$\log|z|$ has no harmonic conjugate in $\Bbb C\setminus\{0\}$ – different proof

There are a number of proofs in the Internet but I am specifically interested in the claim made by John B. Conway on page 43 of his Functions of One Complex Variable I (2nd edition).

Suppose $G$ is a region in the plane and $u:G\to\Bbb R$ is harmonic, if there exists a harmonic function $v:G\to\Bbb R$ such that $f=u+iv$ is analytic in $G$ then $v$ is a harmonic conjugate $\ldots u(z) = \log |z|$ of a harmonic function on the region $G = \mathbb{C} - \{ 0 \}$ has no harmonic conjugate, indeed, if it did then it would be possible to define an analytic branch of the logarithm on $G$, and this cannot be done.

The definition of an analytic function and analytic branch as given in text:

Def 1 A function $f: G \rightarrow \mathbb{C}$ is analytic if it is continuously differentiable on $G$, where $G$ is an open subset of $\mathbb{C}$.

Def 2 If $G$ is an open conncted set such that in $\mathbb{C}$ and $f: G \rightarrow \mathbb{C}$ is a continuous function such that $z = \exp f(z)$ for all $z \in G$ then $f$ is an analytic branch of logarithm.

I could only prove the latter part of the statement that it is not possible to define an analytic branch of the logarithm on $G$, which I provide below in (1), hopefully it is right. The part which I do not understand is how it is possible to define an analytic branch of the logarithm on $G$.

What am I missing? Thanks!

My proof for (1): If $f$ is a branch of logarithm we require that for all $z \in G$, $$ e^{f(z)} =e^{a+ib} = e^{a(z)} e^{ib(z)} = z = |z|e^{i \theta}, $$ where $a(z), b(z) \in \mathbb{R}$ and $\theta = \arg z$. Now $e^{a(z)} = |z| $ and $e^{ib(z)} = e^{i \theta}$. So we must have $a(z) = \log |z|$, and $b(z)= \theta + 2 \pi k(z) \in ( - \pi, \pi ] $ for some $k(z) \in \mathbb{N}$. So we obtain that $$ f(z) = \log |z| +i [ \theta + 2 \pi k(z) ] $$ where $k: G \rightarrow \mathbb{Z}$. But since $f$ is continuous (we assume $f$ is a branch) and that $G$ is connected, $k(G)$ is a connected set. Hence, $k$ is constant function. We now show that there exists a sequence, $z_n \rightarrow z$ such that $f(z_n) \not \rightarrow f(z)$ so $f$ is not continuous.
Consider $z = e^{i \pi} $, and the sequence of points, $z_n = e^{ i (-1)^n (\pi - \frac{1}{n})}$. Then \begin{align*} |f(z) - f(z_n)| & = | \log 1 - \log z_n + i ( \theta - \theta_n)| \\ &= | \log \bigg| \cos ( \pi - \frac{1}{n}) \bigg| + i \bigg( \pi - (-1)^n ( \pi - \frac{1}{n} )\bigg) | \\ &\ge | \pi - (-1)^n (\pi - \frac{1}{n} ) | \\ \end{align*} which does not converge.

Solution 1:

You're proof looks mostly sound, but I think you're making it more complicated than it needs to be. In your proof, you are showing the imaginary part of $f$ is discontinuous, which is the right idea. Your $\theta$ is really $\text{Arg}(z)$, which is a discontinuous function of $z$ (this should be pretty clear: as you approach the negative axis from above, $\text{Arg}(z)$ goes to $\pi$, whereas as you approach from below, it goes to $-\pi$. Now you have $f(z) = \log{|z|} + i(\text{Arg}(z) + 2\pi k(z))$. Your argument about $k(z)$ being constant needs to be refined a little bit, but all you need to show is that no integer-valued function makes $\text{Arg}(z) + 2\pi k(z)$ continuous, but basically the same principle applies. If $\text{Arg}(z) + 2\pi k(z)$ is continuous, then $k(z)$ has to be discontinuous on the negative half-axis (to counterbalance $\text{Arg}(z)$'s discontinuity there), and then you just have to show that $k(z)$ must be discontinuous somewhere else too.