Answer with explanation:
[tex]z=2^{\frac{1}{3}}\\\\z^3=2\\\\z^3=2 \times (1 + 0 i)\\\\z^3=2 \times [\cos 0^{\circ} + i \sin 0^{\circ}]\\\\z^3=2 \times [\cos (2 k\pi+ 0^{\circ}) + i \sin(2 k\pi+0^{\circ})],where,k=0,1,2,\\\\z=[2\times[ (\cos (2 k\pi+ 0^{\circ}) + i \sin(2 k\pi+0^{\circ})]]^{\frac{1}{3}}\\\\z=2^{\frac{1}{3}}\times[\cos \frac{2 k\pi+ 0^{\circ}}{3} + i \sin\frac{(2 k\pi+0^{\circ})}{3}]\\\\Z_{0}=2^{\frac{1}{3}},for, k=0\\\\z_{1}=2^{\frac{1}{3}}\times[\cos \frac{(2\pi)}{3} + i \sin\frac{(2\pi)}{3}][/tex]
[tex]z_{1}=2^{\frac{1}{3}}\times[\cos(120^{\circ})+i\sin(120^{\circ})],\text{Used Demoiver's theorem}[\cos A + i\sin A]^k=\cos Ak + i\sin Ak[/tex]
Option A: →Cube root of 2 [cos (120 degrees) + i sin (120 degrees)]
