Question

If $w=\cos (\pi /n)+i\sin (\pi /n)$, then value of $1+w+$ $w^2+\cdots +w^{n-1}$ is

• A. $1+i$
• B. $1+i\tan (\pi /2n)$
• C. $1+i\cot (\pi /2n)$
• D. none of these

Answer 1

Answer: (C)

$1+w+w^2+\dots +w^{n-1}$
$=\frac{1-w^n}{1-w}=\frac{1-\cos \pi +i\sin \pi }{1-\cos (\pi /n)-i\sin (\pi /n)}$
$=\frac{2}{2{\sin }^2(\pi /2n)-2i\sin (\pi /2n)\cos (\pi /2n)}$
$=\frac{2}{-2i\sin (\pi /2n)[\cos (\pi /2n)+i\sin (\pi /2n)]}$
$=\frac{\cos (\pi /2n)-i\sin (\pi /2n)}{-i\sin (\pi /2n)}=1+i\cot \left(\frac{\pi }{2n}\right)$