Answer
$Z_{0}=\sqrt{\frac{3}{2}}+i\frac{\sqrt{2}}{2}\;\;\;\;\;\;\;\;\;and\;\;\;\;\;\;\;\;\;Z_{1}=-\sqrt{\frac{3}{2}}-i\frac{\sqrt{2}}{2}$
Work Step by Step
$\{2\;[cos(\frac{\pi}{3})+isin(\frac{\pi}{3})]\;\}^{\frac{1}{2}}\;=\;\;\;\;\;\;\;\;\;\;\sqrt{2}[\;cos(\frac{\pi}{3}+2\pi n)+isin(\frac{\pi}{3}+2\pi n)\;]^{\frac{1}{2}}\;\;=\;\\\\$
$\sqrt{2}\;[cos(\frac{\pi+6\pi n}{6})+isin(\frac{\pi+6\pi n}{6})]\;\;\\\\$
At n=0;
$z_{0}=\sqrt{2}\;[cos(\frac{\pi}{6})+isin(\frac{\pi}{6})]\;=\;\sqrt{2}\;[\frac{\sqrt{2}}{3}+i\frac{1}{2}]\;=\;\sqrt{\frac{3}{2}}+i\frac{\sqrt{2}}{2}$
At n=1;
$z_{1}=\sqrt{2}\;[cos(\frac{7\pi}{6})+isin(\frac{7\pi}{6})]\;=\;\sqrt{2}\;[-\frac{\sqrt{2}}{3}-i\frac{1}{2}]\;=\;-\sqrt{\frac{3}{2}}-i\frac{\sqrt{2}}{2}$
