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Complex Numbers — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsComplex Numbers2 Marks
MCQ
$\sum_{k=1}^6\left(\sin \frac{2 \pi k}{7}-i \cos \frac{2 \pi k}{7}\right)=$
  • A
    $-1$
  • B
    $0$
  • ✓
    i
  • D
    $- i$
✓

Answer

Correct option: C.
i
(C)
$\sum_{ k =1}^6\left(\sin \frac{2 \pi k }{7}- i \cos \frac{2 \pi k }{7}\right)$
$=- i \sum_{ k =1}^6\left(\cos \frac{2 \pi k }{7}+ i \sin \frac{2 \pi k }{7}\right)$
$=-i\left[\left(\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7}\right)+\left(\cos \frac{4 \pi}{7}+i \sin \frac{4 \pi}{7}\right)\right.$ $\left.+\ldots .+\left(\cos \frac{12 \pi}{7}+i \sin \frac{12 \pi}{7}\right)\right]$
$=-i\left[\left(\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7}\right)+\left(\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7}\right)^2\right.$ $\left.+\ldots .+\left(\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7}\right)^6\right]$
$=- i \left[\frac{x\left(x^6-1\right)}{x-1}\right]$, where $x=\cos \frac{2 \pi}{7}+ i \sin \frac{2 \pi}{7}$
$=- i \left[\frac{x^7-x}{x-1}\right]=- i \left[\frac{1-x}{x-1}\right]$
$\ldots\left[\begin{array}{rl}\because x^7 & =\left(\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7}\right)^7 \\ & =\cos 2 \pi+i \sin 2 \pi=1\end{array}\right]$
$= i \left(\frac{x-1}{x-1}\right)= i$

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