isin\theta )^{n}=\cos (n\theta )+isin(n\theta )$ .
The result is trivial for $n=1$ and can be proven for positive integers $n$ by mathematical induction, using the expansions of $cos?(n\theta +\theta )$ and $sin?(n\theta +\theta )$ . It can then be extended to the negative integers using complex division and $(cos\theta +isin\theta )^{-n}=\frac{1}{(cos\theta +isin\theta )^{n}}$ . The proof is left as an exercise to the reader.

The most obvious application of De Moivre’s theorem is in calculating integer powers of complex numbers. For any complex number $z=rcis\theta$ , De Moivre’s theorem tells us that $z^{n}=r^{n}(cis\theta )^{n}=r^{n}cis(n\theta )=r^{n}\cos (n\theta )+r^{n}sin?(n\theta )$ . This means that integer powers of a complex number can be easily calculated using De Moivre’s theorem and the modulus-argument form of the number. The following example illustrates how to calculate the integer power of a complex number.

Example 21

Express in modulus-argument form:

(i) $-1+i$

(ii) $(-1+i)^{n}$ , where $n$ is a positive integer.

Solution 21

(i) $\vert -1+i\vert =\sqrt{(-1)^{2}+1^{2}=\sqrt{2}}$
$arg(-1+i)=cos^{-1}(\frac{-1}{\sqrt{2}})=\frac{3\pi }{4}$ (where we use $cos^{-1}$ since we see that $arg?(-1+i)$ is in the 2^nd quadrant)

$-1+i=\sqrt{2}cis(\frac{3\pi }{4})$ (ii) $(-1+i)^{n}= (\sqrt{2}cis(\frac{3\pi }{4}))^{n}$

$= (\sqrt{2})^{n} (cis(\frac{3\pi }{4}))^{n}$ $=(\sqrt{2})^{n}(cis(\frac{3n\pi }{4}))$ by de Moivre’s theorem

De Moivre’s theorem is actually a lot more powerful than it looks. As was just demonstrated, directly comparing the two sides of the equation provides a direct formula for powers of complex numbers. However, the real and imaginary components can also be compared separately. Doing so provides an efficient way of calculating $sin?(n\theta )$ and $cos?(n\theta )$ as polynomials in $cos\theta$ and $sin\theta$ . The following example illustrates how this is done.

Example 22

se De Moivre’s thorem to express $cos4\theta$ and $\sin 4\theta$ in terms of $cos\theta$ and $sin\theta$ . Hence show that $\tan 4\theta =\frac{4 tan\theta -4 tan^{3}\theta }{1-6tan^{2}\theta +tan^{4}\theta }$

Solution 22

Since we want to express $cos4\theta$ and $sin 4\theta$ in terms of $cos\theta$ and $sin\theta$ , it makes sense to use De Moivre’s theorem for $n=4$ .
$cos4\theta +isin4\theta =(cos\theta )+isin\theta )^{4}$ (by De Moivre’s theorem)

$=cos^{4}\theta +4icos^{3}\theta sin\theta -6cos^{2}\theta sin^{2}\theta -4icos\theta sin^{3}\theta +sin^{4}\theta$

$=(\cos ^{4}\theta -6cos^{2}\theta sin^{2}\theta +sin^{4}+\theta )+i(4cos^{3}\theta sin\theta -4cos\theta sin^{3}\theta )$

Comparing real and imaginary components:

$cos4\theta =cos^{4}\theta -6cos^{2}\theta sin^{2}\theta +sin^{4}\theta sin4\theta =4cos^{3}\theta sin\theta -4cos\theta sin^{3}\theta$

$tan4\theta =\frac{sin4\theta }{cos4\theta }$ $=\frac{4cos^{3}\theta sin\theta -4cos\theta sin^{3}\theta }{cos^{4}\theta -6cos^{2}\theta sin^{2}\theta +sin^{4}\theta }$ Since we want our answer to be in terms of $tan\theta$ we note that $tan\theta =\frac{sin\theta }{cos\theta }$ so we divide through by $cos^{4}\theta$ to put all the powers of $cos\theta$ on the bottom of each term.

