1.2 DeMoivre's theorem and Euler's formula

1.

By Euler's formula, $e^{i\theta} = \cos{\theta} + i\sin{\theta}$. Thus for all real numbers $n$,

$$(\cos{\theta} + i\sin{\theta})^{n} = (e^{i\theta})^{n} = e^{in\theta} = \cos{n\theta} + i\sin{n\theta}$$

2.

(a) If it contains exponentiation, it is recommended to change the inside of parentheses to polar form once.

$$\frac{1 - i}{\sqrt{2}} = \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}$$

Then $r = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1, \theta = \tan^{-1}{-1} = -\frac{\pi}{4}$. Thus

$$\frac{1 - i}{\sqrt{2}} = e^{-\frac{\pi}{4}i}$$

This shows that

$$\left(\frac{1 - i}{\sqrt{2}}\right)^{7} = (e^{-\frac{\pi}{4}i})^{7} = e^{-\frac{7\pi}{4}i} = e^{\frac{\pi}{4}i}$$

$$= \cos{\frac{\pi}{4}} + i \sin{\frac{\pi}{4}} = \frac{1 + i}{\sqrt{2}}$$

(b) If it contains exponentiation, it is recommended to change the inside of parentheses to polar form once.

$$\sqrt{3} - i = 2e^{-\frac{\pi}{6}i}$$

Then

$$\left(\sqrt{3} - i \right)^{6} = (2e^{-\frac{\pi}{6}i})^{6} = 64e^{-\pi i}$$

$$= 64[\cos{\pi} - i \sin{\pi}] = -64$$

(c) If it contains exponentiation, it is recommended to change the inside of parentheses to polar form once.

$$\frac{(1 + \sqrt{3}i)^{6}}{(-1 + i)^{10}} = \frac{(2e^{\pi i/3})^6}{(\sqrt{2}e^{3\pi i/4})^{10}}$$

$$= \frac{2^{6} e^{2\pi i}}{2^{5} e^{15\pi i/2}} = 2 e^{2\pi i - 15\pi i/2} = 2e^{-11\pi i/2} = 2e^{\pi i/2} = 2i$$

3.

(a) Use the root formula of the quadratic equation

$z^{2} = i = e^{\pi i/2}$ implies $\cos{\frac{\pi/2 + 2k\pi}{2}} + i\sin{\frac{\pi/2 + 2k\pi}{2}} (k = 0,1)$

(b) Use the root formula of the quadratic equation

$z^{3} = -1 = e^{\pi i}$ implies $\cos{\frac{\pi + 2k\pi}{3}} + i\sin{\frac{\pi + 2k\pi}{3}} (k = 0,1,2)$

(c) Use the root formula of the quadratic equation

$z^{4} = -1 + \sqrt{3} i = 2e^{2\pi i/3}$ implies $2^{1/4}[\cos{\frac{2\pi/3 + 2k\pi}{4}} + i\sin{\frac{2\pi/3 + 2k\pi}{4}}] (k = 0,1,2,3)$

4.

(a) To convert from polar coordinates to Cartesian coordinates, use $x = r\cos{\theta}, y = r\sin{\theta}$.

$$e^{i\frac{3}{4}\pi} = \cos{3\pi/4} + i \sin{3\pi/4} = -\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$$

(b) To convert from polar coordinates to Cartesian coordinates, use $x = r\cos{\theta}, y = r\sin{\theta}$.

$$e^{-i\frac{1}{6}\pi} = \cos{\pi/6} - i \sin{\pi/6} = \frac{\sqrt{3}}{2} - i \frac{1}{2}$$

(c) To convert from polar coordinates to Cartesian coordinates, use $x = r\cos{\theta}, y = r\sin{\theta}$.

$$e^{2 + i\pi} = e^2 e^{i \pi} = e^{2}[\cos{\pi} + i \sin{\pi} ]= -e^{2}$$

(d) To convert from polar coordinates to Cartesian coordinates, use $x = r\cos{\theta}, y = r\sin{\theta}$.

$$e^{2 - i\frac{3}{2}\pi} = e^2 e^{-i\frac{3}{2}\pi} = e^{2}[\cos{3\pi/2} - i \sin{3\pi/2}]= e^{2}i$$

5.

(a) To convert from Cartesian coordinates to polar coordinates, we use $r = \sqrt{x^2 + y^2}$ , $\theta = \tan^{-1}{\frac{y}{x}}$ for $- \pi < \theta \leq pi$.

$$-2 = -2 + 0i = \sqrt{2^2 + 0^2}e^{\pi i} = 2e^{\pi i}$$

(b) To convert from Cartesian coordinates to polar coordinates, use $r = \sqrt{x^2 + y^2}$ , $\theta = \tan^{-1}{\frac{y}{x}}$ for $- \pi < \theta \leq pi$.

$$i = 0 + i = \sqrt{0 + 1^2}e^{\pi i/2} = e^{\pi i/2}i$$

(c) To convert from Cartesian coordinates to polar coordinates, use $r = \sqrt{x^2 + y^2}$ , $\theta = \tan^{-1}{\frac{y}{x}}$ for $- \pi < \theta \leq pi$.

$$1 + i = \sqrt{1^2 + 1^2}e^{\pi i/4} = \sqrt{2}e^{\pi i/2}$$

(d) To convert from Cartesian coordinates to polar coordinates, use $r = \sqrt{x^2 + y^2}$ , $\theta = \tan^{-1}{\frac{y}{x}}$縺溘□縺97, $- \pi < \theta \leq pi$.

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

6. First if $\vert z\vert = 1$ then we show $z = e^{i\theta}$.

Let $z = re^{\theta}$. Then since $\vert z\vert = 1$, we have

$$\vert re^{i\theta}\vert = \vert r\vert\vert e^{i\theta}\vert = \vert r\vert = 1$$

$r \geq 0$ implies $r = 1$ and $z = e^{\theta}$.

Next, if $z = e^{i\theta}$, then show $\vert z\vert = 1$.

$$\vert z\vert = \vert e^{i\theta}\vert = \vert\cos{\theta} + i\sin{\theta}\vert = \sqrt{\cos{\theta}^{2} + \sin{\theta}^{2}} = 1$$