If z=cos 6^{circ}+i sin 6^{circ},then sum_{n= | vedclass

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(D) Given $z = \cos 6^{\circ} + i \sin 6^{\circ} = e^{i 6^{\circ}}$.We need to find $\sum_{n=1}^{20} \operatorname{Im}(z^{2n-1}) = \operatorname{Im} \

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$\frac{3}{4 \sin 6^{\circ}}$

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(D) Given $z = \cos 6^{\circ} + i \sin 6^{\circ} = e^{i 6^{\circ}}$.We need to find $\sum_{n=1}^{20} \operatorname{Im}(z^{2n-1}) = \operatorname{Im} \left( \sum_{n=1}^{20} z^{2n-1} ight)$.The sum is a geometric progression: $S = z + z^3 + z^5 + \dots + z^{39}$.Here,the first term $a = z$ and the common ratio $r = z^2 = e^{i 12^{\circ}}$.The sum of $20$ terms is $S = z \frac{(z^2)^{20} - 1}{z^2 - 1} = z \frac{z^{40} - 1}{z^2 - 1}$.Substituting $z = e^{i 6^{\circ}}$,we get $S = e^{i 6^{\circ}} \frac{e^{i 240^{\circ}} - 1}{e^{i 12^{\circ}} - 1}$.Using the identity $e^{i 	heta} - 1 = e^{i 	heta/2} (e^{i 	heta/2} - e^{-i 	heta/2}) = e^{i 	heta/2} (2i \sin(	heta/2))$,we have:$S = e^{i 6^{\circ}} \frac{e^{i 120^{\circ}} (2i \sin 120^{\circ})}{e^{i 6^{\circ}} (2i \sin 6^{\circ})} = e^{i 120^{\circ}} \frac{\sin 120^{\circ}}{\sin 6^{\circ}}$.$S = (\cos 120^{\circ} + i \sin 120^{\circ}) \frac{\sin 120^{\circ}}{\sin 6^{\circ}} = \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} ight) \frac{\sqrt{3}/2}{\sin 6^{\circ}}$.$S = -\frac{\sqrt{3}}{4 \sin 6^{\circ}} + i \frac{3}{4 \sin 6^{\circ}}$.Taking the imaginary part,$\operatorname{Im}(S) = \frac{3}{4 \sin 6^{\circ}}$.

If $z=\cos 6^{\circ}+i \sin 6^{\circ}$,then $\sum_{n=1}^{20} \operatorname{Im}\left(z^{2 n-1}\right)=$

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