Question

If $z=\cos 6^{\circ}+i \sin 6^{\circ}$ then $\displaystyle\sum^{20}_{n=1} 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 \displaystyle\sum_{n=1}^{20} z^{2 n-1}=z+z^{3}+z^{5}+\ldots+z^{39}$
Now, $z=\cos 6^{\circ}+i \sin 6^{\circ}=e^{i 6^{\circ}}$
$\therefore \displaystyle\sum_{n=1}^{20} z^{2 n-1}=e^{i 6^{\circ}}+e^{i 18^{\circ}}+e^{i 30^{\circ}}+\ldots+e^{i(39 \times 69}$
$=e^{i 6^{\circ}}\left[ 1 +e^{i 12}+e^{i 24}+\ldots+e^{i 228}\right]$
$=e^{i 6}\left[\frac{e^{i 240}-1}{e^{i 12}-1}\right]$
$=\frac{e^{i 6}\left[-\frac{1}{2}-\frac{\sqrt{3}}{2} i-1\right]}{e^{i 6}\left(2 i \sin 6^{\circ}\right)}=\frac{-\frac{3}{2}-\frac{\sqrt{3}}{2} i}{2 i \sin 6^{\circ}}$
$=-\frac{\sqrt{3}}{4 \sin 6^{\circ}}+\frac{3}{4 \sin 6^{\circ}} i$
$\therefore \displaystyle\sum_{n=1}^{20} Im\left(z^{2 n-1}\right)=\frac{3}{4 \sin 6^{\circ}}$