How $[\sqrt[5]3 \cdot (\frac{\sqrt{3}}2+\frac{i}2)]^{10}$ is $[\sqrt[5]{3}\cdot cis(30^{\circ})]^{10}$ in polar form?

Question

Evaluate $[ \sqrt[5]{3} ( \frac{ \sqrt{3}}{2}+ \frac{i}{2} ) ]^{10}$.

I calculated and as a result:

$r^{2}=x^{2}+y^{2}$

$r\;=\;\sqrt{{(\frac{\sqrt3}2)}^2+{(\frac i2)}^2}=\sqrt{\frac12}\\$

$x\;=\;r\;\cos\;\theta\\\theta\;=\;r\;\mathrm{cos}^{-1}\;(x)\;=\;\sqrt{\frac12}\cos^{-1}(\frac{\sqrt3}2)=\frac{\sqrt3\sqrt{\mathrm\pi}}6\\\lbrack\sqrt[5]3\;\sin(\frac{\sqrt3\sqrt{\mathrm\pi}}6)]^{10}$

The result is not the same as $[\sqrt[5]{3}\cdot cis(30^\circ )]^{10}$

Answer 1

You made some miscalculations. To get $\dfrac{\sqrt3}2+i\dfrac12$ in polar form,

$r=\sqrt{\left(\dfrac{\sqrt3}2\right)^2+\left(\dfrac12\right)^2}=1,$

and $\cos\theta=\dfrac{\sqrt3}2$ means $\theta=30^\circ.$

Can you take it from here?