Question

If $z=\cos 6^{\circ }+i\sin 6^{\circ }$ then $\sum _{n=1}^{20}Im(z^{2n-1})=$

• A. $0$
• B. $-1$
• C. $\frac{-3}{4\sin 6^{\circ }}$
• D. $\frac{3}{4\sin 6^{\circ }}$

Answer 1

Answer: (D)

$\because \sum _{n=1}^{20}{z}^{2n-1}=z+z^3+z^5+\dots +z^{39}$
Now, $z=\cos 6^{\circ }+i\sin 6^{\circ }=e^{i{6}^{\circ }}$
$\therefore \sum _{n=1}^{20}{z}^{2n-1}=e^{i{6}^{\circ }}+e^{i18^{\circ }}+e^{i30^{\circ }}+\dots +e^{i(39\times 69}$
$=e^{i{6}^{\circ }}\left[1+e^{i12}+e^{i24}+\dots +e^{i228}\right]$
$=e^{i6}\left[\frac{e^{i240}-1}{e^{i12}-1}\right]$
$=\frac{e^{i6}\left[-\frac{1}{2}-\frac{\sqrt{3}}{2}i-1\right]}{e^{i6}\left(2i\sin 6^{\circ }\right)}=\frac{-\frac{3}{2}-\frac{\sqrt{3}}{2}i}{2i\sin 6^{\circ }}$
$=-\frac{\sqrt{3}}{4\sin 6^{\circ }}+\frac{3}{4\sin 6^{\circ }}i$
$\therefore \sum _{n=1}^{20}Im\left(z^{2n-1}\right)=\frac{3}{4\sin 6^{\circ }}$