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Problem statement

Let $\Omega \subset \mathbb C^*$ be open, we call branch of logarithm of $z$ in $\Omega$ to every continuous function $f:\Omega \to \mathbb C$ such that $e^{g(z)}=z$ for all $z \in \Omega$.

$(i)$ Prove that every branch of logarithm is injective and holomorphic on $\Omega$. Let $f_1,f_2$ be two branches of logarithm on $\Omega$. Show that if $\Omega$ is connected and there is $z_0 \in \Omega$: $f_1(z_0)=f_2(z_0) \implies f_1=f_2$ on $\Omega$.

$(ii)$ Prove that if there exists a branch of logarithm on $\Omega$, then $S^1 \not \subset \Omega$.

My attempt at a solution

For $(i)$, I suppose I must find a continuous function $f$ continuous on $\Omega$ and such that $e^{f(z)}=z$. I know that for $z=a+ib \in \mathbb C$, we define $$e^z=e^{a+ib}=e^ae^{ib}=e^a(\cos(b)+i\sin(b))$$ Also, if $e^z=w=|w|\dfrac{w}{|w|}$, then it is clear that $e^a=|w|$ and $e^{ib}=\dfrac{w}{|w|}$.

So, if I define the function $f(z)=log(|z|)+iarg(z)$, then $e^{f(z)}=e^{log(|z|)}e^{iarg(z)}=|z|(\cos(arg(z))+i\sin(arg(z)))=z$.

As $f$ is sum of two continous functions, then $f$ is continuous. Now, is $f$ is well defined? I ask this because for $z \in \mathbb C$, it is the same to describe $z$ by $arg(z)$ or $arg(z)+i2k\pi$.

I don't know how to prove the other part other part of $(i)$ and I also don't know what it means the notation $\mathbb C^*$.

As for $(ii)$ I don't know how to prove that statement but what's more important is that I intuitively don't get why can't $S^1$ be in $\Omega$ for existence of a branch.

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    $\begingroup$ $\mathbb C^*$ means $\mathbb C -(0,0)$. Intuitively, IMHO, the complications with the log have to see , not with the Real part, but with its complex part, which is the argument.Once you wind around the origin, your argument cannot be continuous, since it must jump from $0$ to $2\pi$ upon winding around the origin. The condition of winding around the origin is equivalent to containing a copy of $S^1$. $\endgroup$ – gary Jun 28 '14 at 23:40
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As you correctly observe, defining a branch of log in $\Omega$ comes down to defining arg continuously over all of $\Omega$. For any particular $0 \ne z \in \mathbb{C}$, you have many choices for arg. They are obtained by taking any one choice and then adding or subtracting a multiple of $2\pi$. For example, if $z=1$ then the choices are the integer multiples of $2\pi$.

For (i), the idea is that if $\Omega$ is connected, once you choose a value for arg for any single point $z_0 \in \Omega$, the rest of the values are forced. Thus if $f_1$ and $f_2$ agree at a single point, they agree on all of $\Omega$.

Formally, if arg is continuous in the neighborhood of a point $z_0 \in \Omega$, then there is a neighborhood $N$ of $z_0$ where the image $\arg N$ is contained in an open interval of length < $2\pi$. Thus arg is uniquely determined by its value at $z_0$. Now for any other point $z \in \Omega$, take a path from $z_0$ to $z$. We can find neighborhoods $N_w$ as above for any point $w$ along the path. By compactness of the path, we can find finitely many overlapping $N_w$ that cover the path, so arg is uniquely determined along the entire path to $z$. See picture at: http://en.wikipedia.org/wiki/Monodromy_theorem

For (ii), observe that if $S^1 \subset \Omega$ then when you wind around $S^1$ once counterclockwise, whatever value of arg you started with will increase by $2\pi$ by the time you come back around. Thus you cannot make a consistent choice of arg in $\Omega$.

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  • $\begingroup$ I have some doubts: how can I formally prove that connectedness determines the arg? On the other hand, I've proposed the function $f(z)=log(|z|)+iarg(z)$ I know that $log(x)$ is differentiable for $x>0$, but why can I affirm that $f$ is holomorphic? $\endgroup$ – user100106 Jun 29 '14 at 18:05
  • $\begingroup$ @user100106 See some details for (i) edited in. $\endgroup$ – Ted Jun 29 '14 at 18:41
  • $\begingroup$ @Ted Would u be able to have me with this releated about branch points question as well? Thanks math.stackexchange.com/questions/1905092/… $\endgroup$ – ys wong Aug 27 '16 at 15:21
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The way that branches of logz make more sense to me is that they are local inverses of $e^z$; $e^z$ is not globally-invertible, since it is infinite-to-1 , because it is periodic, but , e.g., by the inverse function theorem, since $d/dz(e^z)=e^z \neq 0$, $e^z$ does have local inverses, which are called branches. Each branch is a rectangle of infinite width, and with height $2\pi^-$ , i.e., the height is of the form $[y, y+2\pi)$, which is made to prevent $e^z$ from realizing a period within the region, i.e., since $e^z=e^{z+2\pi}$ we want an injection, so we want to avoid having $z-$ values that are $2\pi$ appart.

