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I had been given the question as shown in the following, with the answer also given.

EXAMPLE 6.13. Let $\gamma$ be the circle with centre 0 and radius 1 traced anticlockwise. By integrating along $\gamma$, show that there is no function on $\mathbb{C} \backslash\{0\}$ with derivative $z^{-1}$. SOLUTION. We can parametrise $\gamma$ as $z=e^{i t}$ for $0 \leq t \leq 2 \pi$. We have $$ d z=i e^{i t} d t=i z d t $$ hence $$ \int_{\gamma} z^{-1} d z=\int_{0}^{2 \pi} z^{-1} i z d t=\int_{0}^{2 \pi} i d t=2 \pi i \neq 0 $$ Since $\gamma$ is a closed path, it follows that $z^{-1}$ cannot have an anti-derivative on $\mathbb{C} \backslash\{0\}$.

Surely however the anti-derivative of $z^{-1}$ would be $\log(z)$? I have seen a similar question asked, however, I didn't see the answer to this specific question.

Mateusz Del Percio
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    But you cannot define a branch of $\log$ on all of $\mathbb{C}\setminus {0}$. You get local primitives of $z^{-1}$, but not a global primitive. – Daniel Fischer Apr 24 '16 at 14:13
  • The complex logarithm is very, very tricky. First of all it is multivalued, so you need to go into branch selection, and the branch cut usually lies along the negative real axis. – KR136 Apr 24 '16 at 14:13
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    It's impossible to define $\log(z)$ so as to be holomorphic in a deleted neighborhood of the origin. (Proof: See your post.) – David C. Ullrich Apr 24 '16 at 14:13
  • @DanielFischer , In layman terms, does this mean that The Integral can be evaluated over a domain, but does not have a specifically defined antiderivative function F(x)? – Mateusz Del Percio Apr 24 '16 at 14:18

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$\DeclareMathOperator{\Re}{Re}\DeclareMathOperator{\Im}{Im}$Because the integral of the reciprocal about the unit circle is non-zero, the value of the integral $$ w = \int_{1}^{z} \frac{d\zeta}{\zeta} \tag{1} $$ depends not only on $z$, but on the path from $1$ to $z$. You can denote the value of the integral $w = \log z$, but doing so does not make $\log$ single-valued. The green paths below depict two "lifts" of the unit circle, starting at $(1, 0)$ and tracing counterclockwise, ending at $(1, 2\pi i)$; tracing clockwise, ending at $(1, -2\pi i)$.

For each non-zero $z$, there are infinitely many distinct values of $\log z$. Fixing a path $\gamma$ from $1$ to $z$ fixes a value of $$ \log z = \int_{\gamma} \frac{d\zeta}{\zeta}. $$ Every other choice differs by an added integer multiple of $2\pi i$, corresponding to a path that winds about the origin some number of times before following $\gamma$ to $z$.

There's no perfectly adequate picture because the graph of the relation $w = \log z$ naturally sits in the space of ordered pairs $(z, w)$ of complex numbers (a real four-dimensional space), but the projection $$ (z, w) \mapsto (\Re z, \Im z, \Re w + \Im w) $$ (which perhaps is not as widely-known as it should be), captures the qualitative "parking garage" behavior of the Riemann surface of $\log$.

The Riemann surface of log

Andrew D. Hwang
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