Question

If $\omega =\cos \frac{\pi }{n}+i\sin \frac{\pi }{n}$, then value of $1+\omega +{\omega }^2+\dots +{\omega }^{n-1}$ is

• A. $1+i\cot \left(\frac{\pi }{2\pi }\right)$
• B. $1+i\tan \left(\frac{\pi }{n}\right)$
• C. $1+i$
• D. none of these

Answer 1

Answer: (A)

We have
$1+\omega +{\omega }^2+\dots +{\omega }^{n-1}=\frac{1-{\omega }^n}{1-\omega }=\frac{2}{1-\omega }$
as ${\omega }^n=\cos \left(\frac{n\pi }{n}\right)+i\sin \left(\frac{n\pi }{n}\right)=\cos \pi +i\sin \pi =-1$
Now,
$\frac{2}{1-\omega }=\frac{2(1-\bar{\omega })}{(1-\omega )(1-\bar{\omega })}=\frac{2(1-\bar{\omega })}{1-(\omega +\bar{\omega })+\omega \bar{\omega }}$
$=\frac{2(1-\bar{\omega })}{2-2\mathrm{Re}(\omega )}\left[\because \omega \bar{\omega }=|\omega {|}^2=1\right]$
$=\frac{1-\mathrm{Re}(\omega )+i\mathrm{Im}(\omega )}{1-\mathrm{Re}(\omega )}=1+i\frac{\mathrm{Im}(\omega )}{1-\mathrm{Re}(\omega )}$
$=1+i\frac{\sin \left(\frac{\pi }{n}\right)}{1-\cos \left(\frac{\pi }{n}\right)}=1+i-\frac{2\sin \left(\frac{\pi }{2n}\right)\cos \left(\frac{\pi }{2n}\right)}{2{\sin }^2\left(\frac{\pi }{2n}\right)}$
$=1+i\cot (\pi /2n)$