For 2), a standard result is that a branch of logz exists in $\Omega$ when $\Omega$ is simply-connected, and it does not wind around the origin. For one thing, logz is defined as $\int_{\gamma} \frac{dz}{z}$ , where $\gamma \in \Omega$ , and for the integral to be well-defined, we need independence of path, for which we need a simply-connected region. If logz was globally defined, then the integral about a closed curve would be zero. But the integral of $dz/z$ about $S^1$ is known to be $2\pi i$; this will happen whenever you wind around the origin, which is equivalent to having a copy of $S^1$ in your region $\Omega$.

For1), you can just use the fact that in a branch , as you said, you have $e^{logz}=z=log(e^z)$, then, by invertibility, you must have a bijection and, in particular, an injection.

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  • $\begingroup$ There are two things where I still have doubts: in part (i) I still don't know how to prove that if $\Omega$ is connected and there is $z_0 \in \Omega$ such that $f_1(z_0)=f_2(z_0) \implies f_1=f_2$ on $\Omega$. For $(ii)$, I understand and like your argument but here the hypothesis is that $\Omega \subset \mathbb C^*$, so I cannot use that $\Omega$ is simple connected, just the fact that $\Omega \subset \mathbb C^*$ and that $\Omega$ is open. $\endgroup$ – user100106 Jun 29 '14 at 6:10
  • $\begingroup$ For i), if I understood you correctly, you have that branches are local inverses, so these are bijections of $\mathbb R \times [y,y+2\pi)$ with $\mathbb C^*$, so that they are $1-1$. So if $f_1(z)=f_2(z)$, these must both be the same branch of logz. for ii), I assume $S^1$:={$z:|z|=1$}, so that if $\Omega$ contains a copy of $S^1$ , then $\Omega$ must either not be simply-connected, or $\Omega$ must wind around the origin. I hope I understood (and answered) your questions correctly. $\endgroup$ – gary Jun 29 '14 at 6:19
  • $\begingroup$ You are assuming what you are trying to prove. Assuming $e^{log(z)}=log(e^z)$ is assuming they are inverses (i.e. one to one) before you have proven that they are one to one. Same thing with the inverse function theorem. You are assuming that the logarithms are differentiable when that is what the OP is asking to prove. $\endgroup$ – Bobby Ocean Jun 29 '14 at 6:39
  • $\begingroup$ By definition/construction a branch logz of the log is a local inverse of $e^z$, so that $e^{logz}=log(e^z)=z$ within the branch. I don't know where differentiability comes up, but $e^z$ is differentiable, with $d/dz(e^z)=e^z \neq 0$ , so that the inverse function theorem guarantees a local inverse; that local inverse is the branch $logz$. $\endgroup$ – gary Jun 29 '14 at 6:44
  • $\begingroup$ @gary I agree with what you are saying except that what you are saying is not proof. The OP was not told that ``every branch of the lograithm is a local inverse of $e^z$''. I agree that is true, but I would think that is what the OP is suppose to prove (not just state it is so). Likewise, the inverse function theorem states that the local inverse of a non-zero analytic function is differentiable (I agree), but that is not what was being asked. The OP is suppose to prove that statement for the branchs of the logarithm, not just state it is so. Unless I am misunderstanding the OP. $\endgroup$ – Bobby Ocean Jun 29 '14 at 6:53
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Proof of (ii), Suppose there exists a continuous branch of argument on $\Omega$, with $S^1 \subset \Omega$, say $f$.

Consider the function $$g(\theta)=\frac{f(e^{i\theta})+f(e^{-i\theta})}{2\pi}\hspace{5mm} \theta \in [0,2\pi) $$ Then $g$ is continuous on $[0,2\pi)$ (connected set) and $g$ takes only integer values. So, $g$ is constant function. So, $g(0)=g(\pi) \implies f(1)=f(-1)$, which is a contradiction, as $f(1)\in \{2n\pi: n \in \mathbb{Z}\}$ and $f(-1)\in \{2n\pi-\pi: n \in \mathbb{Z}\}$.

So there does not exist a continuous branch of argument on $\Omega$, with $S^1 \in \Omega$, and hence there does not exist a holomorphic branch of logarithm on $\Omega$

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