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
Best Answer
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).