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} \end{array} $

$$ \begin{aligned}=\left[\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^{3}\right]^{\frac{1}{4}} \\\\ &=\left[\cos 3\left(\frac{\pi}{3}\right)+i \sin 3\left(\frac{\pi}{3}\right)\right]^{1 / 3} \\\\ &=[\cos \pi+i \sin \pi]^{1 / 4} \\\\ &=[\cos (2 k \pi+\pi)+i \sin (2 k \pi+\pi)]^{1 / 4} \\\\ &=\cos \left(\frac{2 k \pi+\pi}{4}\right)+i \sin \left(\frac{2 k \pi+\pi}{4}\right) \\ \quad \text { where } k=0,1,2,3 \end{aligned}\\ $$ $$ \text { when } \begin{aligned} \mathrm{k} &=0 ; \mathrm{R}_{1}=\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \\\\ \mathrm{k}=1 ; \mathrm{R}_{2} &=\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4} \\\\ \mathrm{k}=2 ; \mathrm{R}_{3} &=\cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4} \\\\ \mathrm{k}=3 ; \mathrm{R}_{4} &=\cos \frac{7 \pi}{4}+i \sin \frac{7 \pi}{4}\\ \end{aligned} $$ Product of the four values: $$ \begin{array}{l} R_{1} \times R_{2} \times R_{3} \times R_{3} \\\\ =\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)\left(\cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4}\right)\left(\cos \frac{7 \pi}{4}+i \sin \frac{7 \pi}{4}\right) \\\\ =\cos \left(\frac{\pi}{4}+\frac{3 \pi}{4}+\frac{5 \pi}{4}+\frac{7 \pi}{4}\right)+i \sin \left(\frac{\pi}{4}+\frac{3 \pi}{4}+\frac{5 \pi}{4}+\frac{7 \pi}{4}\right) \\\\ =\cos \frac{16 \pi}{4}+i \sin \frac{16 \pi}{4} \\\\ =\cos 4 \pi+i \sin 4 \pi \\\\ =1+i(0) =1 \end{array} $$

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

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

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

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

$$Now,\ Cos\ θ\ =\ \frac{1}{2}$$

$$And, Sin\ θ\ =\ \frac{\sqrt{3}}{2}$$

$$i.e.\ θ\ =\ \frac{\pi}{3}$$

∴ The general Polar form is

$$cos\ (\frac{\pi}{3}\ +\ 2n\pi)\ +\ isin\ (\frac{\pi}{3}\ +\ 2n\pi)$$

Apply De Moivre's Theorem we get,

$$(\frac{1}{2}\ +\ \frac{i\sqrt{3}}{2})^{\frac{3}{4}}\ =\ cos\ \frac{3}{4}\ (\frac{\pi\ +\ 6n\pi}{3})\ +\ isin\ \frac{3}{4}\ (\frac{\pi\ +\ 6n\pi}{3})$$

$$=\ cos\ (\frac{\pi\ +\ 6n\pi}{3})\ +\ isin\ (\frac{\pi\ +\ 6n\pi}{3})\ ........\ (i)$$

Substitute n=0,1,2,3 in equation (1), we get for n=0.

$$=\ cos\ (\frac{\pi\ +\ 6(0)\pi}{3})\ +\ isin\ (\frac{\pi\ +\ 6(0)\pi}{4})\ =\ cos\ (\frac{\pi}{4})\ +\ isin\ (\frac{\pi}{4})$$

For n=1,

$$=\ cos\ (\frac{\pi\ +\ 6(1)\pi}{3})\ +\ isin\ (\frac{\pi\ +\ 6(1)\pi}{4})\ =\ cos\ (\frac{\pi}{4})\ +\ isin\ (\frac{7\pi}{4})$$

Similarly,

For n=2,

$$cos\ (\frac{13 \pi}{4}\ +\ i\ sin\ \frac{13\ \pi}{4})$$

For n=3,

$$cos\ (\frac{19 \pi}{4}\ +\ i\ sin\ \frac{19\ \pi}{4})$$

∴ the product of four values is

$$cos\ (\frac{\pi}{4}\ +\ \frac{7\pi}{4}\ +\ \frac{13\pi}{4}\ +\ \frac{19\pi}{4})\ +\ isin\ (\frac{\pi}{4}\ +\ \frac{7\pi}{4}\ +\ \frac{13\pi}{4}\ +\ \frac{19\pi}{4})$$

$$=\ cos\ (\frac{40\ \pi}{4})\ +\ isin\ (\frac{40\ \pi}{4})$$

$$=\ cos\ (10π)\ +\ isin\ (10π)$$

$$=\ [cos(π)+isin(π)]^{10} =[1+0]^{10} =1$$

Hence, Proved.