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I've been trying to develop this excercise but can not find a way to justify the fact that the residue is -1/2

Could anyone help me please?

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up vote 2 down vote accepted

I like this formula for the residue of a pole of order $n$ at $z=z_0$:

$$ \text{Res}[f,z_0]=\frac{1}{(n-1)!}\lim_{z\rightarrow z_0}\frac{d^{n-1}}{dz^{n-1}}((z-z_0)^nf(z)) $$

Here, your poles are both order 1 (simple poles), so we have:

$$ \text{Res}[f,0]=\lim_{z\rightarrow 0}zf(z)=\lim_{z\rightarrow 0}\frac{z+1}{z-2}=-\frac{1}{2} $$

Note of course that this method is only really practical for low-order poles, and doesn't apply at all to essential singularities such as $\exp(1/z)$ at $z=0$.

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There is another theorem that saidxthat the coefficient is the residue the residue – Miguel Mora Luna Dec 7 '12 at 5:36
Yes, by definition the residue is the coefficient of the $1/z$ term in the Laurent series, but this method tends to be much easier to use. – icurays1 Dec 7 '12 at 5:37
Thanks for your help – Miguel Mora Luna Dec 7 '12 at 5:41
And the residue calculating z = 2, the residue must be the coefficient of 1 / (z-2), for the same reason mentioned above – Miguel Mora Luna Dec 7 '12 at 5:54

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