Laurent series expansion of $\frac1{(x-1)^2(x-3)}$

Question

What is the Laurent series expansion of $\;\dfrac1{\left(x-1\right)^2\left(x-3 \right)}\,$ over $\;0<\left\lvert x-1\right\rvert < 2$ and $0<\left\lvert x-3 \right\rvert < 3$?

Answer 1

As I have mentioned in the comment section,We use Partial Fractions here.

\begin{equation} \frac{1}{(z-1)^2(z-3)} = \frac{-1}{4(z-1)} - \frac{1}{2(z-1)^2} +\frac{1}{4(z-3)} ---(1) \end{equation}

(i) Expanding in the region $0<|z-1|<2$

This means we are trying to expand around the point $z = 1$. Hence we will not disturb the first two terms in equation (1).

For the third term, we do the following.

$\frac{1}{4(z-3)} = \frac{1}{4(1+(z-4))} = \frac{(1+(z-4))^{-1}}{4}$.

(Using the expansion, $(1+x)^{-1} = 1-x+x^2-\ldots$)

$ \implies \frac{(1+(z-4))^{-1}}{4} = 1 - (z-4) +(z-4)^2 - \ldots $

Hence final answer should be $$ \boxed{= (1 - (z-4) +(z-4)^2 - \ldots) +\frac{-1}{4(z-1)} - \frac{1}{2(z-1)^2}} $$

Similarly for, (i) Expanding in the region $0<|z-3|<3$

This means we are trying to expand around the point $z = 3$. Hence we will not disturb the third term in equation (1).

For the first two terms,

$\frac{-1}{4(z-1)} = \frac{-1}{4((z-3)+2)} =\frac{-1}{8(\frac{z-3}{2}+1)}= \frac{-(\left(\frac{z-3}{2}\right)+1)^{-1}}{8} =\frac{-(1-(\frac{z-3}{2}) + \frac{(z-3)^2}{2}-\ldots)}{8} $

And we know that if we differentiate, $\frac{d}{dz}\frac{-1}{4(z-1)} = \frac{1}{4(z-1)^2}$

Hence to find the series of $\frac{1}{4(z-1)^2}$ , we differentiate the series of $\frac{-1}{4(z-1)}$ and multiply $-1$ to it

Add all the results for final answer.