I am trying to show the following statement:

Let $\Omega \subset \mathbb C^*$ be an open subset. We define a branch of logarithm as a continuous function $g:\Omega \to \mathbb C$ such that $e^{g(z)}=z$ for all $z \in \Omega$. Show that if $g$ is a branch of logarithm on $\Omega$, then $\Omega$ can't contain $S^1$.

So here's what I did:

Suppose $S^1 \subset \Omega$, then there is $0<\epsilon<1$ such that $B(1,\epsilon) \subset \Omega$. The functions $$f_1(z)=ln|z|+i\arg_1(z), \space \arg_1(z) \in (-\pi,\pi)$$ $$f_2(z)=ln|z|+i\arg_2(z), \space \space \arg_2(z) \in (0,2\pi)$$

are branches of logarithm in $S_1=B(1,\epsilon)$ and $S_2=B(1,\epsilon) \cap (Im(z)<0)$ respectively.

Now, since $e^{g(1)}=1$, then $g(1)=i2m\pi$ for some $m$ integer. First suppose $m=0$. We have that $g(1)=f_1(1)$, so we must have $g=f_1$ in the ball $B(1,\epsilon)$. And since $g$ is a branch of logarithm in $S_2$ as well, then $$g=f_2+i2k\pi$$for some integer $k$.

Take a sequence $(z_n) \in S_2$ such that $z_n \to 1$. Then we have $g(z_n)=ln|z_n|+i(\arg_2(z_n)+2k\pi)$. It is clear that $\arg_2(z_n) \to 2\pi$.

Since $g$ is continuous, we have $g(z_n) \to g(1)=0$. So we must have $\arg_2(z_n)+2k\pi \to 0$, it follows that $k=-1$.

I wanted to arrive to an absurd but I couldn't. I would appreciate if someone could help me to show the statement.

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    $\begingroup$ For every (sufficiently regular, e.g. piecewise continuously differentiable) closed curve $\gamma$ in $\Omega$, you have $$\int_{\gamma} g'(z)\,dz = 0.$$ $\endgroup$ – Daniel Fischer Oct 7 '15 at 18:46
  • $\begingroup$ Thanks for your suggestion but I haven't covered integration yet, I am supposed to solve this problem without using any complex integration theory. However, when I get to that I'll try to solve the exercise with your approach. $\endgroup$ – user16924 Oct 7 '15 at 18:50
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    $\begingroup$ Oy, logarithms introduced before path integrals? That's unusual. What theory have you to work with? Probably, the method of choice will be to cover $S^1$ by finitely many open disks $D_1,\dotsc, D_n$ contained in $\Omega$, order them so that $1 \in D_1$, and $D_k \cap D_{k+1} \neq \varnothing$ as well as $D_n \cap D_1 \neq \varnothing$, and since the argument (imaginary part of the logarithm) on $D_{k+1}$ is determined by that on $D_k$, conclude that on $D_n \cap D_1$, the arguments would differ by $2\pi$. $\endgroup$ – Daniel Fischer Oct 7 '15 at 18:55
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    $\begingroup$ You did arrive at a desired absurdity for a particular case.In general, with $f_k(z)=i\arg z +2\pi i k$ for $|z|=1$, suppose $g(1)=2\pi i k=f_k(1)$ for some $k\in Z$. Prove that the continuity of $g$ implies that $\sup \{A\in [0.2\pi) : g(\cos A +i\sin A)=f_k(\cos A+i\sin A)\}=2\pi.$ Then show this implies $g(z)$ is discontinuous at $z=1$.Note that we don't need to know that the domain of $g$ is open,only that $g$ is (allegedly) continuous on $S^1$. $\endgroup$ – DanielWainfleet Oct 7 '15 at 19:19
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    $\begingroup$ You don't change the balls, you change the numbering to achieve nonempty intersections. $\endgroup$ – Daniel Fischer Oct 7 '15 at 19:20

Let $\log z$ denote the principal branch. Let $S' = S^1\setminus \{-1\}.$ We have $e^{g(z)} = e^{\log z}$ on $S'.$ Define $h(z) = g(z) - \log z.$ Then $e^{h(z)} \equiv 1$ on $S'.$ Thus $h$ maps $S'$ into $2\pi i \mathbb {Z}.$ But $h$ is continuous on $S',$ a connected set. Thus $h(S')$ is a connected subset of $2\pi i \mathbb {Z},$ which implies $h(S')$ is a single point, i.e., $h$ is constant on $S'.$ So we have $g(z) = \log z + C$ on $S',$ which implies $g$ is not continuous on $S^1,$ contradiction.

  • $\begingroup$ Is the following argument correct to justify that $g$ cannot be continuous on $S^1$? if $g$ was continuous, then $log(z)=g(z)-C$ can be continuously extended to $S^1$, so $\arg(z)=\dfrac{1}{i}(g(z)-C-ln|z|)$ which is defined for $z \in S'$ can be extended continuosly to $z=-1$, which is absurd. $\endgroup$ – user16924 Oct 7 '15 at 20:00
  • $\begingroup$ Yes that's right. $\endgroup$ – zhw. Oct 7 '15 at 20:28

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