Finding Laurent expansion of $\frac{2}{(2z-3)(z-5)}$.

Question

Find the Laurent expansion of $\frac{2}{(2z-3)(z-5)}$ that is valid in the annulus $\frac{7}{2} < |z-5| < \infty$. There is a lot to write out, but here is the basics of the work.

We can leave $\frac{2}{(z-5)}$ as it is. So, we need to expand $\frac{1}{2z-3}$. For the given annulus, I got this to equal $(\frac{1}{(z-5)})(\frac{1}{1-(\frac{-7}{2(z-5)})})$.

Using this gives us a final answer of:

$\frac{2}{(z-5)^2}-\frac{7}{(z-5)^3}+\frac{49}{(z-5)^4} - ...$

There was a lot of computation involved, which would make typing this in full way to long. I included the important steps. Is my solution correct?

Answer 1

Here is one way, let $f(z) = \frac{2}{(2z-3)(z-5)}$, and $g(z) = f(z+5)$. Then we want to expand $f$ on the 'annulus' ${7 \over 2} < |z|$.

We have $g(z) = {1 \over z} { 1 \over z + {7 \over 2}} = {1 \over z^2} { 1 \over 1 + {7 \over 2z}}$.

Since ${ 1 \over 1 + {7 \over 2z}} = \sum_{k=0}^\infty (-1)^k ({7 \over 2})^k {1 \over z^k }$, for ${7 \over 2} < |z|$, you can expand to as many terms as you want.

Then you need to use $f(z) = g(z-5)$.

Answer 2

$\begin{align*}\frac{2}{(2z-3)(z-5)}&=\frac{A}{2z-3}+\frac{B}{z-5}\\ 2&=A(z-5)+B(2z-3) \\ 2&=(A+2B)z-(5A+3B)\\B&=\frac{2}{7} \\ A&=-\frac{4}{7}\end{align*}$

Thus, we have (provided my arithmetic is OK)

$\begin{align*} \frac{-4}{14z-21}+\frac{2}{7(z-5)} &= \frac{-4}{14(z-5)+49}+\frac{2}{7(z-5)} \\ &= \frac{-4}{14(z-5)}\frac{1}{1-\big(\frac{-49}{14(z-5)}\big)}+\frac{2}{7(z-5)}\\&=\frac{-2}{7(z-5)}\bigg(-1+\sum_{n=0}^\infty\Big(\frac{-7}{2(z-5)}\Big)^n\bigg), |z-5|>\frac{7}{2} \\ &=\sum_{n=1}^\infty\Big(\frac{-7}{2}\Big)^{n-1}\big(z-5\big)^{-n-1}, |z-5|>\frac{7}{2}.\end{align*}$