$=\frac{4(\frac{sin\theta }{cos\theta })-4(\frac{sin^{3}\theta }{cos^{3}\theta })}{1-6(\frac{sin^{2}\theta }{cos^{2}\theta })+(\frac{sin^{4}\theta }{cos^{4}\theta })}$
$=\frac{4tan\theta -4tan^{3}\theta }{1-6tan^{2}\theta +tan^{4}\theta }$

$LHS=RHS$ as required.

De Moivre’s theorem is also useful because it provides an efficient way of expressing large powers of $sin\theta$ and $cos\theta$ as a linear combination of the sines and cosines of multiples of $\theta$ . This is because if we let $z=cis\theta$ :

$z^{n}+z^{-n}=(cosn\theta +isinn\theta )+(\cos (-n\theta )+isin(-n\theta ))$
$=(cosn\theta +isinn\theta )+(cosn\theta -isinn\theta )=2cosn\theta$
$z^{n}-z^{-n}=(cosn\theta +isinn\theta )-(\cos (-n\theta )+isin(-n\theta ))$
$=(cosn\theta +isinn\theta )-(cosn\theta -isinn\theta )$
$=2isinn\theta$

The next example demonstrates how De Moivre’s theorem can be used to calculate large powers of $sin\theta$ and $cos\theta$ .

Example 23

If $z=cos\theta +isin\theta$ show that $z^{n}+\frac{1}{z^{n}}=2cosn\theta$ and hence show that

$cos^{4}\theta =\frac{1}{8}(cos4\theta +4cos2\theta +3)$

Solution 23

Let $z=cos\theta +isin\theta$ .

Then $z^{n}+\frac{1}{z^{n}}=(cis\theta )^{n}+(cis\theta )^{-n}$

$=cis(n\theta )+cis(-n\theta )$ (by De Moivre’s theorem)

$=(cosn\theta +isinn\theta )+(cosn\theta -isinn\theta )$ $=2cosn\theta$ Students might then be tempted to go straight for $n=4$ . This, unfortunately, yields two powers of $z$ and only one cosine term. It seems better to express $cos^{4}\theta$ in terms of multiple powers of $z$ that can then be grouped together and converted to cosines of multiples of $\theta$ . The following method is the general method for calculating large powers of $sin\theta$ and $cos\theta$ .

From this, we know that $cos\theta =\frac{1}{2}(z+\frac{1}{z})$

$cos^{4}\theta =\frac{1}{2^{4}} (z+\frac{1}{z})^{4}$ $=\frac{1}{16}(z^{4}+4z^{2}+6+4z^{-2}+z^{-4})$ $=\frac{1}{16}((z^{4}+z^{-4})+4(z^{2}+z^{-2})+6)$ $=\frac{1}{16}(2cos4\theta +8cos2\theta +6)$ $=\frac{1}{8}(cos4\theta +4cos2\theta +3)$

The $N^{th}$ Roots of Unity

The n^th roots of unity are the complex numbers $z$ such that $z^{n}=1$ . These can be determined using de Moivre’s theorem:

Let $z=rcis\theta$ . Then $z^{n}=r^{n}(cis\theta� )^{n}=r^{n}cis(n\theta )=1$ . Since two non-zero complex numbers can only be equal if their moduli are equal, $r^{n}=1$ so that $cis(n\theta )=1$ . This gives $\cos (n\theta )=1$ , $sind(n\theta )=0$ and $n\theta =2\pi k$ , $k?Z$ . This yields $n$ distinct values for $z$ corresponding to $k=0,1,2,3K(n-1)$ . Any other value of $k$ yields a value of $\theta$ that gives the same sine and cosine values as one of the n values of $\theta$ specified above.

Therefore the n n^th roots of unity are given by $z=cis(\frac{2\pi k}{n})[latex],[latex]k=0,1,2K(n-1)$ .

