So I'm working on an example problem at the moment but as much research as I have done in the notes I can't seem to get my head around how power series' work.
The current problem: The function $f: \mathbb{C\ } \backslash \ \{2\}\rightarrow \mathbb{C}$ is given by $$f(z):= \frac{1}{(2-z)^2}$$
I want to derive the power series expansion of $f$ around $a=3$
I have deduced that this means I essentially need to write $f$ in the form $$f(z) = \sum^\infty_{n=0} a_n (z-3)^n$$ where $ (a_n)_{n \in \mathbb{N}_0} $ is a complex sequence.
From recognition from another problem I believe that I can find the expansion for, say $g(z)= \frac{1}{2-z}$ around $a=3$ and then use the fact that the derivative $g'(z)=f(z)$.
This is what I have done so far:
If I say $$g(z)=\frac{1}{2-z}=\frac{1}{2-a-(z-a)}=\frac{1}{2-a-w}$$ where $w=z-a$
Then I can say that $$\frac{1}{(2-a)} \frac{1}{1-\frac{w}{2-a}} = \frac{1}{2-a} \sum_{n=0}^\infty (\frac{w}{2-a})^n = \sum_{n=0}^\infty a_n(z-3)^n$$ where $a=3$
At this point I don't know what the next steps are?
Those are my workings so far and I understand they are probably incorrect, unfortunately Covid-19 means that my lectures for this were cancelled and I am struggling to use the notes to help with my problem, any help would be greatly appreciated
Best Answer
Note that\begin{align}\frac1{2-z}&=\frac1{-1-(z-3)}\\&=-\frac1{1+(z-3)}\\&=-\sum_{n=0}^\infty(-1)^n(z-3)^n\text{ (if $\lvert z-3\rvert<1$)}\\&=\sum_{n=0}^\infty(-1)^{n+1}(z-3)^n\end{align}and that therefore\begin{align}\frac1{(2-z)^2}&=\left(\frac1{2-z}\right)'\\&=\left(\sum_{n=0}^\infty(-1)^{n+1}(z-3)^n\right)'\\&=\sum_{n=1}^\infty(-1)^{n+1}n(z-3)^{n-1}\\&=\sum_{n=0}^\infty(-1)^n(n+1)(z-3)^n.\end{align}