Question

Find all the values ${(\frac{1}{2}+i\frac{\sqrt{3}}{2})}^4$. Hence prove that the product of the four values is $1$.

Answer 1

Given that ${(\frac{1}{2}+\frac{\sqrt{3}}{2})}^{\frac{3}{4}}=r(\cos \theta +i\sin \theta )$
We have to write it in Polar form
i.e., $r=|\frac{1}{2}+\frac{i\sqrt{3}}{2}|=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$
$=\sqrt{\frac{1}{4}+\frac{3}{4}}=1$
$\therefore r=1$
$\cos \theta =\frac{1}{2}$

$\sin \theta =\frac{\sqrt{3}}{2}\Rightarrow \theta =\frac{\pi }{3}$
$\therefore$ The general Polar from is
$\cos (\frac{\pi }{3}+2n\pi )+i\sin (\frac{\pi }{3}+2n\pi )$
Apply De Moivre's Theorem we get
${(\frac{1}{2}+\frac{i\sqrt{3}}{2})}^{\frac{3}{4}}=\left[\cos \frac{3}{4}\left(\frac{\pi +6n\pi }{3}\right)i\sin \frac{3}{4}\left(\frac{\pi +6n\pi }{3}\right)\right]$
$=[\cos \left(\frac{\pi +6n\pi }{4}\right)+i\sin \left(\frac{\pi +6n\pi }{3}\right)]$ ..... (1)
Substitute $n=0,1,2,3$ in equation (1), we get for $n=0$.
$[\cos \left(\frac{\pi +6\left(0\right)\pi }{4}\right)+i\sin \left(\frac{\pi +6\left(0\right)\pi }{4}\right)]=\cos \left(\frac{\pi }{4}\right)+i\sin \left(\frac{\pi }{4}\right)$
For $n=1$,
$[\cos \left(\frac{\pi +6\left(1\right)\pi }{4}\right)+i\sin \left(\frac{\pi +6\left(1\right)\pi }{4}\right)]=\cos \left(\frac{\pi }{4}\right)+i\sin \left(\frac{7\pi }{4}\right)$
III ly
For $n=2$ is $\cos (\frac{13\pi }{4}+i\sin \frac{13\pi }{4})$
For $n=3$ is $\cos (\frac{19\pi }{4}+i\sin \frac{19\pi }{4})$
$\therefore$ the product of four values is
$\cos [\frac{\pi }{4}+\frac{7\pi }{4}+\frac{13\pi }{4}+\frac{19\pi }{4}]+i\sin [\frac{\pi }{4}+\frac{7\pi }{4}+\frac{13\pi }{4}+\frac{19\pi }{4}]$
$=\cos \left(\frac{40\pi }{4}\right)+i\sin \left(\frac{40\pi }{4}\right)$
$=\cos \left(10\pi \right)+i\sin \left(10\pi \right)$
$=\left[\cos \right(\pi )+i\sin (\pi ){]}^{10}=[1+0{]}^{10}=1$
Hence, proved.