Question

Let $z=\cos \theta +i\sin \theta$ .Then the value of $\sum _{m=1}^{15}Im\left(z^{2m-1}\right)$ at $\theta =2^{\circ }$ is

• A. $\frac{1}{\sin 2^{\circ }}$
• B. $\frac{1}{3\sin 2^{\circ }}$
• C. $\frac{1}{4\sin 2^{\circ }}$
• D. $\frac{1}{2\sin 2^{\circ }}$

Answer 1

Answer: (D)

$\frac{1}{2\sin 2^{\circ }}$

$\sum _{m=1}^{15}Im\left(z^{2m-1}\right)=Im\left(\sum _{m=1}^{15}{z}^{2m-1}\right)$
$=Im\left[\frac{z(1-{\left(z^2\right)}^{15})}{1-z^2}\right]=Im\left[\frac{1-z^{30}}{\overline{z}-z}\right]$
$=Im\left[\frac{1}{2i\sin \theta }\{1-\cos 30\theta -i\sin 30\theta \}\right]=\frac{1}{2\sin \theta }=\frac{1}{2\sin 2^0}$