$$|\sqrt{3} – i| = \sqrt{3 + 1} = 2$$ $$\arg(\sqrt{3} – i) = -\frac{\pi}{6}$$ $$\sqrt{3} – i = 2e^{-i\pi/6}$$

$$\sqrt{3} – i = 2e^{i(-\pi/6 + 2\pi k)}, \quad k = 0, 1, 2, 3, 4$$

$$(\sqrt{3} – i)^{2/5} = 2^{2/5} \cdot e^{i\left(-\frac{\pi}{6} + 2\pi k\right) \cdot \frac{2}{5}}$$ $$= 2^{2/5} \cdot e^{i\left(-\frac{\pi}{15} + \frac{4\pi k}{5}\right)}, \quad k = 0, 1, 2, 3, 4$$

$$\text{Product} = (2^{2/5})^5 \cdot \exp\left(i \sum_{k=0}^{4} \left[-\frac{\pi}{15} + \frac{4\pi k}{5}\right]\right)$$ $$= 2^2 \cdot e^{i\left(-\frac{5\pi}{15} + \frac{4\pi(0+1+2+3+4)}{5}\right)}$$ $$= 4 \cdot e^{i\left(-\frac{\pi}{3} + \frac{8\pi}{5}\right)}$$ $$= 4e^{i \cdot 19\pi/15}$$

$$x^2 + x + 1 = 0 \implies \alpha, \beta = \omega, \omega^2$$ $$\text{where } \omega = e^{2\pi i/3}, \quad \omega^3 = 1, \quad 1 + \omega + \omega^2 = 0$$

$$S_1 = -1, \quad S_2 = -1, \quad S_n = -S_{n-1} – S_{n-2}$$

$$S_1 = -1, S_2 = -1, S_3 = 0, S_4 = 1, S_5 = 1, S_6 = 0$$ $$S_7 = -1, S_8 = -1, S_9 = 0, S_{10} = 1, S_{11} = 1, S_{12} = 0$$

$$\sum_{k=1}^{12} S_k^2 = (-1)^2 + (-1)^2 + 0^2 + 1^2 + 1^2 + 0^2 + (-1)^2 + (-1)^2 + 0^2 + 1^2 + 1^2 + 0^2$$ $$= 1 + 1 + 0 + 1 + 1 + 0 + 1 + 1 + 0 + 1 + 1 + 0$$

$$\sum_{n=0}^{\infty} \left(\frac{i}{3}\right)^n = \frac{1}{1 – \frac{i}{3}} = \frac{3}{3-i}$$

$$\frac{3}{3-i} \cdot \frac{3+i}{3+i} = \frac{3(3+i)}{9+1} = \frac{3(3+i)}{10}$$

$$1 + \sqrt{3}i = 2e^{i\pi/3}, \quad 1 – \sqrt{3}i = 2e^{-i\pi/3}$$

$$(1+\sqrt{3}i)^{2023} = 2^{2023}e^{i \cdot 2023\pi/3}$$ $$(1-\sqrt{3}i)^{2023} = 2^{2023}e^{-i \cdot 2023\pi/3}$$

$$\text{Sum} = 2^{2023}\left(e^{i \cdot 2023\pi/3} + e^{-i \cdot 2023\pi/3}\right) = 2^{2024}\cos\left(\frac{2023\pi}{3}\right)$$

$$2023 = 3 \times 674 + 1 \implies \frac{2023\pi}{3} = 674\pi + \frac{\pi}{3}$$ $$\cos\left(\frac{2023\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$

$$i^0 = 1, \quad i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1 \text{ (cycle repeats)}$$

$$\text{Even powers: } i^{2k} \text{ where } k = 0, 1, 2, \ldots, 1012$$ $$\text{Sequence: } 1, -1, 1, -1, 1, -1, \ldots$$

$$\text{Number of terms} = 1013 \text{ (from } k = 0 \text{ to } 1012)$$ $$\text{Sum} = \underbrace{(1 + (-1)) + (1 + (-1)) + \cdots}_{\text{506 pairs}} + 1 = 1$$

$$\alpha, \beta = \frac{2 \pm \sqrt{4-8}}{2} = 1 \pm i$$

$$1 + i = \sqrt{2}e^{i\pi/4}, \quad 1 – i = \sqrt{2}e^{-i\pi/4}$$

$$(1+i)^{2020} = (\sqrt{2})^{2020}e^{i \cdot 2020\pi/4} = 2^{1010}e^{i \cdot 505\pi}$$ $$(1-i)^{2020} = 2^{1010}e^{-i \cdot 505\pi}$$

$$e^{i \cdot 505\pi} = e^{i\pi} = -1 \quad (\text{since } 505 \text{ is odd})$$ $$(1+i)^{2020} = -2^{1010}, \quad (1-i)^{2020} = -2^{1010}$$

