## A Cute Complex Problem

Question:

If $w=\cos{\frac{\pi}{n}}+i\sin{\frac{\pi}{n}}$, then find the value of $1+w+w^{2}+w^{3}+\ldots+w^{n-1}$.

Solution:

We have $S=1+w+w^{2}+w^{3}+\ldots+w^{n-1}=\frac{1-w^{n}}{1-w}$.

But, $w^{n}=\cos{\frac{n\pi}{n}}+i\sin{\frac{n\pi}{n}}=-1$

Thus, $S=\frac{2}{1-w}$

but, $1-w=1-\cos{\frac{\pi}{n}}-i\sin{\frac{\pi}{n}}$ which equals

$2\sin^{2}{\frac{\pi}{2n}}-2i\sin{\frac{\pi}{2n}}\cos{\frac{\pi}{2n}}$

that is, $-2i\sin{\frac{\pi}{2n}}[\cos{\frac{\pi}{2n}}+i\sin{\frac{\pi}{2n}}]$.

Thus, $S=\frac{-2}{2i\sin{\frac{\pi}{2n}}}[\cos{\frac{\pi}{2n}}+i\sin{\frac{\pi}{2n}} ]^{-1}=1+i\cot{\frac{\pi}{2n}}$.

Hope you are finding it useful,

More later,

Nalin Pithwa

This site uses Akismet to reduce spam. Learn how your comment data is processed.