Getting Residues Complex Analysis

[cathode-ray]

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.

 

[hgfalling]

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}

 

[cathode-ray]

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