Answer to Question #230184 in Complex Analysis for Anand

Question

Say true or false, with a short proof.

All the cube roots of I in C are z₁ = cos(π/2)+ i sin(π/2) , z₂ = cos (π/6)+ isin(π/6) and z₃ = cos (5π/6)+i sin (5π/6)

Answer 1

The polar form of $i$ is $1(\cos(\dfrac{\pi}{2})+i\sin(\dfrac{\pi}{2})).$

According to the De Moivre's Formula, all cubic roots of a complex number $1(\cos(\dfrac{\pi}{2})+i\sin(\dfrac{\pi}{2}))$ are given by $1^{1/3}(\cos(\dfrac{\pi/2+2\pi k}{3})+i\sin(\dfrac{\pi/2+2\pi k}{3})), k=0,1,2.$

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

$$=\cos(\dfrac{\pi}{6})+i\sin(\dfrac{\pi}{6})$$

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

$$=\cos(\dfrac{5\pi}{6})+i\sin(\dfrac{5\pi}{6})$$

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

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

Therefore the statement is False.