Question

If $Z_r=\sin \frac{2\pi r}{11}-i\cos \frac{2\pi r}{11}$ then ${\sum }_{r=0}^{10}{Z}_r=$

• A. $-1$
• B. $0$
• C. $i$
• D. $-i$

Answer 1

Answer: (B)

$0$

$Z_r=sin\frac{2\pi r}{11}-icos\frac{2\pi r}{11}$

$=-i(cos\frac{2\pi r}{11}+isin\frac{2\pi r}{11})$

$=-i{e}^{ir\theta }$ ...$[\theta =\frac{2\pi }{11}]$

${\sum }_0^{10}{Z}_r={\sum }_0^{10}-i{e}^{ir\theta }$

$=-i{\sum }_0^{10}{e}^{ir\theta }$

$=-i(1+e^{i\theta }+e^{2i\theta }+...+e^{10i\theta })$

$=-i\left(\frac{1-e^{11i\theta }}{1-e^{i\theta }}\right)$

$=-i\left(\frac{1-e^{2\pi i}}{1-e^{i\frac{2\pi }{11}}}\right)$ ...[substituting the value of $\theta$]

$=-i\left(\frac{1-1}{1-e^{i\frac{2\pi }{11}}}\right)$

$=0$

So, the answer is option (B)