Good day to you all,
next week I will write my exam in complex analysis and I have some problems with finding the laurent series. I know there are some tricks, sometimes you can use the taylor series to find the laurent series or you use the geometric series. But there are some functions where I am not able to find the correct laurent series, maybe I am thinking too complicated or there is a theorem or trick which I am missing. My problem are the following functions:
- $$f(z)=\frac{e^{z}}{(z+1)^{2}}$$ around z=-1
I know the series for $$e^{z}=\sum \limits_{n=0}^{\infty}\frac{z^{n}}{n!}$$ but what will I do afterwards?
$$f(z)=\frac{z}{(z-1)*(z-2)}$$ I think I need to use the geometric series and with a small calculus I have $$f(z)=\frac{-1}{z-1}+\frac{2}{z-2}$$ but then there is the next problem, the geometric series converge just for $$|z|<1$$ but the function is defined for all z in C, except 0,1, and 2. Or do I have to do a case differentation? For this excercise our tutor just would like to have the residuum (no problem) and the the two terms of the laurent series with the lowest order.
the last one is $$f(z)=\frac{z*sin(z)}{z^{3}-z}$$ Like in the first excercise I know the series for sin(z), but I dont know what I can do afterwards. This function is also definied for all z in C except 0,1 and -1. Although I am bit confused because when you try to transfer the function to $$f(z)=\frac{sin(z)}{z^{2}-1}$$, then 0 is not a problem anymore. Like in the other exercise I just need the residuums (no problem, except for z = 0) and the two terms of the laurent series with the lowest order.
I thank you all in advance and I hope you can understand everything (english is not my mother tongue).
Kind regards,
David
A few ideas:
The Taylor series for $e^z$ at $z=-1$ is easy enough: $$e^{-1}\sum_{n\ge0}\dfrac {(z+1)^n}{n!}$$.
Meanwhile $$\dfrac 1{(z-1)^2}=\dfrac 1{(z+1-2)^2}=\dfrac 12\dfrac 1{(1-\dfrac {z+1}2) ^2}=\dfrac 12(\sum_{n\ge0} (\dfrac {z+1}2)^n)^2=\dfrac 12(1+\dfrac {(z+1)}2+\dfrac 34(z+1)^2+\dots) $$.
So now you have to multiply.
On $$f(z)=\dfrac {-1}{z-1}+\dfrac 2{z-2}$$
You can do as follows for $\mid z\mid\lt1$: $$\dfrac {-1}{z-1}=\dfrac 1{1-z}=\sum_{n\ge0}z^n$$. And
$$\dfrac 2{z-2}=\dfrac1{1-\dfrac z2}=\sum_{n\ge0}(\dfrac z2)^n $$.
So we get $$f(z)=\sum_{n\ge0} z^n+(\dfrac z2)^n=\sum_{n\ge0}(1+2^{-n})z^n$$, and the residue at $0$ is zero.
To get the residues at $1,2$ get the Laurent series centered there.
For $f(z)=\dfrac {\sin z}{z^2-1}$, take your series for $\sin z$ and multiply by $$-\dfrac 1{1-z^2}=-\sum_{n\ge0}z^{2n}$$. So the residue at $0$ will be $0$ (since you are multiplying two Taylor series).