Answer to Question #275261 in Complex Analysis for jungkooki

Question

Find z1/n; for n=3, z = 1-i in C (, the Argand Plane).

Answer 1

The polar form of $1−i=\sqrt{2}(\cos(-\dfrac{\pi}{4}+i\sin(-\dfrac{\pi}{4})).$

According to the De Moivre's Formula, all 3-th roots of a complex number 

$\sqrt{2}(\cos(-\dfrac{\pi}{4}+i\sin(-\dfrac{\pi}{4}))$ are given by

$$(\sqrt{2})^{1/3}(\cos(\dfrac{-\pi/4+2\pi k}{3})+i\sin(\dfrac{-\pi/4+2\pi k}{3})),$$ $$k=0,1,2$$

$k=0:$

$$(2)^{1/6}(\cos(\dfrac{-\pi/4+2\pi (0)}{3})+i\sin(\dfrac{-\pi/4+2\pi (0)}{3}))$$

$$=(2)^{1/6}(\cos(\dfrac{\pi}{12})-i\sin(\dfrac{\pi}{12}))$$

$k=1:$

$$(2)^{1/6}(\cos(\dfrac{-\pi/4+2\pi (1)}{3})+i\sin(\dfrac{-\pi/4+2\pi (1)}{3}))$$

$$=(2)^{1/6}(\cos(\dfrac{7\pi}{12})+i\sin(\dfrac{7\pi}{12}))$$

$k=2:$

$$(2)^{1/6}(\cos(\dfrac{-\pi/4+2\pi (2)}{3})+i\sin(\dfrac{-\pi/4+2\pi (2)}{3}))$$

$$=(2)^{1/6}(\cos(\dfrac{5\pi}{4})+i\sin(\dfrac{5\pi}{4}))$$

$$=(2)^{-1/3}(1+i)$$