Laurent series expansion of $f(z) = \cfrac{z^2-1}{z^3} - z + 1$ at the origin

Question

I started complex analysis a few weeks ago and we have arrived at the Laurent series! There was this exercise that I was having some trouble with!

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Determine the Laurent expansion of the function $$f(z) = \cfrac{z^2-1}{z^3} -z + 1$$ at $z_0 = 0$ and characterize the singularity.

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I "simplified" the expression and found that $z_0$ is a pole of order 3! What im confused on how to do is how do I find the residue point $\mbox{Res}(f,0)$?

Can you help me find it in this case and maybe if possible how do i do that in general! I am kinda confused at that point!

Answer 1

Remember that the residue is just the coefficient of the simple pole. $$f(z)=\frac{z^2-1}{z^3}-z+1=\frac{1}{z}-\frac{1}{z^3}-z+1$$ So the residue $Res_{z=0} f(z)=1$ the coefficient of $\frac{1}{z}$.

Answer 2

The Laurent series is $f(z)=\frac 1 z -\frac1 {z^{3}} -z+1$ and the residue is the coefficient of $\frac 1 z$ which is $1$.