What happens to coefficients of complex fourier series when the function $f$ is odd? We know $f(x)=\sum_{n=-\infty}^{\infty} c_ne^{\frac{i\pi nx}{l}}$ and $c_n =\frac{1}{2l}\int_{-l}^{l}f(x)e^{-\frac{i\pi nx}{l}}dx$.
I can rewrite $ e^{\frac{i\pi nx}{l}} $ in terms of $\sin$ and $\cos$ and I assume $\cos$ terms would vanish but I am not sure if it helps. Can someone help on how to proceed?
Your intuition is correct!
Notice that $$\begin{split}c_n &=\frac{1}{2l}\int_{-l}^{l}f(x)e^{-\frac{2i\pi nx}{l}}dx\\ &=\frac{1}{2l}\int_{-l}^{l}f(-u)e^{\frac{2i\pi nu}{l}}du \text{ (change the variable $u=-x$)}\\ &=-\frac{1}{2l}\int_{-l}^{l}f(u)e^{\frac{2i\pi nu}{l}}du \text{ ($f$ is odd)}\\ &=-c_{-n} \end{split}$$ This, in particular, implies that $c_0=0$. But also, the Fourier series of $f$ can be rewritten as $$\sum_{n=-\infty}^{\infty} c_ne^{\frac{i\pi nx}{l}}=\sum_{n\geq 1}c_n\left(e^{\frac{2i\pi nt}{l}}-e^{-\frac{2i\pi nt}{l}}\right)=2i\sum_{n\geq 1}c_n\sin\left(\frac{2i\pi nt}{l}\right)$$