$N^{th}$ Roots of Unity: $z=cis(\frac{2\pi k}{n})$ where $k=0, 1, 2, k(n-1)$

This shows that n^th roots of unity can be expressed on the Argand diagram as $n$ vectors of length 1 radiating from the origin, spaced equally around the unit circle. This means that they form the vertices of a regular n-gon. Students will often be asked to demonstrate their understanding of this through graphing the n^th roots on an Argand diagram. For example, the cube roots of unity can be represented as the vertices of an equilateral triangle as follows:

Students must include:

Indication of angles between adjacent vectors.
Indication that each vector is of length 1 (either using equal length markings or by lightly drawing in the unit circle)
The relevant marking $z_{i}=cis (\frac{2i\pi }{n})$ at each point.

The roots can be found in the form $a+ib$ by converting from modulus-argument form.

The n^th roots of $-1$ can similarly be determined using de Moivre’s theorem:
Let $z=rcis\theta$ . Then $z^{n}=r^{n}(cis\theta )^{n}=r^{n}cis(n\theta )=-1.$ .
$z^{n}=r^{n}(cis\theta )^{n}=r^{n}cis(n\theta )=-1.$ , $\cos (n\theta )=-1$ and $\sin (n\theta )=0$ and $\theta =\frac{(2k+1)\pi }{n}$ , $k?Z$ .

$N^{th} Roots of -1:$

$z=cis(\frac{(2k+1)\pi }{n})$

Where $k=0, 1, 2, (n-1)$

Again, the roots must be equally spaced around the unit circle. However, they look a little different to the n^th roots of unity. One way to think about it is that whereas n^th roots of unity begin at 1 and move around the circle anticlockwise with a step angle of $\frac{2\pi }{n}$ , n^th roots of unity begin at -1 and move around the circle anticlockwise with a step angle of $\frac{2\pi }{n}$ . Another way to think about it is that each n^th root of -1 is of the form $z=cis(\frac{(2k+1)\pi }{n})=cis(\frac{2k\pi }{n})cis(\frac{\pi }{n})$ , which is one of the n^th roots of unity rotated anticlockwise about the origin by an angle of $\frac{\pi }{n}$ .

Polynomials and the $N^{th}$ Roots of $\pm 1$

One of the more difficult applications of complex number theory involves using the algebra and ideas expressed in de Moivre’s theorem and the derivation of expression for the n^th roots of $\pm 1$ to solve polynomials. These questions can usually be broken down into little steps by using complex number theory and ideas involved in solving polynomials, such as sum and product of roots. The nice thing is that HSC questions will often help students to separate the question into simple steps, or even present the steps and links as part of the problem statement. However, since this is not always the case, students should acquaint themselves with the various processes involved in breaking down questions.

Often, smaller steps can be obtained by considering the type of information given to them in the problem statement and the type of information that they needs to be extract. The following examples illustrate how this can be done.

Example 24

showing that the n^th roots of unity can be written as powers of one of its roots, dividing through by $z-1$ to get an equation whose roots are all complex and then using a relation to do with polynomials or complex numbers with magnitude 1 to get some new information.

Let $w=cos(\frac{2\pi }{9})+isin(\frac{2\pi }{9})$

(i) Show that $w^{k}$ is a solution of $z^{9}-1=0$ , where $k$ is an integer.

(ii) Prove that $w+w^{2}+w^{3}+w^{3}+w^{4}+w^{5}+w^{6}+w^{7}+w^{8}=-1$

(iii) Hence show that $\cos (\frac{\pi }{9})\cos (\frac{2\pi }{9})\cos (\frac{4\pi }{9})=\frac{1}{8}$

Solution 24

(i) To show that $w^{k}$ is a solution of $z^{9}-1=0$ , all we have to do is plug it in and show that $LHS=RHS$ .

$LHS= (w^{k})^{9}-1$ $=(w^{9})^{k}-1$ $=( (cis\frac{2\pi }{9})^{9})^{k}-1$ $= (cis\frac{2\pi \times 9}{9})^{k}-1$ by De Moivre’s Theorem

$=(cis2\pi )^{k}-1$ $=1^{k}-1$ $=0$ $=RHS$ Hence $w^{k}$ , where $k$ is an integer, is a solution of $z^{9}-1=0$ .

