Question

Which of the following is a solution of 2z3+16i=0 ?

Answer 1

$$2z^3+16i=0$$

$$z^3=-8i$$

The polar form of $- 8 i$  is $8(\cos(-\dfrac{\pi}{2})+i\sin(-\dfrac{\pi}{2})).$

$k=0:$

$$\sqrt[3]{8}(\cos(\dfrac{-\pi/2+2\pi(0)}{3})+i\sin(\dfrac{-\pi/2+2\pi(0)}{3}))$$

$$=\sqrt{3}-i$$

$k=1:$

$$\sqrt[3]{8}(\cos(\dfrac{-\pi/2+2\pi(1)}{3})+i\sin(\dfrac{-\pi/2+2\pi(1)}{3}))$$

$$=i$$

$k=2:$

$$\sqrt[3]{8}(\cos(\dfrac{-\pi/2+2\pi(2)}{3})+i\sin(\dfrac{-\pi/2+2\pi(2)}{3}))$$

$$=-\sqrt{3}-i$$

The solutions are $\{-\sqrt{3}-i, i, \sqrt{3}-i\}.$