Chapter 7: Problem 18

Show that for $g$ and $h$ analytic functions at $z_{0}$, with $g\left(z_{0}\right) \neq 0, h\left(z_{0}\right)=0$, and $h^{\prime}\left(z_{0}\right) \neq 0$ $$ \operatorname{Res}\left[\frac{g(z)}{h(z)} ; z_{0}\right]=\frac{g\left(z_{0}\right)}{h^{\prime}\left(z_{0}\right)} $$

Short Answer

Expert verified

Under the given conditions where $h(z_0) = 0$, $h'(z_0) ≠ 0$ and $g(z_0) ≠ 0$, it's been demonstrated that the residue of a function $f(z) = g(z)/h(z)$ at a point $z_0$ is given by $Res[f(z); z_0] = g(z_0)/h'(z_0)$.

Step by step solution

Identify the functions and their properties

Firstly, identify that $g$ and $h$ are analytic functions at $z_0$ which means that they are complex differentiable at point $z_0$. Additionally, $g(z_0)$ is not equal to zero and $h(z_0)$ is equal to zero. The derivative of $h$ at $z_0$ is not equal to zero.

Use definition of Residue

The definition of the residue at $z_0$ for a function which can be expressed as a ratio of two functions $g(z)$ and $h(z)$ is given by $Res[f(z); z_0] = lim_{z -> z_0} (z - z_0)f(z)$. Apply this definition by substituting $f(z) = g(z)/h(z)$. We have $Res[f(z); z_0] = lim_{z -> z_0} (z - z_0)\frac{g(z)}{h(z)}$. Since $h(z_0) = 0$, this limit forms an indeterminate form.

L'Hopital's Rule

To resolve the indeterminate form, we can make use of L'Hopital's rule. Applying the rule, we get $Res[f(z); z_0] = lim_{z -> z_0} \frac{g(z)}{h'(z)}$. Evaluating this limit at $z_0$, we obtain $Res[f(z); z_0] = \frac{g(z_0)}{h'(z_0)}$ which proves the theorem.