(ii) Method 1:
From part (i), we know that $w^{k}$ is a solution to the equation $z^{9}-1=0$ for all integers $k$ . Furthermore, if we take $k=0,1,2K 8$ then $arg(w^{k})=karg(w)=\frac{2k\pi }{9}$ so that $0\le arg(w^{k})<2\pi$ and there are $9$ distinct solutions to the equation. These are therefore the 9 roots of the equation $P(z)=z^{9}-1=0$ . Since the sum of the roots is the coefficient of $z$ in $P(z)$ , this means that the sum of the roots is 0.
That is, $w^{0}+w^{1}+w^{2}+w^{3}+w^{4}+w^{5}+w^{6}+w^{7}+w^{8}=0$

$[latex]w+w^{2}+w^{3}+w^{4}+w^{5}+w^{6}+w^{7}+w^{8}=-1$
Method 2:
From part (i), we know that W is a solution to the equation $z^{9}-1=0$ .
This means that $(w-1)(1+w+w^{2}+w^{3}+w^{4}+w^{5}+w^{6}+w^{7}+w^{8})=0$ .
Since $w$ is complex and $w\ne 1$ , $w-1\ne 0$ so that $1+w+w^{2}+w^{3}+w^{4}+w^{5}+w^{6}+w^{7}+w^{8}$ , which implies that $w+w^{2}+w^{3}+w^{4}+w^{5}+w^{6}+w^{7}+w^{8}=-1$ .
Both these methods can be used for this question. The first approach is usually used when something is known about all the roots, so that it is possible to sum them up. The second is used when something is known about one root and so that one of the factors in the factorised expression can be cancelled.

(iii) We want to relate cosines, which are real numbers, with powers of $w=cos\frac{2\pi }{9}+isin\frac{2\pi }{9}$ , which are complex numbers. To do this, we use one of the niftier consequences of de Moivre’s theorem:
$(cis\theta )^{n}+(cis\theta )^{-n}=2cosn\theta$ and $(cis\theta )^{n}-(cis\theta )^{-n}=2isinn\theta$ .
So now we know that if we can express the left hand side of required result in terms of cosines of multiples of $\theta$ , where $w=cis\theta$ i.e. $\theta =\frac{2\pi }{9}$ , we can then express those as powers of $w$ and use the results from the previous parts. This gives us three steps.
Step 1: Express $LHS$ as cosines of multiples of $\theta =\frac{2\pi }{9}$

$LHS=cos\frac{\pi }{9}\cos \frac{2\pi }{9}\cos \frac{4\pi }{9}=-cos\frac{8\pi }{9}\cos \frac{2\pi }{9}\cos \frac{4\pi }{9}=-cos4\theta cos\theta cos2\theta$

Step 2: Evaluate cosines of multiples of $\theta$ in terms of $w=cis\theta$ .

$LHS=-cos\theta cos2\theta cos4\theta =-\frac{1}{8}(w+w^{-1})(w^{2}+w^{-2})(w^{4}+w^{-4})=-\frac{1}{8}w^{-7}(w^{2}+1)(w^{4}+1)(w^{8}+1)$ Step 3: Simplify this expression using results from previous parts until we get to the required result

$LHS=-\frac{1}{8}w^{-8}(w^{2}+1)(w^{4}+1)(w^{9}+w)=-\frac{1}{8}w^{9}w^{-8}(w^{2}+1)(w^{4}+1)(1+w)$ What we’ve done here is make the result easier to expand (we think that we will have to expand, since part (ii) looks very much like an expanded result).
In the first equality we multiply through by enough powers of $w$ to make each factor have only non-negative powers. Of course, this means that we have to divide through by them again, and hence we get $w^{-8}$ at the front.
In the second equality we use the fact from part (i) that $w^{9}=1$ to make the powers in the factors as small as possible, and to make the power out front a positive power.

$LHS=-\frac{1}{8}w(1+w^{2}+w^{4}+w^{6})(1+w)=-\frac{1}{8}(w+w^{2}+w^{3}+w^{4}+w^{5}+w^{6}+w^{7}+w^{8})=-\frac{1}{8}(-1)=\frac{1}{8}$ Now it’s just a matter of expanding and hoping for the best. We expand and lo and behold, we get the long side of the equation in part (ii) and we can just replace it with $-1$ and everything falls out.

