How Is the Residue Calculated for 1/(z^2+4)^2 at z=2i?

[bugatti79]

Folks,

I am trying to understand calculating residues.

http://www.wolframalpha.com/input/?i=residue+of+1/(z^2+4)^2+at+z=2i

How is that answer determined? I mean (2i)^2=-4 and hence denominator is 0...?
Thanks

 

[DonAntonio]

Folks,

I am trying to understand calculating residues.

http://www.wolframalpha.com/input/?i=residue+of+1/(z^2+4)^2+at+z=2i

How is that answer determined? I mean (2i)^2=-4 and hence denominator is 0...?
Thanks

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

 

[bugatti79]

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

Very good, thank you. Just have 2 queries

1) shouldn't $/dz^2$ 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

 

[DonAntonio]

Very good, thank you. Just have 2 queries

1) shouldn't $/dz^2$ be just $/dz$?



*** Of course, thanx. The "n" in \,\,\frac{1}{(n-1)!}\,\,and\,\,\frac{d^n}{dz^n}\,\, must be the same


2) Why did you choose pole $z-2i$ instead of $z+2i$ for the limit..?



*** Because this is what they do in that link to Wolfram you gave: the pole at \,\,z=2i\,\,, and thus the limit

must be taken as z approaches this point. ***


Is it to do with $0 < |z-z_0|<R$


*** No, and I don't have much of an idea what you mean by this within this context.

DonAntonio


Thanks

...

 

[bugatti79]

...

Ok that makes sense, thank you.