Compute the residue of a function

[docnet]

Homework Statement

Find $Res(h,z_0)$

Relevant Equations

$h=\frac{f}{g}$

There is a typo. It should say $h=\frac{f}{g}$.

Attempt: $f$ and $g$ are holomorphic on $\Omega$. Homomorphic functions form a $\mathcal{C}^*$ algebra, so $h$ is holomorphic on $\Omega$ where $g\neq 0$.

If $z_0$ is a removal singularity of $h$, then $Res(h,z_0)=0$ by Morera's theorem.

Assume $f(z_0)\neq 0$, then $z_0$ is a second order pole of h since $g''(x_0)\neq 0$.
$$Res(h,z_0)=lim_{z\rightarrow z_0}\frac{d}{dz}\Big[\frac{f}{g}(z-z_0)^2\Big]$$
This method produces indeterminate forms, even after applying the L'Hopital's rule.

Let $\mathcal{C}=\{z:|z-z_0|=1/2\}$. By Residue theorem,
$$Res(h,z_0)=\frac{1}{2\pi i}\int_\mathcal{C}\frac{f}{g}dz$$
This integral cannot be computed since $f$ and $g$ are not given.

Can anyone think of a clever method that could solve this problem?

 

[Office_Shredder]

You can just start writing out series. The Taylor series for f is known. For $1/g$,, you know if looks like $1/((z-z_0)^2(a+b(z-z_0)+c(z-z_0)^2+...)$ with a nonzero.

You can basically just bash this with the Taylor series of $1/(1-x)$. Let $x=\frac{1}{a}(-b(z-z_0)-c(z-z_0)^2+...)$ and write
$$\frac{1}{g}= \frac{1}{(z-z_0)^2 a (1-x)}$$

Now apply the normal Taylor expansion to $1/(1-x)$, and after the first couple terms you should get stuff that is high order in $(z-z_0)$ and hence can be ignored.

 

[docnet]

You can just start writing out series. The Taylor series for f is known. For $1/g$,, you know if looks like $1/((z-z_0)^2(a+b(z-z_0)+c(z-z_0)^2+...)$ with a nonzero.

You can basically just bash this with the Taylor series of $1/(1-x)$. Let $x=\frac{1}{a}(-b(z-z_0)-c(z-z_0)^2+...)$ and write
$$\frac{1}{g}= \frac{1}{(z-z_0)^2 a (1-x)}$$

Now apply the normal Taylor expansion to $1/(1-x)$, and after the first couple terms you should get stuff that is high order in $(z-z_0)$ and hence can be ignored.

Thank you so much for your reply. I will try this method.