# Complex Analysis Residue Query

1. May 5, 2012

2. May 5, 2012

### DonAntonio

The point $\,\,z=2i\,\,$ is a pole of order 2 of $\,\,\frac{1}{(z^2+4)^2}=\frac{1}{(x-2i)^2(z+2i)^2}\,\,$ . Thus, from the well-known

formula that stems from Laurent series, the residue is given by
$$\lim_{z\to 2i}\frac{1}{(2-1)!}\frac{d}{dz^2}\left(\frac{(z-2i)^2}{(z^2+4)^2}\right)=\lim_{z\to 2i}-\frac{2}{(z+2i)^3}=-\frac{i}{32}$$

DonAntonio

3. May 5, 2012

### bugatti79

Very good, thank you. Just have 2 queries

1) shouldnt $/dz^2$ be just $/dz$?

2) Why did you choose pole $z-2i$ instead of $z+2i$ for the limit..?
Is it to do with $0 < |z-z_0|<R$

Thanks

4. May 5, 2012

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5. May 5, 2012

### bugatti79

Ok that makes sense, thank you.