# Getting Residues Complex Analysis

Hi everyone!

I still didn't fully understand how to get the residues of a complex function. For example the function $$f(z)=\frac{1}{(z^{2}-1)^{2}}$$ in the region $$0<|z-1|<2$$ has a pole of order 2. So the residue of f(z) in 1 should be given by the limit:

$$\lim_{z \to 1}(z-1)^{2}f(z)=1/4$$​

But when I get the Laurent serie:

$$\sum_{n=1}^{+\infty} (-1)^{n+1}n\frac{(z-1)^{n-3}}{2^{n+1}}$$​

i don't know how to get the residue directly from it.

Hi.

The formula for the residue of a function at $z_0$, a pole of order $n$ is:

$$res(f;z_0) = \lim_{z \rightarrow z_0} \frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}} (z-z_0)^n f(z)$$

Here the factorial term is just 1! = 1. So what we need is the derivative of $(z-z_0)^2 f(z)$.

So we get:
$$\lim_{z\rightarrow z_0} \frac{d}{dz} \frac{1}{(z+1)^2} = \lim_{z\rightarrow z_0} \frac{-(2z+2)}{(z+1)^4} = -\frac{1}{4}$$

Now since you have the Laurent series (I didn't check this), you can pull the residue right off the series. The residue of a function at a point $z_0$ is the coefficient of the term $(z-z_0)^{-1}$ in its Laurent series expansion around $z_0$.

The way you have written the expansion means that that term is when n=2.

So the residue is the coefficient of the term when n=2, which is
$$\frac {(-1)^{2+1} (2)}{2^{2+1}} = -\frac{1}{4}$$

Thanks! Your explanation really helped me. Now it makes sense.