Residue

Recall, from the Laurent series, that we have a term for $a_{-1}$:

$$a_{-1}=\frac{1}{2\pi i}{\oint }_{C}f(z)dz=\frac{1}{2\pi i}{\oint }_C\frac{f(z)}{(z-z_0{)}^{-1+1}}dz$$

This term is called the residue of $f(z)$ at $z_0$ and has several important implications for complex analysis.

$$a_{-1}=\text{Res}(f(z);z_0)$$

A problem we face is that we need good ways to find residues — we can’t just keep expanding the Laurent series all the time. There are formulas below.

Computations

An immediately important application of the residue is in Cauchy’s residue theorem. Formulas to compute residues depend on how “bad” $f(z)$ is at $z_0$.

At a simple pole, we have:

$$\underset{z\to z_0}{\lim }(z-z_0)f(z)=a_{-1}=\text{Res}(f(z);z_0)$$

Also for a simple pole, if $f(z)=\frac{g(z)}{h(z)}$, then:

$$\text{Res}(f(z);z_0)=\frac{g(z_0)}{h^{\prime }(z_0)}$$

If $z_0$ is a pole of order $m$, then:

$$\underset{z\to z_0}{\lim }\left[\frac{1}{(m-1)!}\right(\frac{d}{dz}{)}^{m-1}(z-z_0{)}^{m}f(z)]=\text{Res}(f(z);z_0)$$

The steps for this formula are simple:

• Evaluate the factorial.
• Simplify the function behind the derivative, then differentiate.
• Substitute the limit value into what’s leftover.

Derivation

For the simple pole:

$$f(z)=\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+\dots$$ $$(z-z_0)f(z)=a_{-1}+a_0(z-z_0)+\dots$$

Then we take the limit, as above.