Using complex variables to find sums of Fourier series Use complex variables to find the sum of the Fourier Series:
$$\sin(\theta) + \frac{\sin(2\theta)}{2^{2}} + \frac{\sin(3\theta)}{2^{3}}+\cdots$$
where $\theta$ is a real variable.
 A: $$\sum_{n=1}^{+\infty}\frac{\sin(n\theta)}{2^n}=\text{Im}\sum_{n\geq 1}\left(\frac{e^{i\theta}}{2}\right)^n =\text{Im}\left(\frac{e^{i\theta}}{2-e^{i\theta}}\right)=\text{Im}\left(\frac{e^{i\theta}(2-e^{-i\theta})}{5-4\cos\theta}\right)$$
hence:
$$\sum_{n=1}^{+\infty}\frac{\sin(n\theta)}{2^n}=\frac{2\sin\theta}{5-4\cos\theta}.$$
A: We have 
$$\begin{align}
f(\theta)&=\sin\theta + \sum_{k=2}^{\infty}\frac{\sin k\theta}{2^k} \tag 1\\\\
&=\sin\theta+\text{Im}\left( \sum_{k=2}^{\infty}\left(\frac{e^{i\theta}}{2}\right)^k\right) \tag 2\\\\
&=\sin\theta+\text{Im}\left(\frac{e^{i2\theta}}{2-e^{i\theta}}\right) \tag 3\\\\
&=\sin\theta+\frac12 \frac{2\sin 2 \theta-\sin \theta}{5-4\cos \theta} \tag 4\\\\
&=\frac12 \frac{9 \sin \theta-2\sin 2\theta}{5-4\cos \theta}\tag 5
\end{align}$$
In going from $(1)$ to $(2)$, we note that the imaginary part of $e^{i\theta}=\cos \theta+i\sin \theta$ is $\sin \theta$ and thus write $\sin \theta=\text{Im}(e^{ \theta})$.  We also make use of the fact that $e^{ik\theta}=\left(e^{i\theta}\right)^k$.
In going from $(2)$ to $(3)$, we made use of the sum for a geometric series $\sum_{k=m}^{\infty}r^k=\frac{r^m}{1-r}$ with $r=\frac{e^{i\theta}}{2}$.
In going from $(3)$ to $(4)$, we multiplied both numerator and denominator by the complex conjugate of the denominator, thereby rendering the denominator real and equal to the square of the magnitude of $2-e^{i\theta}$.

NOTE:
We remark that if the first term of the series were $\frac{\sin \theta}{2}$ rather than $\sin \theta$, then $(5)$ would be 
$$\begin{align}
\frac12 \frac{9 \sin \theta-2\sin 2\theta}{5-4\cos \theta}-\frac 12 \sin \theta &=\frac12 \frac{9\sin \theta-2\sin 2 \theta-(5\sin \theta -2 \sin 2\theta)}{5-4\cos \theta}\\\\
&=\frac{2\sin \theta}{5-4\cos \theta}
\end{align}$$
A: You should be able to do this with the observation that $$\sin(n\theta)=\Im(e^{in\theta})$$
And that the imaginary part is a linear operator.
