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:


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 $(1+z)$?

  • 2
    $\begingroup$ Hint 1: $1+z=(1+2i)+(z-2i).$ Hint 2: $e^z=e^{2i}e^{z-2i}.$ $\endgroup$ Commented Jun 24, 2011 at 18:08
  • $\begingroup$ About $2$, yes, a very nice one, without those unpleasant negative powers. $\endgroup$ Commented Jun 24, 2011 at 18:22

2 Answers 2


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...


It may be easier to consider $f(z + 2i)$ and then consider the Laurent expansion about the origin, then shift it by $2i$. In other words, you have $$f(z + 2i) = {1 + z + 2i \over (z + 2i)^2 + 4} + e^{z + 2i}$$ $$= {z + 1 + 2i \over z^2 + 4iz} + e^{2i} e^z$$ The right hand term has Taylor series $e^{2i} \sum_{n=0}^{\infty}{z^n \over n!}$, so it will contribute $e^{2i} \sum_{n=0}^{\infty}{(z - 2i)^n \over n!} = \sum_{n=0}^{\infty}{e^{2i}(z - 2i)^n \over n!} $ to the Laurent series. The first term is $(1 + {1 + 2i \over z}){1 \over z + 4i}$, and you can expand the ${1 \over z + 4i}$ in the usual way by considering the $|z| < 4$ and $|z| > 4$ cases and using geometric series. One you're done, shift everything by $2i$ like above for the exponential series.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .