Solution:

Let $a+i b=\frac{1}{2}+i \frac{\sqrt{3}}{2}=r(\cos \theta+i \sin \theta)$....(1)

Here $a=\frac{1}{2} \quad \& \quad b=\frac{\sqrt{3}}{2}$

Modulus:

$$ r=\sqrt{a^{2}+b^{2}}=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=\sqrt{\frac{1}{4}+\frac{3}{4}}=\sqrt{1}=1 $$

Arugument:

$$ \theta=\tan ^{-1}\left(\frac{b}{a}\right)=\tan ^{-1}\left[\frac{\sqrt{3} / 2}{1 / 2}\right]=\tan ^{-1}(\sqrt{3})=\frac{\pi}{3} $$

$ \therefore$ (1) becomes,

$ \begin{array}{l} \frac{1}{2}+i \frac{\sqrt{3}}{2}=1\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right) \\\\ \Rightarrow\left(\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)^{3 / 4}=\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^{3 / 4} …

Given that $\left(\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)^{3 / 4}$.

i.e. $\left(\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)^{3 / 4}$ = r(cos θ + isin θ)

Let's change it into Polar Form, so we have

$$r = \frac{1}{2}\ +\ \frac{i \sqrt{3}}{2}$$

$$r\ =\ \sqrt{(\frac{1}{2})^{2}\ +\ (\frac{\sqrt{3}}{2})^2}$$

$$r\ =\ \sqrt{\frac{1}{4}\ +\ \frac{3}{4}}\ =\ 1$$

$$Now,\ Cos\ θ\ =\ \frac{1}{2}$$

$$And, …