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

Answer 1

Answer: (C)

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