Question

a solution of 2z3+16i=0

Answer 1

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

$$z^3=-8i$$

$$z=2\sqrt[3]{-i}$$

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

$k=0:$

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

$$=\dfrac{\sqrt{3}}{2}-\dfrac{i}{2}$$

$k=1:$

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

$$=i$$

$k=2:$

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

$$=-\dfrac{\sqrt{3}}{2}-\dfrac{i}{2}$$

$$z_1=\sqrt{3}-i, z_2=2i,z_3=-\sqrt{3}-i,$$