This problem demonstrates a type of problem that many students will shy away from: one involving not 1, not 2, but n variables. The trick is to deal with each variable separately, or to work patiently through sums, preferably using summation notation. Students will see that this question, as scary as it looks, is actually a lot easier than the previous one.

Let the points $A_{1}$ , $A_{2}$ …, $A_{n}$ represent the n^th roots of unity, $w_{1}$ , $w_{2}$ , …, $w_{n}$ , and suppose $P$ represents any complex number $z$ such that $\vert z\vert =1$ .

(i) Prove that $w_{1}+w_{2}+\ldots +w_{n}=0$ .

(ii) Show that $PA_{i}^{2}=(z-w_{i})(\overline{z}-\overline{w_{i}})$ for $i=1,2,K ,n$ .

(iii) Prove that $\sum_{i=1}^{n}{PA_{i}^{2}=2n}$

Solution 25

(i) We want to know the value of $w_{1}+w_{2}+\ldots +w_{n}$ , where $w_{1}$ , $w_{2}$ , …, $w_{n}$ are the n^th roots of unity. This in itself already tells us something: they are the n solutions to some equation, and we want to find their sum. If this equation is a polynomial, we know right away what the sum of the solutions is. So let’s consider this equation.
Since $w_{1}$ , $w_{2}$ , …, $w_{n}$ are the n^th roots of unity, they are the n solutions to the equation $z^{n}=1\to z^{n}-1=0$ . This means that their sum is given by the coefficient of the linear term in $z$ , which is 0.

$w_{1}+w_{2}+\cdots +w_{n}=0$ (ii) To determine what $PA_{i}^{2}$ is, we first have to take a look at what is meant by these points. $P$ represents the complex number $z$ . $A_{i}$ represents the complex number $w_{i}$ . Therefore by the triangle method for vector subtraction, $\overrightarrow{PA_{i}}$ represents $w_{i}-z$ and $PA_{i}^{2}=|w_{i}-z|^{2}$ .
$PA_{i}^{2}=|w_{i}-z|^{2}$ (since $\overrightarrow{PA_{i}}$ represents $w_{i}-z$ )
$=(z-w_{i})(z-w_{i})$ (we separate it out because then one factor is in the RHS and the other is nearly)
$=(z-w_{i})(\overline{z}-\overline{w_{i}})$ (one more property of conjugates gives us the expression on the RHS)

(iii) Before we dive into summation, let’s see if we can simplify each term in the expression any more.

$PA_{i}^{2}=(z-w_{i})(\overline{z}-\overline{w_{i}})$
$=z\overline{z}-zw_{i}-\overline{z}w_{i}+w_{i}\overline{w_{i}}$

$=|z|^{2}-z\overline{w_{i}}-\overline{z}w_{i}+|w_{i}| ^{2}$ $=2-zw_{i}-\overline{z}w_{i}$
Summing over all n,

$\sum_{i=1}^{n}{PA_{i}^{2}}=\sum_{i=1}^{n}{(2-z\overline{w_{i}}-\overline{z}w_{i})}$ $=\sum_{i=1}^{n}{2-z\sum_{i=1}^{n}{\overline{w_{i}}-\overline{z}}\sum_{i=1}^{n}{w_{i}}}$ (since $z$ is independent of $i$ )
$=2n-z\overline{(\sum_{i=0}^{n}{w_{i}})}-z(0)$ (since sum of conjugates is conjugate of sum, and the sum is 0 by part (i))
$=2n-z\overline{(0)}$ (again since the sum is 0)

$=2n$

Sketching Curves and Regions in the Complex Plane

Sometimes it is useful to represent the complex solutions to a given equation or inequality in graphical form. In particular, there are a few standard curves that students are expected to know how to sketch. In the HSC, the complex equations describing these curves are often combined or turned into inequalities to create more involved questions. However, these curves and regions are quite simple to sketch if students have a firm understanding of the elementary curves that they are comprised of.

