Find the Laurent expansion of $\displaystyle f(z)=\frac{1}{(z-4)(z-1)}$ in the annulus ${z: 1<|z|<4}$. Find $\displaystyle a_{-1}, a_{-10},\text{ and }a_{10}$
I am looking for feedback on whether or not I'm doing this correctly.
I started with a partial fraction decomposition
$\displaystyle f(z)=\frac{1}{(z-4)(z-1)} = \frac{1}{3}\left[\frac{1}{z-4}-\frac{1}{z-1}\right]$
For the first term, we see $\displaystyle \frac{|z|}{4}<1$ so we need to manipulate it into that form. For the second term, we see $\displaystyle \frac{1}{|z|}<1$ so we manipulate into that form.
Next, the partial fraction decomposition manipulated takes the form:
$\displaystyle f(z)=\frac{1}{3}\left[\frac{-1}{4}\left(\frac{1}{1-\frac{z}{4}}\right)-\frac{1}{z}\left(\frac{1}{1-\frac{1}{z}}\right)\right]$
re-writing this in series form with the one-third distributed we find:
$\displaystyle f(z)=-\frac{1}{12}\left[1+\frac{z}{4}+\left(\frac{z}{4}\right)^{2}+\left(\frac{z}{4}\right)^{3}+\dots\right]-\frac{1}{3z}\left[1+\frac{1}{z}+\left(\frac{1}{z}\right)^{2}+\left(\frac{1}{z}\right)^{3}+\dots\right]$
Then condensing into series form we find the Laurent expansion to be:
$\displaystyle f(z)=\sum_{k=0}^{\infty}\left(-\frac{1}{3}\right)\frac{z^{k}}{4^{k+1}}+\sum_{k=-1}^{-\infty}\left(-\frac{1}{3}\right)z^{k}$
Next, for the coefficients we see that $\displaystyle \forall~k\geq1\text{, }a_{-k}=-\frac{1}{3}\Rightarrow a_{-1}\text{, }a_{-10}=-\frac{1}{3}$ from the second series. Then, for $a_{10}$ from the first series we use $k=10$ to find $\displaystyle a_{10}=-\frac{1}{3\cdot4^{11}}$
Any feedback would be greatly appreciated.