Laurent series in complex functions

[y.moghadamnia]

hey there,
I just studied the whole taylor and laurent series, and I think I mixed them up alittle.so here's what I know:
- if we have a contor in which our f(z) is analytic completely, we can expand it in taylor series.
- if we have singularities, we can expand the functions around the singularities in laurent series.
now, suppose we have a funcion that we want to expand around some point. what formula exactly we should use? the integrals? the $$\Sigma$$ s?
suppose for example the function f(z)=exp(z).
what is the taylor expansion of that?
- can anyone give me some hard example and solve it?

 

[HallsofIvy]

Well, for f(x)= exp(z), you just said it didn't you? exp(z) is analytic for all z so it can be exended in a Taylor's series for any a: around z= a,
$$e^z= \sum_{n=0}^\infty \frac{e^a}{n!}(z- a)^n$$

The function $f(x)= e^z/(z- 1)$ is analytic everywhere except at z= 1. To find its Laurent series, about z= 1, take the Taylor's series for $e^z$ around z= 1,
$$\sum_{n=0}^\infty \frac{e}{n!}(z- 1)^n$$
and divide each term by z- 1:
$$\sum_{n=0}^\infty \frac{e}{n!}(z- 1)^{n-1}$$
which can be written as
$$\sum_{m=-1}^\infty \frac{e}{((m+1)!}(z-a)^m$$
which is a Laurent series because it contains negative powers,
by letting m= n- 1.

 

[y.moghadamnia]

thanx for the nice example which I completely understood, but I still feel like my knowledge on this subject is a bit bald! can u suggest me anything good to read? I have had churchill complex analysis and it was o.k, but I think I need more examples.

 

[mathwonk]

i like the complex books by frederick greenleaf, by henri cartan, and by serge lang.