When determining the nature of the following standard curves, a number of algebraic and geometric approaches can be taken.If using an algebraic approach, students should ensure that they are able to obtain a Cartesian equation for the locus, and able to translate this equation into a geometric description.If using the geometric approach, students should ensure that they are able to translate the geometric properties describing the locus into algebraic form, through a Cartesian equation.

Solutions to the Equations $Re(z-z_{0})=c$ and $Im(z-z_{0})=k$

Algebraic approach

The solutions of these equations are given by lines parallel to the y axis and x axis respectively:
If $Re(z-z_{0})=c$ , $Re(z)Re(z_{0})=c\to x=Re(z_{0})+c$ .If $Im(z-z_{0})=k$ .

$Im(z)Im(z_{0})=k\to x=Im(z_{0})+k$

solutions to the equations of sketching curves

solution to the equation two of sketching curves

Solutions to Equations of the Form $\vert z-z_{1}\vert =|z-z_{2}|$

Algebraic approach

Since the equations involve the subtraction of complex numbers,the simpler algebraic approach should involve expressing the complex numbers as the sum of their real and imaginary parts.

Let $z=x+iy$ , $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$ .

$\vert z-z_{1}\vert =\vert z-z_{2}\vert$

$\vert z-z_{1}\vert ^{2}=|z-z_{2}| ^{2}$

$(x-x_{1})^{2}+(y-y_{1})^{2}+(x-x_{2})^{2}+(y-y_{2})^{2}$

$x^{2}-2x_{1}x+x_{1} ^{2}+y^{2}-2y_{1}y+y_{1} ^{2}=x^{2}-2x_{2}x+x_{2} ^{2}+y^{2}-2y_{2}y+y_{2} ^{2}$

$2x(x_{2}-x_{1})+2y(y_{2}-y_{1})+(x_{1} ^{2}-x_{2} ^{2})+(y_{1} ^{2}-y_{2} ^{2})=0$ which is of the form

$ax+by+c=0$ , a straight line!

Geometric approach

This seems like a remarkably simple result for some not-so-simple equations. However it is quite easily explained using a geometric interpretation. The solutions describe the locus of points that are equidistant from two fixed points in the plane: the point $A$ representing $z_{1}$ and the point $B$ representing $z_{2}$ . If a possible value of $z$ is represented by the point $P$ , then $\Delta PAB$ is isosceles with $PA=PB$ so that the altitude from $P$ to $AB$ bisects $AB$ . That is, $P$ lies on the perpendicular bisector of $AB$ , which is a straight line. Furthermore, any point on the perpendicular bisector satisfies $PA=PB$ so that the locus of points is the perpendicular bisector.

To plot the perpendicular bisector of $AB$ , all that is required is to find a point on the line and the gradient of the line.
A point on the line is given by the midpoint of $AB$ . The gradient is the one perpendicular to the gradient of $AB$ .
These give $y-(\frac{y_{1}+y_{2}}{2})=-(\frac{x_{2}-x_{1}}{y_{2}-y_{1}})(x-(\frac{x_{1}+x_{2}}{2}))$ .

Students must include:

Equation of line in Cartesian form
At least two points on the line (e.g. intercepts)

geometric approach of solutions to equations

Solutions to Equations of the Form $\vert z-z_{0}\vert =R$

Algebraic approach

Again, since the equations involve the subtraction of complex numbers, the simpler algebraic approach should involve expressing the complex numbers as the sum of their real and imaginary parts.

Let $z=x+iy$ and $z_{0}=x_{0}+iy_{0}$ .

$\vert z-z_{0}\vert =R$

$\vert z-z_{0}\vert ^{2}=R^{2}$

$(x-x_{0}) ^{2}+(y-y_{0}) ^{2}=R^{2}$

The locus of $z$ therefore traces out a circle in the complex plane center $(x_{0},y_{0})=z$ and radius $R$ .

Geometric approach

The solutions describe the locus of points that are a fixed distance $R$ from the point representing the complex number $z_{0}$ . Thus the locus of $z$ traces out a circle in the complex plane center $(x_{0},y_{0})=z_{0}$ and radius $R$ . This has equation $(x-x_{0}) ^{2}+(y-y_{0}) ^{2}=R^{2}$ .

