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If $\theta=\frac{\pi}{6}$, then the $10^{\text {th }}$ term of the series $1+(\cos \theta+i \sin \theta)^1+(\cos \theta+i \sin \theta)^2+\ldots$ is
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$-i$
$T_{10}=(\cos \theta+i \sin \theta)^9$
$\begin{aligned} & =\cos 9 \theta+i \sin 9 \theta=\cos \frac{9 \pi}{6}+\sin \frac{9 \pi}{6} \\ & =\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2}=0+i(-1)=-i\end{aligned}$
$\begin{aligned} & =\cos 9 \theta+i \sin 9 \theta=\cos \frac{9 \pi}{6}+\sin \frac{9 \pi}{6} \\ & =\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2}=0+i(-1)=-i\end{aligned}$
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