# [SOLVED]Residue Theorem

#### dwsmith

##### Well-known member
Feb 1, 2012
1,673
I have read the chapter on Residue Theorem in Complex Analysis by Serge Lang but don't quite understand how to do the problems.

Can someone walk me through the problem (see below) so I can see a better example?

Find the residue at 0 for
$$\frac{e^z}{z^3}$$

I see we have pole of order 3 at zero.
Do we start by writing the Laurent series?

#### Fantini

MHB Math Helper
Feb 29, 2012
342
Laurent series is one way, yes. In this case, do $$\frac{1}{z^3} \cdot e^z$$ and expand $$e^z$$ so we have $$f(z) = \frac{1}{z^3} \cdot \sum_{n=0}^{\infty} \frac{z^n}{n!} \implies f(z) = \sum_{n=0}^{\infty} \frac{z^{n-3}}{n!}.$$ The coefficient of $$\frac{1}{z}$$ is the residue. There are other ways though.

#### dwsmith

##### Well-known member
Feb 1, 2012
1,673
The coefficient of $$\frac{1}{z}$$ is the residue. There are other ways though.
What do you mean by this?

so $\dfrac{1}{2}$?

Last edited:

#### Fantini

MHB Math Helper
Feb 29, 2012
342
Yes, the residue in this case would be $$\frac{1}{2}$$. When it's easy to do so, expanding in Laurent series and finding the coefficient is a good method.

By using partial fractions you can also use Cauchy's formula for each, as another way.

Last edited:

#### HallsofIvy

##### Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Another way to get the same thing: if f(z) has a pole of order n at $z_0$ then it can be written as a Laurent series: $f(z)= a_{-n}(z-z_0)^{-n}+ \cdot\cdot\cdot+ a_{-1}(z- z_0)^{-1}k+ a_0+ a_1(z- z_0)+ \cdot\cdot\cdot$. Of course, that means that $(z- z_0)^nf(z)= a_{-n}a+ \cdot\cdot\cdot+ a_{-1}z^{n-1}+ a_0(z- z_0)^n+ a_1(z- z_0)^{n-1}+ \cdot\cdot\cdot$, analytic at $z= z_0$.

That is, $a_{-1}$ is the coefficient of $(z- z_0)^{n-1}$ in the Taylor's series expansion of $(z-z_0)^nf(z)$. Of course, that is $\frac{1}{(n-1)!}\frac{d^{n-1}(z-z_0)^nf(z)}{dz^{n-1}}$.
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