How to find residue of a function

Question

Find the residue of $f(z)=\frac{z^2}{z^2-1}$ at $z=\alpha$ my attempt: $ f(z)=\frac {z^2}{({z-\alpha+\alpha})^2-1}$ now take limit as $\lim\limits_{z\to{\alpha}}(z-\alpha)f(z) $ I do understand anything. Pls explain me right answer

Answer 1

• About your attempt

In my opinion, you should try not to let denominator is zero when you take limit.

• Answer

$f(z) = \frac{z^2}{z^2-1} = \frac{z^2}{(z-1)(z+1)}$

so, the function $f(z)$ has a pole of order 1 at $z = \pm1$
Let $Res(\alpha)$ be the residue of function $f(z)$ at $z = \alpha$,

$Res(1) = \lim_{z\to 1}(z-1)f(z) = \lim_{z\to 1}\frac{z^2}{z+1} = \frac{1}{2}$
$Res(-1) = \lim_{z\to -1}(z+1)f(z) = \lim_{z\to -1}\frac{z^2}{z-1} = -\frac{1}{2}$

P.S Perhaps this answer is not detailed , because I major in engineering and learned complex analysis for myself.

Answer 2

Like unknown said, the residue is $0$ unless $\alpha=1$ or $\alpha=-1$. In this case, then

$$f(z)=\frac{z^2}{(z+1)(z-1)}$$

$z=\pm1$ are both poles of order $1$

$$\operatorname{Res}(f,1)=\frac{1}{0!}\lim_{z\to1}\left(\frac{z^2}{z+1}\right)=\frac{1}{2}$$

$$\operatorname{Res}(f,-1)=\frac{1}{0!}\lim_{z\to-1}\left(\frac{z^2}{z-1}\right)=-\frac{1}{2}$$

If wanting to split into a Laurent series, which I think is what's desired, then

$$f(z)=\frac{z^2}{(z+1)(z-1)}=1+\frac{1}{2}\frac{1}{z-1}-\frac{1}{2}\frac{1}{z+1}$$

Here the residues appear again

(The rest is quite unnecessary as we already know the residues)

Centered around $z_0=\alpha$,

$$f(z)=1+\frac{1}{2}\frac{1}{(z-\alpha)-1+\alpha}-\frac{1}{2}\frac{1}{(z-\alpha)+1+\alpha}$$

$$f(z)=1+\frac{1}{2}\frac{1}{(\alpha-1)\left(1-\frac{z-\alpha}{1-\alpha}\right)}-\frac{1}{2}\frac{1}{(\alpha+1)\left(1-\frac{z-a}{-1-\alpha}\right)}$$

$$f(z)=1+\frac{1}{2(\alpha-1)}\sum_{n=0}^\infty\left(\frac{-1}{\alpha-1}\right)^n(z-\alpha)^n-\frac{1}{2(\alpha+1)}\sum_{n=0}^\infty\left(\frac{-1}{\alpha+1}\right)^n\left(z-\alpha\right)^n$$

Which is unbounded if $\alpha=\pm1$ but indeterminant as $z\to\alpha$; there is residue at $z=\pm1$.

The series is the original function, so in evaluating $f(z)$, the value of $\alpha$ has no effect, unless, as it appears at the surface, $\alpha=\pm1$ and $z\neq\alpha$ just by the way it has been constructed.

This is the same as saying that $\displaystyle f(z)=\frac{z^2}{z^2-1}$ is independent of $\alpha$ and unbounded if $z=\pm1$, so the residue is only due to $z=\pm1$