# Complex analysis - branch cut of log

I am trying to show that there is no continuous branch of log on and open set containing the origin.

wlog I am taking the contour integral around the unit circle.

I know that an antiderivative of a continuous function $f$ is a holomorphic function $F$ such that $F' = f$, So using the Fundamental theorem of calculus;

If $f(z) = \frac{1}{z}$, then $$\int_{\gamma}f(z)dz = 2\pi i$$ where $\gamma$ is the contour around the unit circle. Using my previous definition, I know that $f(z)$ is either not continuous or doesn't have a holomorphic antiderivative on this domain containing the origin.

$f(z) = \frac{1}{z} = e^{-\log(z)}$ and so if this is not continuous, then there is no continuous branch of log on this domain. I know that the antiderivative is locally $log(z)$, but how can I say more than it is just not holomorphic in the later case in order to complete the proof?

Looking for an answer to fill in any gaps in my proof and to finish it off.

If $A$ is an open subset of $\mathbb C$ and $0\in A$, then there is no function $L\colon A\longrightarrow\mathbb C$ (continuous or not) such that, for each $z\in A$, $L(z)$ is a logorithm of $z$. That's so simply because $0$ has no logarithm.

Another question is: is there a continuous function $L\colon A\setminus\{0\}\longrightarrow\mathbb C$ such that $(\forall z\in A\setminus\{0\}):e^{L(z)}=z$. No, there is not.

Suppose there was. Fix $r>0$ such that $A\supset\{z\in\mathbb{C}\,|\,|z|=r\}$. We can suppose, without loss of generality, that $r=1$.

Each $z\in S^1(=\{z\in\mathbb{C}\,|\,|z|=1\})$ different from $-1$ can be written in one and only one way as $e^{i\theta}$, with $\theta\in(-\pi,\pi)$. Define, for such a $z$, $l(z)=i\theta$. Then $l$ is a continuous functions and $(\forall z\in S^1\setminus\{-1\}):e^{l(z)}=z$.

Now, consider the function $L-l\colon S^1\setminus\{-1\}\longrightarrow\mathbb C$. It's a continuous function. Furthermore,$$(\forall z\in S^1\setminus\{-1\}):e^{(L-l)(z)}=e^{L(z)-l(z)}=\frac{e^{L(z)}}{e^{l(z)}}=\frac zz=1.$$Therefore, for each $z\in S^1\setminus\{-1\})$, ${(L-l)(z)}=2n\pi i$ for some integer $n$. But $S^1\setminus\{-1\})$ is connected and therefore $(L-l)\bigl(S^1\setminus\{-1\})\bigr)$ is connected too. So, there is an integer $n$ such that$$(\forall z\in S^1\setminus\{-1\})):L(z)=l(z)+2n\pi i.$$However, since $-1\in A\setminus\{0\}$ , $\lim_{z\to-1}L(z)=L(-1)$. But the limit $\lim_{z\to-1}l(z)+2n\pi i$ doesn't exist: the limit is $(2n+1)\pi i$ if we approach $-1$ from the upper half of the unit circle and it is $(2n-1)\pi i$ if we approach $-1$ from the lower half of the unit circle.

We've reached a contradiction. Since we assumed that there is a continuous map $L\colon A\setminus\{0\}\longrightarrow\mathbb C$ such that, for each $z\in A\setminus\{0\}$, $L(z)$ is a logarithm of $z$, no such map can exist.

• I have produced this proof before. The purpose of my question was to construct an slightly alternative proof to this one – KingJ May 26 '18 at 14:24