Question

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

Answer 1

Step 1: Find the value of $\sum _{m=1}^{15}Im\left(z^{2m-1}\right)$

Given that,

$z=\cos \left(\theta \right)+i\sin \left(\theta \right)$

We can write the above equation as

$z=\cos \left(\theta \right)+i\sin \left(\theta \right)=e^{i\theta }$

We also have,

$\begin{array}{rcl}\sum _{m=1}^{15}Im\left(z^{2m-1}\right)& =& \sum _{m=1}^{15}Im\left({\left(e^{i\theta }\right)}^{2m-1}\right)\\ & =& \sum _{m=1}^{15}Im\left({\left(e\right)}^{i\left(2m-1\right)\theta }\right)\end{array}$

Solving the above equation,

$\begin{array}{rcl}\sum _{m=1}^{15}Im\left({\left(e\right)}^{i\left(2m-1\right)\theta }\right)& =& \sin \theta +\sin 3\theta +\sin 5\theta +\dots ..+\sin 29\theta \\ & =& \frac{\sin \left(\frac{\theta +29\theta }{2}\right)\sin \left(\frac{15\times 2\theta }{2}\right)}{\sin \left(\frac{2\theta }{2}\right)}\\ & =& \frac{\sin 15\theta \sin 15\theta }{\sin \theta }\\ & =& \frac{1}{4}\sin 2^0\end{array}$

Therefore, the value of $\sum _{m=1}^{15}Im\left(z^{2m-1}\right)$ is $\frac{1}{4}\sin 2^0$.