Question

Use De Moivre’s Theorem to determine the cube root of Z and leave your answer in polar

form with the angle in radians

(a) Z = 1+i√3

Answer 1

a) The polar form of $1+i\sqrt{3}$ is

$$2\big(\cos(\dfrac{\pi}{3})+i\sin(\dfrac{\pi}{3})\big)$$

We have that $r=2, \theta=\dfrac{\pi}{3}, n=3.$

According to the De Moivre's Formula

$k=0:$

$$\sqrt[3]{2}\big(\cos(\dfrac{\pi/3+2\pi(0)}{3})+i\sin(\dfrac{\pi/3+2\pi(0)}{3})\big)$$

$$=\sqrt[3]{2}\big(\cos(\dfrac{\pi}{9})+i\sin(\dfrac{\pi}{9})\big)$$

$k=1:$

$$\sqrt[3]{2}\big(\cos(\dfrac{\pi/3+2\pi(1)}{3})+i\sin(\dfrac{\pi/3+2\pi(1)}{3})\big)$$

$$=\sqrt[3]{2}\big(\cos(\dfrac{7\pi}{9})+i\sin(\dfrac{7\pi}{9})\big)$$

$$=\sqrt[3]{2}\big(-\cos(\dfrac{\pi}{9})+i\sin(\dfrac{\pi}{9})\big)$$

$k=2:$

$$\sqrt[3]{2}\big(\cos(\dfrac{\pi/3+2\pi(2)}{3})+i\sin(\dfrac{\pi/3+2\pi(2)}{3})\big)$$

$$=\sqrt[3]{2}\big(\cos(\dfrac{13\pi}{9})+i\sin(\dfrac{13\pi}{9})\big)$$

$$=\sqrt[3]{2}\big(-\cos(\dfrac{4\pi}{9})-i\sin(\dfrac{4\pi}{9})\big)$$