$$\alpha + \beta = 2\sqrt{3}, \quad \alpha\beta = 4$$

$$\alpha^2 + \beta^2 = (\alpha+\beta)^2 – 2\alpha\beta = 12 – 8 = 4$$ $$\alpha^3 + \beta^3 = (\alpha+\beta)^3 – 3\alpha\beta(\alpha+\beta) = 24\sqrt{3} – 24\sqrt{3} = 0$$

$$\alpha^6 + \beta^6 = (\alpha^3 + \beta^3)^2 – 2(\alpha\beta)^3$$ $$= 0^2 – 2(4)^3 = -128$$

$$1 + i = \sqrt{2}e^{i\pi/4}, \quad 1 – i = \sqrt{2}e^{-i\pi/4}$$

$$(1+i)^n + (1-i)^n = 2^{n/2}\left(e^{in\pi/4} + e^{-in\pi/4}\right) = 2^{(n+2)/2}\cos\left(\frac{n\pi}{4}\right)$$

$$\text{For purely imaginary result: } \cos\left(\frac{n\pi}{4}\right) = 0$$ $$\text{and amplitude matches } -i$$

$$\cos\left(\frac{n\pi}{4}\right) = 0 \text{ when } \frac{n\pi}{4} = \frac{\pi}{2} + k\pi$$ $$n = 2 + 4k \quad \text{or} \quad n \equiv 2 \pmod{4}$$

$$\cos\frac{\pi}{2} = 0, \quad \sin\frac{\pi}{2} = 1$$

$$\frac{(\cos\theta + i\sin\theta)^{2020}}{(\sin\theta + i\cos\theta)^{2021}} = \frac{(i)^{2020}}{(1+0i)^{2021}} = \frac{1}{1} = 1$$

$$\left(\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta-i\sin\theta}\right)^{2021} = \left(\frac{1+i}{1-i}\right)^{2021} = (i)^{2021} = -i$$

$$\text{Sum} = 1 + (-i) = 1 – i$$ $$x + y = 1 + (-1) = 0$$

$$i + i^2 + i^3 + i^4 = i – 1 – i + 1 = 0$$ $$\text{Period of sum = 4}$$

$$i^2 + i^3 + \cdots + i^{4000} = (i^2 + i^3 + i^4 + i^5) + (i^6 + i^7 + i^8 + i^9) + \cdots$$

$$4000 – 2 + 1 = 3999 \text{ terms total}$$ $$3999 = 4 \times 999 + 3$$

$\text{Complete groups} = 999 \times 0 = 0$ $\text{Remaining: } i^{3998} + i^{3999} + i^{4000}$ $= i^2 + i^3 + i^4 = -1 – i + 1 = -i$

$\frac{(1+i)^{2011}}{(1-i)^{2009}} = \left(\frac{1+i}{1-i}\right)^{2009} \cdot (1+i)^2$

$\frac{1+i}{1-i} = \frac{(1+i)^2}{(1-i)(1+i)} = \frac{1 + 2i – 1}{2} = \frac{2i}{2} = i$

$2009 = 4 \times 502 + 1 \implies i^{2009} = i$

$(1+i)^2 = 1 + 2i – 1 = 2i$

$i^{2009} \cdot (1+i)^2 = i \cdot 2i = 2i^2 = -2$

$\frac{\sqrt{3}}{2} + \frac{i}{2} = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = e^{i\pi/6}$

$\left(e^{i\pi/6}\right)^n = 1 \implies e^{in\pi/6} = 1$

$e^{in\pi/6} = 1 \implies \frac{n\pi}{6} = 2\pi k, \quad k \in \mathbb{Z}$ $n = 12k$

$x^6 = 1 \implies \alpha = e^{2\pi i k/6}, \quad k = 1, 2, 4, 5 \text{ (non-real)}$

$\alpha^{-1} = \alpha^5, \quad \alpha^4 = \bar{\alpha}^2 \text{ (for appropriate } \alpha\text{)}$

$\alpha^2 = e^{i2\pi/3}, \quad \alpha^4 = e^{i4\pi/3}, \quad \alpha^{-1} = e^{-i\pi/3}$

$\alpha^2 + 1 = e^{i2\pi/3} + 1 = -\frac{1}{2} + i\frac{\sqrt{3}}{2} + 1 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$ $|\alpha^2 + 1| = 1$

$\alpha = \frac{1+\sqrt{3}i}{2} = e^{i\pi/3}$

$\alpha^2 – \alpha + 1 = 0$ $\implies x^2 – x + 1 \text{ is a factor of } x^4 – x^2 + x – 1$

$x^4 – x^2 + x – 1 = (x^2 – x + 1)(x^2 + x – 1)$

$x = \frac{-1 \pm \sqrt{1+4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$