Mathematics - Complex Number Question with Solution | TestHub

Given that $z=\cos \theta +i\sin \theta =e^{i\theta }$ $$\begin{array}{rl}\therefore \sum _{m=1}^{15}\mathrm{Im}\left(z^{2m-1}\right)& =\sum _{m=1}^{15}\mathrm{Im}{\left(e^{i\theta }\right)}^{2m-1}=\sum _{m=1}^{15}\mathrm{Im}{e}^{i(2m-1)\theta }\\ & =\sin \theta +\sin 3\theta +\sin 5\theta +\dots +\sin 29\theta \\ & =\frac{\sin \left(\frac{\theta +29\theta }{2}\right)\sin \left(\frac{15\times 2\theta }{2}\right)}{\sin \left(\frac{2\theta }{2}\right)}=\frac{\sin (15\theta )\sin (15\theta )}{\sin \theta }=\frac{1}{4\sin 2^{\circ }}\end{array}$$