Let $f(z)=\frac{1}{z(z+1)^2}$
Find the Laurent series expansion for f centered at 0 and which converges on ${0 < |z| < 1}$
In my teachers solutions, he has written out the answer as follows:
I don't really understand why he incorporates the derivative? I have a feeling it's because I don't truly understand Laurant series, but I am hoping someone here can help. I've spent a few hours looking into it and can't figure it out.
Thank you.
There are two "tricks" your teachers are using here. The first is leaving out the factor $\frac1z$ because it gives the pole at $z=0$, where we want to center the series. I would have written instead that $$zf(z)=\frac1{(z+1)^2},$$ to get $$zf(z)=1-2z+3z^2-4z^3+5z^4+\cdots,$$ (this step will be clear later) and get to $$f(z)=\frac1z-2+3z-4z^2+5z^3+\cdots,$$ which is the desired Laurent series.
Now, the trick that confused you: the derivative was used as a shortcut to prove that $$\frac1{(z+1)^2}=1-2z+3z^2-4z^3+5z^4+\cdots,$$ but of course, this could have been done by looking for the derivatives of every order at $z=0$ (the standard procedure).
Here's how they simplified the procedure. The idea is that since this series is not as obvious or well known as $$\frac1{z+1}=\frac1{1-(-z)}=1-z+z^2-z^3+\cdots,$$ we can use the fact that the derivative of the expression on the left side is $-\frac1{(z+1)^2} $ to argue that $$\frac1{(z+1)^2}=\frac d{dz}\left(-\frac1{z+1}\right)=\frac d{dz}\big(-(1-z+z^2-z^3+\cdots)\big)=$$$$=1-2z+3z^2-4z^3+5z^4+\cdots,$$ as we wanted.