Solve: $z=\sqrt{1+\sqrt3i}$

Question

Solve: $z=\sqrt{1+\sqrt3i}$

$r=\sqrt{\sqrt{3}^2+(1)^2}=\sqrt{4}=2$
$\theta=tan^{-1}(\frac{\sqrt{3}}{1})=\frac{\pi}{3}$

0: $\sqrt{z}=2*[cos(\frac{\pi}{3})+isin(\frac{\pi}{3})]=1+\sqrt{3}i$

Is it right?

Answer 1

No, it is not correct. An answer will be$$\sqrt2\left(\cos\left(\frac\pi6\right)+i\sin\left(\frac\pi6\right)\right)=\sqrt{\frac32}+\frac i{\sqrt2}.$$Another one will be $-\sqrt{\frac32}-\frac i{\sqrt2}$, of course.