Here is the question: By Laurent expansions, there exist constants $a_n$ such that for $1<|z|<4$, $\frac{1}{z^2-5z+4}=\sum_{n=-\infty}^{n=\infty}a_nz^n$. Find $a_{-10}$ and $a_{10}$.
My idea:
Using partial fraction decomposition, we find that $\frac{1}{z^2-5z+4}=\frac{\frac{1}{3}}{z-4}-\frac{\frac{1}{3}}{z-1}$. Using Laurent expansions, we know that $\frac{\frac{1}{3}}{z-4}=\frac{\frac{1}{3}}{-1-\frac{1}{4}z}=-\frac{1}{3}\sum_{n=0}^\infty \frac{1}{4^{n+1}z^n}$, and that $\frac{\frac{1}{3}}{z-1}=-\frac{1}{3}\sum_{n=0}^{\infty}\frac{1}{z^n}$. Now, I am sort of stuck on where to go from here. I feel like I've done the majority of the problem, but is it really just a matter of playing with indices now? Moreover, I should ask if I made any mistakes, and if there is anything that ought to be considered for the rest of the problem. Furthermore, just out of curiosity, I was wondering if there is any way to maybe do this simply through integration? Any thoughts, suggestions, tips, etc. are greatly appreciated! Thank you!
You are in the right direction. However, since you are working in $1<|z|<4$, so for $$\underbrace{\frac{1}{z-1}=\frac{1}{z\left(1-\frac{1}{z}\right)}}_{\because \, \left|\frac{1}{z}\right|<1}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{1}{z^n}=\sum_{n=0}^{\infty}\frac{1}{z^{n+1}}.$$ And $$\underbrace{\frac{1}{z-4}=\frac{1}{4\left(\frac{z}{4}-1\right)}}_{\because \,\left|\frac{z}{4}\right|<1}=\frac{-1}{4}\sum_{n=0}^{\infty}\left(\frac{z}{4}\right)^n=-\sum_{n=0}^{\infty}\frac{z^n}{4^{n+1}}$$ Now you can have \begin{align*} \frac{1}{z^2-5z+4}&=\frac{1}{3}\left[\frac{1}{z-4}-\frac{1}{z-1}\right]\\ &=\frac{-1}{3}\left[\sum_{n=0}^{\infty}\frac{z^n}{4^{n+1}}+\sum_{n=0}^{\infty}\frac{1}{z^{n+1}}\right] \end{align*} Observe that the first series will give all non-negative powers, whereas the second series will give the negative powers.
For example, $a_{-10}=\frac{-1}{3}$.