$f(z)=\frac{1}{z}$ has an antiderivative on any simply connected domain

Question

Prove that the function $f(z)=\frac{1}{z}$ has an antiderivative on any simply connected domain of $\Bbb C$ which does not contain zero.

Also prove that this function does not have an antiderivative on its entire domain.

I feel as if I have to use a lot of topology to prove these facts and I am not as proficient in that area as I would like to be.

Answer 1

With your hypotheses, use residue thm to show that if you integrate $1/z$ on any simple closed curve (in your simply connected domain which avoids origin), it integrates to zero. Pick a point $c$ in your domain. Set $f(z) = f(c) + \text{ integral of } (1/z)dz$ along any path from $c$ to $z$. Previous sentence implies this is independent of path, hence well-defined. Fundamental theorem of calc for rest

Answer 2

Let $U$ be a simply-connected domain of $\Bbb C$ which does not contain zero. Fix $z_0 \in U$. Since $U$ is open and connected, it is path connected. Hence, given $z\neq z_0$, there is a path $\gamma$ from $z_0$ to $z$. Define

$$F(z) := \int_{\gamma} \frac{dw}{w},$$

Since $f(w) = 1/w$ is analytic in the simply-connected set $U$, the integral defining $F$ is path-independent, and hence $F$ is well defined. Since $f$ is continuous on the open set $U$, for a given $\epsilon > 0$ and $z\in U$, there corresponds $\delta > 0$ such that the open disk $D(z;\delta) \subseteq U$ and for all $z'$, $|z' - z| < \delta$ implies $|f(z') - f(z)| < \epsilon$. Now if $|z' - z| < \delta$, then the straight line segment $[z,z']$ lies inside $D(z;\delta)$. So

$$F(z') - F(z) = \int_{\gamma + [z,z']} \frac{dw}{w} - \int_\gamma \frac{dw}{w} = \int_{[z,z']} \frac{dw}{w},$$

and consequently

\begin{align} |F(z') - F(z) - f(z)(z' - z)| &= \left|\int_{[z,z']} \frac{dw}{w} - \int_{[z,z']} f(z) dw\right|\\ & = \left|\int_{[z,z']} [f(w) - f(z)]\, dw\right|\\ &< \epsilon |z - z'|. \end{align}

Since the inequality holds for all $z'$ with $|z' - z| < \delta$ and $\epsilon$ was arbitrary, $F$ is differentiable with $F'(z) = f(z) = 1/z$.

To prove the second part, suppose $f$ is analytic in the entire domain and consider integrating $f(z)$ over the unit disc. Show that the integral is $2\pi i$. On the other hand, since $f$ in analytic in the unit disc, by Cauchy's theorem, the integral is zero. This is a contradiction.