Students must include:

Coordinates of center of circle.
Some indication of the size of the radius (e.g. using intercepts, arbitrary point on circle, radius marking)

Solutions to Equations of the Form $arg(z-z_{0})=\theta$

The graph is a ray starting with an open circle on the point $A$ representing the complex number $z_{0}$ and pointing in a direction that is $\theta$ anticlockwise from the positive x-axis.

Students must include:

Open circle at the point A representing $z_{0}$
Coordinates of A.
An indication that the angle between the ray AP and the positive $x$ -axis is $\theta$ .

solution to equation of only one required

Geometric approach

Consider the vector representation of complex numbers, where $\theta$ is the angle from the positive x-axis to the ray $OP$ . Since $z-z_{0}$ is the vector pointing from the point $A$ representing $z_{0}$ to the point $P$ representing $Z$ , if this makes an angle of $\theta$ with the positive x-axis then the ray $AP$ is $\theta$ radians anticlockwise from the positive x-axis. Similarly if $z$ is represented by an arbitrary point $P$ on the ray with tail $A$ and angle $\theta$ from positive x-axis, then $arg(z-z_{0})=\theta$ as required. At $P=A$ or $z=z_{0}$ $arg?(z-z_{0})$ is undefined so that there is an open circle at that point.

Equations involving both $z$ and $\overline{z}$

Students will often be asked to graph equations involving both $z$ and $\overline{z}$ . Cartesian equations can usually be obtained by substituting $z=x+iy$ and $\overline{z}=x-iy$ .

Worked Examples involving Sketching Curves and Regions in the Complex Plane

Sketching the solutions to inequalities is not any more difficult than sketching the solutions to equalities. The equality case creates a curve, or boundary, that splits the complex plane into two sections. The inequality sign means that everything on one side of the boundary is part of the region and everything on the other side is not.

The following examples illustrate how compositions of elementary curves can be used to graph regions in the complex plane that are described by a number of equations and inequalities.

Example 26

Sketch the region where the inequalities $|z-3+i|\le 5$ and $|z+1|\le |z-1|$

Solution 26

The trick with sketching composite regions is to sketch each region separately, and see which parts of the plane lie in both regions.

$|z-3+i|\le 5|$ solution 26 of region in the complex lane

$\vert z-3+i\vert =5$ describes the set of points 5 units from the point $(3,-1)$ . This is a circle in the complex plane with center $(3,-1)[latex]� and radius 5.describes the set of points less than or equal to 5 units from the point. This is a disc in the complex plane with center (3,-1) and radius 5.<em>To check which region should be shaded in, pick a point on one side of the boundary. In this case, the point (3,-1) inside the disc yields [latex]|z-3+i|=0 \le 5$ so that it is in the shaded region.
Students should note the common mistake of taking the center to be the point (3,-1). Remember that it is usually the quantity $z-z_0$ that is involved in these equations.

$|z+1|\le |z-1|[latex]<a href="https://dc.edu.au/wp-content/uploads/solution-twenty-six-two-of-region-in-the-complex-lane.png"><img class="alignnone size-full wp-image-2867" src="https://dc.edu.au/wp-content/uploads/solution-twenty-six-two-of-region-in-the-complex-lane.png" alt="solution twenty six two of region in the complex lane" width="269" height="265" /></a></td> <td width="447">[latex]\vert z+1\vert =|z-1|$ describes the set of points equidistant from the points $(-1,0)$ and $(1,0)$ . This is the perpendicular bisector of $(-1,0)$ and $(1,0)$ . It is the line through $(0,0)$ perpendicular to the x-axis, which is the y-axis. $|z+1|\le |z-1|$ describes the set of points closer to $(-1,0)$ than $(1,0)$ . This is the region to the left of the y-axis.
To check which region should be shaded in, pick a point on one side of the boundary. In this case, the point $(2,0)$ on the right of the boundary yields $\vert z+1\vert =3>1=|z-1|$ so that it is not in the shaded region.

