Find the taylor series at $z=3$ of $f(z)=\frac{z(z-3)}{(z-1)}$

Question

Find the Taylor series of $f(z)$ centered at $z=3$ and its convergence radius.

$$f(z)=\frac{z(z-3)}{(z-1)}$$

I wanted to find the Taylor series of f without using $f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n !}(x-a)^{n}$ as it is quite time consuming to do it this way.

Example: $f(z)=\frac{z-1}{z^2}$, its Taylor series can be found using the geometric series through some manipulation. So I wished to know if there is some kind of different “trick” to solve the original problem or is it analogous to the example.

Answer 1

Note that\begin{align}\frac z{z-1}&=1+\frac1{z-1}\\&=1+\frac1{2+z-3}\\&=1+\frac12\cdot\frac1{1+\frac{z-3}2}\\&=1+\frac12-\frac{z-3}{2^2}+\frac{(z-3)^2}{2^3}-\cdots\\&=\frac32-\frac{z-3}{2^2}+\frac{(z-3)^2}{2^3}-\cdots\end{align}and therefore the Taylor series of $\dfrac{z(z-3)}{z-1}$ about $3$ is$$\frac32(z-3)-\frac{(z-3)^2}{2^2}+\frac{(z-3)^3}{2^3}-\cdots$$

Answer 2

Hint

To make life simpler, let $z=t+3$ to make $$\frac{z(z-3)}{(z-1)}=t+1-\frac{2}{t+2}=t+1-\frac{1}{1+\frac t 2 }$$ and work around $t=0$