Taylor-Laurent series expansions

Question

I'm having some issues finding how to series expand some complex functions that my professor gave past years in exams. For example, in this exercise, it is asked to find the first two terms of the Taylor series of the function $$f(z) = \frac{\sqrt{z}}{(z+i)^{2}}$$ considering that we take $\sqrt{z} = \sqrt{|z|} + e^{(i\operatorname{arg}z )/ 2}$. I tried by expanding $\sqrt{z+i}$ at $z=0$ and from this I got: $$ f(z) \simeq \frac{i + \frac{1}{2}z + \left(\frac{i}{2} - \frac{i}{2}\right)}{(z+i)^2}= \frac{\frac{i}{2} + \frac{1}{2}(z+i)}{(z+i)^2} = \frac{i}{2(z+i)^2} + \frac{1}{2(z+i)^2} $$ but from what wolfram alpha tells me, this is wrong.

Another example would be expanding the following: $$ g(z) = \frac{e-e^{iz}}{z+ia} $$ at $z=-ia$. To do this, he uses (I got solutions for this problem exactly) the following procedure:

$\begin{align*} g(z) &= \frac{e-e^{i(z+ia-ia)}}{z+ia} \\ &= \frac{e - e^a e^{i(z+ia)}}{z+ia} \\ &= \frac{e-e^a \sum_{n=0}^{+\infty}\frac{i^n}{n!}(z+ia)^n}{z+ia} \\ &= \frac{e-e^a}{z+ia} - e^a \sum_{n=1}^{+\infty} \frac{i^n}{n!}(z+ia) ^{n-1} \\ &= \frac{e-e^a}{z+ia} - e^a \sum_{p=0}^{+\infty} \frac{i^{p+1}}{(p+1)!} (z+ia)^p \end{align*}$

My question with this is: is there a way to alway compute this series expansions without having to think at tricks to achieve that like adding and subtracting $\pm ia$ on the exponential? I mean, when I see it, the procedure is obvious, but during a test I don't know if I would be capable of thinking of something like this. Thank you in advance and sorry for the stupid question.

Answer 1

There are other methods. In the case where you are trying to calculate coefficients of a Taylor series at a given point, you can express the coefficients in terms of derivatives of the function. You can also calculate the coefficients of Taylor or Laurent series at a given point by evaluating certain line integrals.

However, these methods typically require quite a bit of work. If you are given a function which is given in terms of another function which has a nice power series expansion and you are asked to calculate the Laurent series expansion at a given point, it really does reduce your work if you are able to make use of power expansions of other functions and are able to add and subtract certain times at suitable places in order to do this.

Let's take the function $g$ as an example. First of all, you want to think about what kind of terms will be involved in the series. For a power series expansion at $-ia$, you will need to find a series involving terms of the form $(z - (-ia))^{n} = (z+ia)^{n}$ as $n$ ranges over $\mathbb{Z}$. Then you want to look at your function $g$ and see which parts of the function have a nice series expansion. As $g(z) = \frac{e - e^{iz}}{z + ia}$ for $z\in \mathbb{C}\setminus \{-ia\}$, most of the function is a rational polynomial in $z+ia$ apart from the $e^{iz}$ term. So if you can express this $e^{iz}$ term in the form $e^{i(z+ia)}$, then you can obtain the desired expansion. All that needs to be done to do this is note that $e^{iz} = e^{i(z+ia) + a} = e^{a}e^{i(z+ia)}$ for each $z\in\mathbb{C}$. Then you can use the series expansion of $e^{i(z+ia)}$ and rearrange the expression to obtain the desired result.

So the main idea is to work out what terms you are looking for in the Laurent expansion and which functions can be adjusted to use their power series expansion. Some good examples of functions where the power series expansion can be used in such a manner are the exponential map, the sine and cosine maps, the map $z\mapsto \sqrt{1+z}$ for $|z|<1$ and the map $z\mapsto (1-z)^{-1}$ for $|z|<1$.

When you are not sure what to do, try thinking about what information you have at the moment, what you want to end up with and what objects you have to work with at the moment. That makes it more clear to connect the pieces of information and work out what needs to be done from there.