Given an function, how can I find its Laurent series representation?

Answer 1

Usually it is done by using known power series expansions and modifying them.

For example, we know that

e^z=1+z+(1/2)z^2+(1/6)z^3+...

So the expansion for e^(1/z) will be
...(1/6)z^(-3)+(1/2)z^(-2)+
z^(-1)+1.

If you then wanted to consider z^2 e^(1/z), you would simply
multiply this by z^2.

Another way that is common is to start with
1/(1-z)=1+z+z^2+z^3+..
and make appropriate modifications.
For example, the expansion for 1/(2+z) around z=2 will go as follows:
1/(2+z)=1/[4+(z-2)]
=(1/4)* 1/[1+( (z-2)/4)]
=(1/4)*[1-[(z-2)/4] +[(z-2)/4]^2 -[(z-2)/4]^3 +...]
This will converge as long as |z-2|<4. To find what happens for
|z-2|>4, do this:
1/(2+z)=1/(4+(z-2))
=1/(z-2) * [1/(1+ 4/(z-2) )]
=1/(z-2)*[ 1- 4/(z-2) +[4/(z-2)]^2 -[4/(z-2)]^3 +...]
=1/(z-2) -4/(z-2)^2 +16/(z-2)^3 .....

Answer 2

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Answer 3

Very carefully.

or go here for some help:
http://en.wikipedia.org/wiki/Laurent_series