MHB Residue Theorem: Find Residue at 0 for e^z/z^3

[Dustinsfl]

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]

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.

 

[Dustinsfl]

The coefficient of \( \frac{1}{z} \) is the residue. There are other ways though.

What do you mean by this?

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

 

[Fantini]

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.

 

[HallsofIvy]

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 $\frac{1}{(n-1)!}\frac{d^{n-1}(z-z_0)^nf(z)}{dz^{n-1}}$.