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  4. The value of displaystyle∑k=110( sin (2 π k/11)-i cos (2 π k/11)) is

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Q. The value of $\displaystyle\sum_{k=1}^{10}\left(\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11}\right)$ is

Complex Numbers and Quadratic Equations

Solution:

We have,
$\displaystyle\sum_{k=1}^{10}\left(\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11}\right)$
$=\displaystyle\sum_{k=1}^{10}\left(-i^{2} \sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11}\right)$
$=-i \displaystyle\sum_{k=1}^{10}\left(\cos \frac{2 \pi k}{11}+i \sin \frac{2 \pi k}{11}\right)=-i \sum_{k=1}^{10} e^{i \frac{2 \pi k}{11}}$
$=-i\left[\displaystyle\sum_{k=0}^{10} e^{i \frac{2 \pi k}{11}}-1\right]$
$=-i($ sum of $11\text{th}$ roots of unity $-1)$
$=-i(0-1)=i$