Branches of the complex logarithm

Question

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.

Answer 1

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 $\Omega$.

Answer 2

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.

Answer 3

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 $\Omega$