Cauchy Integral Theorem for punctured domains comprehension (seems too powerful)

145 Views Asked by At

We are supposed to calculate a few integrals where the integrand is a quotient of holomorphic functions. Wikipedia says (and there is a proof on this site) that the quotient $(f/g)$ is holomorphic wherever $g\ne 0$.

We also have the following lemma from class:

Let $G$ be a simply connected domain, $c \in G$, and let $f \in C(G)$ be holomorphic on $G \setminus \{c\}$. Then for all closed piecewise $C^1$-paths $\gamma$ in $G$ $$\int_\gamma f(z)dz = 0$$

This seems way too strong a formula (as far as I understand it). For instance the integrals

  1. $$\int_{\partial D} \frac{\cosh(z)}{4z^3-z}dz$$
  2. $$\int_{\partial D} \frac{\cos(\pi z)}{(z-2)^3}dz$$

would evaluate to zero, as long as the closed disks $D$ contain at most one of the roots of the denominator function:

  1. $\{-0.5,0,0.5\}$
  2. $\{2\}$

respectively. This root would then be our $c$, $D$ our connected domain $G$, and f would be holomorphic on $G\setminus\{c\}$.

I have a strong feeling this is wrong, as it seems too good to be true. But many properties of holomorphic functions seem too good to be true, that's why I'm asking.

1

There are 1 best solutions below

0
On BEST ANSWER

Not quite. In order to apply that theorem in order to compute $\displaystyle\int_{\partial D}\frac{\cosh(z)}{4z^3-z}\,\mathrm dz$, the disk $D$ must contain none of the roots of $4z^3-z$. Otherwise, $\dfrac{\cosh(z)}{4z^3-z}\notin C(D)$, since, for each $z_0\in\left\{-\frac12,0,\frac12\right\}$, the limit $\displaystyle\lim_{z\to z_0}\frac{\cosh(z)}{4z^3-z}$ doesn't exist.