Show that the function $f(z) = 1/(z-4)$ has an antiderivative in a domain that is containing C, where C is the contour that denotes the unit circle, $|z| = 1$.
The function $f(z) = 1/(z-4)$ is continuous on the domain $D = \{ z\in \mathbb{C}: z \neq4\}$, and this domain contains the unit circle. However, how would I go about finding the antiderivative? Is it some form of a complex logarithm? My thought process would that the antiderivative would be $F(z) = log(z-4) = lnr + i\theta$ where $|z-4|>0$ and $0<\theta<2\pi$. Is this on the right track?
I would appreciate it if anyone could give any help or hints on this problem.
$f(z)=-\frac 1 4 \frac 1 {1-\frac z 4}=-\frac 1 4\sum\limits_{k=0}^{\infty} (\frac z 4)^{n}$ for $|z| <4$. So an anti-derivative valid in $|z|<4$ is $\sum\limits_{k=0}^{\infty} -\frac 1 {(n+1)}(\frac z 4)^{n+1}$.
[You are not asked to find an anti-derivative explicitly. It is enough to prove existence and if you know what simply connceted regions are there is a one line answer to this question].