Laurent Series and residue of $\frac{z}{(z-1)(z-3)}$ around z = 3

Question

As mentionned in the title, I'd like to get the function's Laurent series and after its residue, I have tried to separate the two denominators to get a partial fraction but I still have a z at numerator that's bothering me ...

So far : $$ \frac{z}{(z-1)(z-3)} = \frac{-z}{2(z-1)} + \frac{3z}{(z-3)} $$

I can obviously try $\omega = z-3$ and change my variable but I don't feel it'll will work ... Because I will still have this z on top of my fraction ...

Could you help me ? Thank you very much ;)

Answer 1

$z=3$ is a simple pole so $Res(f)_{z=3}=\lim_{z\to 3} f(z)={3\over 2}$

Answer 2

Hint: $$ \frac{3z}{z - 3} = \frac{3(z - 3) + 9}{z - 3} $$ \begin{align*} \frac{-z}{2(z - 1)} &= \frac{1}{2}\frac{-(z - 3) - 3}{(z - 3) + 2}\\ &= -\frac{1}{2}\frac{z - 3}{(z - 3) + 2} - \frac{1}{2}\frac{3}{(z - 3) + 2}\\ &= -\frac{z - 3}{4}\cdot\frac{1}{(z - 3)/2 + 1} - \frac{3}{4}\cdot\frac{1}{(z - 3)/2 + 1} \end{align*} I think you should be able to expand these with the help of well-known power series...

Answer 3

$$\frac1{z-1}=\frac1{z-3+2}=\frac12\frac1{1+\frac{z-3}2}=\frac12\left(1-\frac{z-3}2+\frac{(z-3)^2}4-\ldots\right)=$$

$$=\frac12-\frac{z-3}4+\ldots$$

So

$$\frac z{(z-1)(z-3)}=\frac12\left(\frac3{z-3}-\frac1{z-1}\right)=\frac32\frac1{z-3}-\frac14+\frac{z-3}8+\ldots$$

so the residue is $\,3/2\,$ .