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I'm having trouble proving or disproving this question. I kind of expect that there is no such function since the definition of the logarithm of a function $\phi$ that is analytic on a simply connected open set $U$ is

$$L_\phi(z)=\int^z_{z_0}\frac{\phi'(\zeta)}{\phi(\zeta)}d\zeta + w_0.$$

With $w_0$ being a complex number such that $e^{w_0}=f(z_0)$. But in our case we are looking for an analytic function on $\mathbb{C}\setminus[-1,1]$ which is not simply connected.

But on the other hand if I let $\phi(z)=\frac{z+1}{z-1}$ and $\gamma$ be a path in $\mathbb{C}\setminus[-1,1]$ from $z_0$ to $z$ and define $f(z)=L_\gamma\phi(z)$ as

$$L_\gamma\phi(z)=\int^z_{z_0,\gamma}\frac{\phi'(\zeta)}{\phi(\zeta)}d\zeta + w_0.$$

Then for any two different paths in $\mathbb{C}\setminus[-1,1]$ connecting $z_0$ and $z$ the results of $f(z)$ will differ by an integer multiple of $2\pi i$. So even though $f$ may not be well defined the exponential of $f$ is well defined since

$$e^{L_{\gamma_1}\phi(z)}=e^{L_{\gamma_2}\phi(z)+m2\pi i}=\phi(z)=\frac{z+1}{z-1}.$$

These have been my attempts so far, but I am not sure if I am on the right track or completely missing something.

Gehaktmolen
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    To what set does $z \mapsto \frac{z+1}{z-1}$ map $\mathbb{C}\setminus [-1,1]$? – Daniel Fischer Jun 21 '15 at 20:08
  • I believe $\mathbb{C}\setminus\mathbb{R}_{\leq0}$, right? – Gehaktmolen Jun 21 '15 at 20:12
  • Yes, that's right. And what follows about the existence of $f$? – Daniel Fischer Jun 21 '15 at 20:13
  • I'm not sure what you mean by that – Gehaktmolen Jun 21 '15 at 20:15
  • If the image of $\phi$ is contained in a domain where a branch of the logarithm exists ... – Daniel Fischer Jun 21 '15 at 20:18
  • So we define $f(z)=\log(z)$ on $\mathbb{C}\setminus\mathbb{R}_{\leq0}$ and since the image of $\phi$ under $\mathbb{C}\setminus[-1,1]$ is precisely that and that the logarithm is analytic for a simply connected open set we have that $f(\phi(z))$ is analytic on $\mathbb{C}\setminus[-1,1]$ and satisfies $e^{f(z)}=(z+1)/(z-1)$? – Gehaktmolen Jun 21 '15 at 20:23
  • Yes. And alternatively, going further along the way you started on in your question, if you compute $\frac{\phi'(z)}{\phi(z)}$ for $\phi(z) = \frac{z+1}{z-1}$, you'll note that the integral over every closed curve in $\mathbb{C}\setminus [-1,1]$ vanishes. – Daniel Fischer Jun 21 '15 at 20:27
  • Thank you very much, looking at it this way was much easier than considering the integral definition of the logarithm... – Gehaktmolen Jun 21 '15 at 20:29
  • But the integral is more generally applicable. It's a special case when we can just compose with a branch of $\log z$. – Daniel Fischer Jun 21 '15 at 20:35
  • If I would be to use the integral I would just show that the integral of any closed curve in $\mathbb{C}\setminus[-1,1]$ is zero and that you are just left with a path from $z_0$ to $z$ not winding around $[-1,1]$ thus making the integral path independent? But then I'm stuck on how to show that it is analytic? – Gehaktmolen Jun 21 '15 at 20:43
  • Since the integral is path-independent, we have $$\frac{f(z_2) - f(z_1)}{z_2 - z_1} = \frac{1}{z_2 - z_1}\int_0^1 \frac{\phi'(z_1 + t(z_2-z_1))}{\phi(z_1 + t(z_2-z_1))}\cdot (z_2-z_1),dt,$$ provided $z_2$ is close enough to $z_1$ that the straight line segment connecting the two doesn't pass through $[-1,1]$. – Daniel Fischer Jun 21 '15 at 20:48
  • $\log (z\pm 1)$ has a branch point at $\mp 1$. For $f(z)=\log \frac{z+1}{z-1}$ we may cut $[-1,1]$ from the plane rendering $f$ analytic in the plane equipped with the aforementioned cut. – Mark Viola Jun 21 '15 at 21:25

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As stated in comments, the answer is affirmative: $(1+z)/(1-z)$ maps the unit disk onto the open right half-plane, which is within the domain of the principal branch of the logarithm. Using this branch, we get $\log\frac{1+z}{1-z}$ as a holomorphic function in the unit disk.