1. Tardigrade
  2. Question
  3. Mathematics
  4. If ω= cos (π/n)+i sin (π/n), then value of 1+ω+ω2+ ldots+ωn-1 is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

Complex Numbers and Quadratic Equations

Solution:

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