how do i start with this laurent series $f(z) = {\sin z \over z-3i}+ e^{3iz}$ around the point $z_0 = 3i$ I have no idea how to start it. Can someone help me please? 
$f(z) = {\sin z \over z-3i}+ e^{3iz}$
 A: Hint.
Around $z = 3i$ you can have two different sections. 
First
Let's consider just the interesting term of the function, namely 
$$\frac{1}{z-3i}$$
And let's do what follows:
$$\frac{1}{z-3i} = \frac{1}{z\left(1 - \frac{3i}{z}\right)} = \frac{1}{z}\sum_{k = 0}^{+\infty} \left(\frac{3i}{z}\right)^k = \frac{1}{z}\sum_{k = 0}^{+\infty} i^k\left(\frac{3}{z}\right)^k$$
And this was the Geometric Series, and it holds in the region $\boxed{z > 3}$.
Second
Again let's consider just the same fraction and write
$$\frac{1}{z-3i} = \frac{1}{3i\left(\frac{z}{3i} - 1\right)} = \frac{-i}{3\left(\frac{z}{3i} - 1\right)} = \frac{i}{3}\frac{1}{\frac{z}{3i} - 1} = \frac{i}{3}\sum_{k = 0}^{+\infty}\left(\frac{z}{3i}\right)^k = \frac{i}{3}\sum_{k = 0}^{+\infty} (-i)^k \left(\frac{z}{3}\right)^k$$
And this was again the Geometric Series and this time it holds in the region: $\boxed{z < 3}$.

Consider also that your function has a residue. Indeed at the point $z = 3i$ we have:
$$\text{Res}[f(z), z=3i] = \lim_{z\to 3i} (z-3i)\cdot \frac{\sin(z)}{z-3i} + e^{3iz} = \sin(z=3i) = \sin(3i) = \boxed{i\sinh(3)}$$
The residue is the coefficient of $z^{-1}$ (or also $a_{-1}$) in the Laurent expansion.

You shall be able to proceed, right?
