Writing a function $f : [-\pi,\pi) \to \mathbb{R}$ as $\sum c_k e^{ikx}$ where $c_k$ is to be found I have a function on $[-\pi, \pi)$ defined as:
$$
f(x) = \begin{cases}
-1 & \mbox{if} \;x \in [-\pi,0) \\
1 & \mbox{if} \;x \in [0,\pi) \\
\end{cases}
$$
And I have to write it in the form $f(x) = \sum_{k=-\infty}^\infty c_k e^{ikx}$, where I have to find the values of $c_k$.

My first thought was to split it up into $\sum c_k(\cos kx + i \sin kx)$, but I don't know where to go from there. If I had to find the Fourier series for this function, i.e. finding the values of $a_n$ and $b_n$ in $\sum a_k \cos kx + b_k \sin kx$, then I'd see that the function was odd and throw away all of the cos terms by saying $a_k = 0$. But I can't do that when both the cos and sin bits have the same coefficient $c_k$, so I'm a bit stuck. 
 A: The exponential Fourier coefficients are given by:
$$c_k=\frac1{2\pi}\int_{-\pi}^{\pi}f(t)\mathrm{e}^{-ikt}\,\mathrm{d}t.$$
With this we get:
$$c_k=\begin{cases}i\dfrac{(-1)^k-1}{k\pi}&\text{if $k\neq0$}\\0&\text{if $k=0$}.\end{cases}$$
Since $f$ is piecewise of class $C^1$ we obtain, by Dirichlet Theorem, that
$$\forall x\in[-\pi,\pi),\ f^{\mathrm{reg}}(x)=\sum_{k\in\mathbb{Z}}c_k\mathrm{e}^{ikx},$$
where $f^{\mathrm{reg}}$ is the regular part of $f$:
$$\forall x\in[-\pi,\pi),\ f^{\mathrm{reg}}(x)=\begin{cases}0&\text{if $x\in\{-\pi,0\}$}\\-1&\text{if $x\in(-\pi,0)$}\\1&\text{if $x\in(0,\pi)$.}\end{cases}$$

Another method using what you tried: indeed, $f$ is essentially odd, hence all the coefficients $a_k$ (using your notation) are nil. We only need to determine the $b_k$: for $k\in\mathbb{N}^*$,
$$b_k=\frac1\pi\int_{-\pi}^{\pi}f(t)\sin(kt)\,\mathrm{d}t=\frac2\pi\int_0^{\pi}\sin(kt)\,\mathrm{d}t=-2\frac{(-1)^k-1}{k\pi}.$$
By Dirichlet Theorem, we have:
$$\forall x\in[-\pi,\pi),\ f^{\mathrm{reg}}(x)=\sum_{n=1}^{+\infty}b_n\sin(nx).$$
Now, using the fact that:
$$\sin(x)=\frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i},$$
the general term of our Fourier series reads as:
$$b_n\sin(nx)=i\frac{(-1)^n-1}{n\pi}\mathrm{e}^{inx}+i\frac{(-1)^{-n}-1}{-n\pi}\mathrm{e}^{-inx}.$$
We're hence tempted to set:
$$\forall n\in\mathbb{Z},\ c_n=\begin{cases}0&\text{if $n=0$}\\i\dfrac{(-1)^n-1}{n\pi}&\text{if $n\neq0$}\end{cases}$$
and we hence obtain:
$$\forall x\in[-\pi,\pi),\ f^{\mathrm{reg}}(x)=\sum_{n\in\mathbb{Z}}c_n\mathrm{e}^{inx}.$$

Notes.


*

*By convention, the sum on $\mathbb{Z}$ is defined by:
$$\sum_{n\in\mathbb{Z}}u_n=\lim_{N\to+\infty}\sum_{n=-N}^{N}u_n.$$

*The sum of the Fourier series is $f^{\mathrm{reg}}$, and not $f$ as you claim.