Taking the intersection of these two regions yields:

taking the intersection of these two regions yields

Example 27

On an Argand diagram, shade in the region determined by the inequalities $2\le Imz\le 4$ and $\frac{\pi }{6}\le argz\le \frac{\pi }{4}$ .

Solution 27

	$2\le Imz\le 4$
Note the open circle at $(0,0)$	$\frac{\pi }{6}\le argz\le \frac{\pi }{4}$
	$2\le Imz\le 4$ and $\text{[math]}$ \frac{\pi }{6}\le argz\le \frac{\pi }{4} $\text{[math]}$

complex number 4 marks(SAQ)

Arithmetic of Complex Numbers

Addition and Subtraction

Example 1

Solution 1

Multiplication of Complex Numbers

Example 2

Solution 2

Complex Conjugates and their Properties

What are Complex Numbers and why do we need them?

For a given complex number , its complex conjugate is denoted by and is given by:

Example 3

Solution 3

Example 4

Solution 4

Division of Complex Numbers

Example 5

Solution 5

Equality of Complex Numbers

Finding the Square Roots of Real and Complex Numbers

Example 6

Solution 6

Using Complex Numbers to Solve Equations

Solving Quadratic Equations with Real Coefficients

Example 7

Solution 7

Solving Quadratic Equations with Complex Coefficients

Example 8

Solution 8

The Argand Diagram

Example 9

Arithmetic of Complex Numbers

Addition and Subtraction

Example 1

Solution 1

Multiplication of Complex Numbers

Example 2

Solution 2

Complex Conjugates and their Properties

Example 3

Solution 3

Example 4

Solution 4

Division of Complex Numbers

Example 5

Solution 5

Equality of Complex Numbers

Finding the Square Roots of Real and Complex Numbers

Example 6

Solution 6

Using Complex Numbers to Solve Equations

Solving Quadratic Equations with Real Coefficients

Example 7

Solution 7

Solving Quadratic Equations with Complex Coefficients

Example 8

Solution 8

The Argand Diagram

Example 9

Example 10

Solution 10

Example 11

Solution 11

Multiplication and Division of Complex Numbers and Properties of the Modulus and Argument

Example 12

Solution 12

Example 13

Solution 13

Example 14 (HSC 1997 Q2d)

Solution 14

Example 15

Solution 15

Geometric Relationships between points on an Argand Diagram

Relationship between Points Representing and

Relationship between Points Representing and ( is real)

Relationship between Points Representing and

Example 16

Solution 16

Representing Complex Numbers as Vectors on an Argand Diagram

Constructing Vector Representations of Complex Numbers

For a given complex number $z=x+iy$ , its complex conjugate is denoted by $\bar{z}$ and is given by:

Relationship between Points Representing $z$ and $\overline{z}$

Relationship between Points Representing $z$ and $cz$ ( $c$ is real)

Relationship between Points Representing $z$ and $iz$

Constructing the Point Representing $z_{1}+z_{2}$ from the Points Representing $z_{1}$ and $z_{2}$

Constructing Vectors Representing $z_{1}-z_{2}$ and $z_{1}-z_{2}$ from Vectors Representing $z_{1}$ and $z_{2}$

Constructing the Point Representing $z_{1}z_{2}$ from the Points Representing $z_{1}$ and $z_{2}$

In an Argand Diagram, $B$ , $C$ and $D$ represent the complex numbers $a$ , $\beta$ , $\gamma$ and $\delta$ respectively.

The $N^{th}$ Roots of Unity

Polynomials and the $N^{th}$ Roots of $\pm 1$

showing that the n^th roots of unity can be written as powers of one of its roots, dividing through by $z-1$ to get an equation whose roots are all complex and then using a relation to do with polynomials or complex numbers with magnitude 1 to get some new information.

Solutions to the Equations $Re(z-z_{0})=c$ and $Im(z-z_{0})=k$

Solutions to Equations of the Form $\vert z-z_{1}\vert =|z-z_{2}|$

Solutions to Equations of the Form $\vert z-z_{0}\vert =R$

Solutions to Equations of the Form $arg(z-z_{0})=\theta$

Equations involving both $z$ and $\overline{z}$