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 ;)

DonAntonio · Answer

$$frac1{z-1}=frac1{z-3+2}=frac12frac1{1+frac{z-3}2}=frac12left(1-frac{z-3}2+frac{(z-3)^2}4-ldotsright)=$$

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

So

$$frac z{(z-1)(z-3)}=frac12left(frac3{z-3}-frac1{z-1}right)=frac32frac1{z-3}-frac14+frac{z-3}8+ldots$$

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

Stahl · Answer

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}cdotfrac{1}{(z - 3)/2 + 1} - frac{3}{4}cdotfrac{1}{(z - 3)/2 + 1}
end{align*}
I think you should be able to expand these with the help of well-known power series...

Marso · Answer

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

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

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