High school math problem: Trouble with complex representation 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!
 A: You're on the right track (or at least on a right track). After expanding the sine, put the $r^n$ on the numerator to get
$$\sum_{n>0\text{ odd}} \frac{r^n(e^{ix})^n-r^n(e^{-ix})^n}{2i} = \sum_{n\text{ odd}}\frac{z^n-\overline z^n}{2i}$$
You can then split that into two series
$$\frac{1}{2i}\left(\sum_{n\text{ odd}}z^n - \sum_{n\text{ odd}}\overline z^n\right)$$
and then get rid of the oddness restriction on $n$ by pulling out one of the factors in each series:
$$\frac{1}{2i}\left(z\sum_{k=0}^\infty (z^2)^k - \overline z\sum_{k=0}^{\infty} (\overline z^2)^k\right)$$
Can you take it from there?
A: Note: This is related to the Poisson Kernel.  Knowledge of this can shorten the solution.
Your sum is $$\sum_{n=0}^{\infty}r^{2n+1}\sin((2n+1)x)=\frac{1}{2i}\sum_{n=0}^{\infty}r^{2n+1}\left(e^{i(2n+1)x}-e^{-i(2n+1)x}\right).$$  This is then $$\frac{1}{2i}\sum_{n=0}^{\infty}r^{2n+1}e^{i(2n+1)x}-\frac{1}{2i}\sum_{n=0}^{\infty}r^{2n+1}e^{-i(2n+1)x}=\frac{re^{ix}}{2i}\sum_{n=0}^{\infty}\left(re^{ix}\right)^{2n}-\frac{re^{-ix}}{2i}\sum_{n=0}^{\infty}\left(re^{-ix}\right)^{2n}.$$  Summing the geometric series, this becomes $$\frac{re^{ix}}{2i}\frac{1}{1-r^{2}e^{2ix}}-\frac{re^{-ix}}{2i}\frac{1}{1-r^{2}e^{-2ix}}=\frac{1}{2i}\left(\frac{re^{ix}-re^{-ix}-r^{3}e^{-ix}+r^{3}e^{ix}}{1-2r^{2}\cos\left(2x\right)+r^{4}}\right) $$
$$=\frac{r(r^{2}+1)\sin(x)}{1-2r^{2}\cos\left(2x\right)+r^{4}}.$$
A: As commented by Heike 
$$
r^{n}\sin nx=\frac{z^{n}-\overline{z}^{n}}{2i}.
$$
This result can be derived by subtracting the complex conjugates
$$
\begin{eqnarray*}
z^{n} &=&r^{n}e^{inx}=r^{n}\cos nx+ir^{n}\sin nx \\
\overline{z}^{n} &=&r^{n}e^{-inx}=r^{n}\cos nx-ir^{n}\sin nx.
\end{eqnarray*}
$$
The given series can be rewritten as
$$
\begin{eqnarray*}
S &:&=\sum_{n>0\;\text{odd}}r^{n}\sin (nx)=\sum_{k=1}^{\infty }r^{2k-1}\sin
(\left( 2k-1\right) x) \\
&=&\sum_{k=1}^{\infty }\frac{z^{2k-1}-\overline{z}^{2k-1}}{2i} \\
&=&\frac{1}{2i}\sum_{k=1}^{\infty }z^{2k-1}-\frac{1}{2i}\sum_{k=1}^{\infty }
\overline{z}^{2k-1}
\end{eqnarray*}
$$
because
$$
r^{2k-1}\sin \left( \left( 2k-1\right) x\right) =\frac{z^{2k-1}-\overline{z}
^{2k-1}}{2i}.
$$
Since the sums of the two geometric series are 
$$
\sum_{k=1}^{\infty }z^{2k-1}=\frac{z}{1-z^{2}}\qquad (\text{ratio }z^{2},
\text{ first term }z=re^{ix},\ r<1)
$$
and
$$
\sum_{k=1}^{\infty }\overline{z}^{2k-1}=\frac{\overline{z}}{1-\overline{z}
^{2}}\qquad (\text{ratio }\overline{z}^{2},\text{ first term }\overline{z})
$$
we get
$$
S=\frac{1}{2i}\frac{z}{1-z^{2}}-\frac{1}{2i}\frac{\overline{z}}{1-\overline{z
}^{2}}.
$$
