- Thread starter
- #1

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?

- Thread starter dwsmith
- Start date

- Thread starter
- #1

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?

- Feb 29, 2012

- 340

- Thread starter
- #3

What do you mean by this?The coefficient of \( \frac{1}{z} \) is the residue. There are other ways though.

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

Last edited:

- Feb 29, 2012

- 340

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.

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

Last edited:

- Jan 29, 2012

- 1,151

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