Finding the Laurent series While studying i got this exercise and would like some pointers where i'm going wrong.
Find the Laurent Series at $z_0=2i$
$$f(z) = \frac{1+z}{z^2+4}+e^z$$
I've tried the following:
$$f(z)=(1+z)\frac{1}{(z+2i)(z-2i)}+e^z$$
and the series for
$$\frac{1}{(z+2i)(z-2i)} = \frac{1}{z-2i}\frac{1}{4i}\frac{1}{1+(\frac{z-2i}{4i})}
 = \displaystyle\sum\limits_{k=0}^{+\infty} \frac{(-1)^k}{(4i)^{k+1}}(z-2i)^{k-1}$$
after this i'm stumped in trying to put everthing has a power of $z-2i$
$$f(z) = (1+z)\displaystyle\sum\limits_{k=0}^{+\infty} \frac{(-1)^k}{(4i)^{k+1}}(z-2i)^{k-1}+\displaystyle\sum\limits_{k=0}^{+\infty}\frac{z^k}{k!}$$
I have several questions regarding this:
1 - should i try to put $\displaystyle \frac{1+z}{z^2+4}$ has a sum of simple fractions?
2 - is there even a Laurent series for $e^z$ around $z_0=b$ since $e^z$ doesn't have any singularities?
3 - What should i do to that $(1+z)$?

There are a couple of different ways that you can think about this. First of all, I suspect you already immediately know how to find the expansion of $e^z$ around whatever point you want. [as an aside, you ask if there exists a Laurent series for $e^z$ around a point even though it doesn't have any singularities - it does. I suspect you think that Laurent expansions must include terms of negative degree - but that's not the case]. So I'll not worry about that term, and I will instead consider only the $\dfrac{1+z}{z^2 + 4}$ term.
But I see that you've already gone through the derivation of the series for $\dfrac{1}{(z+2i)(z-2i)}$ around $z - 2i$. That's the hard part. When you look at the $1+z$ term, you need to extract a $z-2i$ from it. So - add and subtract $2i$ (as Jyrki said). Finally, since you've noticed that $e^z$ doesn't have any singularities, there is no principal part. So there's only that Taylor-series style expansion...
