A Cute Complex Problem


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}.


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


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

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

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

%d bloggers like this: