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What is $\sum\limits_{n>0,\text{ odd}} r^n \sin(nx)$ in terms of $z=re^{ix}$? I tried to write $\sin(nx)={e^{inx}-e^{-inx}\over 2i}$ but then I have a sign problem because the $n$ on the associated $r$ is always $>0$.

Thanks in advance!

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    For $r,x\in\mathbb{R}$, $\bar z = r e^{-i x}$ so $r^n \sin(n x)=r^n(e^{inx}-e^{-inx})/2i=(z^n-\bar{z}^n)/2i$.2011-11-02
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    @QED Careful, that is not quite right unless $r=1$. You need $\overline{z}$ instead of $z^{-1}$.2011-11-02
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    This question is probably difficult for high school!2011-11-02

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