# Holomorphic Logarithms and Roots

In this post we’ll make sense of a holomorphic square root and logarithm. Wrote this up because I was surprised how hard it was to find a decent complete explanation.

Let ${f : U \rightarrow \mathbb C}$ be a holomorphic function. A holomorphic ${n}$th root of ${f}$ is a function ${g : U \rightarrow \mathbb C}$ such that ${f(z) = g(z)^n}$ for all ${z \in U}$. A logarithm of ${f}$ is a function ${g : U \rightarrow \mathbb C}$ such that ${f(z) = e^{g(z)}}$ for all ${z \in U}$. The main question we’ll try to figure out is: when do these exist? In particular, what if ${f = \mathrm{id}}$?

## 1. Motivation: Square Root of a Complex Number

To start us off, can we define ${\sqrt z}$ for any complex number ${z}$?

The first obvious problem that comes up is that there for any ${z}$, there are two numbers ${w}$ such that ${w^2 = z}$. How can we pick one to use? For our ordinary square root function, we had a notion of “positive”, and so we simply took the positive root.

Let’s expand on this: given ${ z = r \left( \cos\theta + i \sin\theta \right) }$ (here ${r \ge 0}$) we should take the root to be

$\displaystyle w = \sqrt{r} \left( \cos \alpha + i \sin \alpha \right).$

such that ${2\alpha \equiv \theta \pmod{2\pi}}$; there are two choices for ${\alpha \pmod{2\pi}}$, differing by ${\pi}$.

For complex numbers, we don’t have an obvious way to pick ${\alpha}$. Nonetheless, perhaps we can also get away with an arbitrary distinction: let’s see what happens if we just choose the ${\alpha}$ with ${-\frac{1}{2}\pi < \alpha \le \frac{1}{2}\pi}$.

Pictured below are some points (in red) and their images (in blue) under this “upper-half” square root. The condition on ${\alpha}$ means we are forcing the blue points to lie on the right-half plane.

Here, ${w_i^2 = z_i}$ for each ${i}$, and we are constraining the ${w_i}$ to lie in the right half of the complex plane. We see there is an obvious issue: there is a big discontinuity near the point ${z_5}$ and ${z_7}$! The nearby point ${w_6}$ has been mapped very far away. This discontinuity occurs since the points on the negative real axis are at the “boundary”. For example, given ${-4}$, we send it to ${-2i}$, but we have hit the boundary: in our interval ${-\frac{1}{2}\pi \le \alpha < \frac{1}{2}\pi}$, we are at the very left edge.

The negative real axis that we must not touch is is what we will later call a branch cut, but for now I call it a ray of death. It is a warning to the red points: if you cross this line, you will die! However, if we move the red circle just a little upwards (so that it misses the negative real axis) this issue is avoided entirely, and we get what seems to be a “nice” square root.

In fact, the ray of death is fairly arbitrary: it is the set of “boundary issues” that arose when we picked ${-\frac{1}{2}\pi < \alpha \le \frac{1}{2}\pi}$. Suppose we instead insisted on the interval ${0 \le \alpha < \pi}$; then the ray of death would be the positive real axis instead. The earlier circle we had now works just fine.

What we see is that picking a particular ${\alpha}$-interval leads to a different set of edge cases, and hence a different ray of death. The only thing these rays have in common is their starting point of zero. In other words, given a red circle and a restriction of ${\alpha}$, I can make a nice “square rooted” blue circle as long as the ray of death misses it.

So, what exactly is going on?

## 2. Square Roots of Holomorphic Functions

To get a picture of what’s happening, we would like to consider a more general problem: let ${f: U \rightarrow \mathbb C}$ be holomorphic. Then we want to decide whether there is a ${g : U \rightarrow \mathbb C}$ such that

$\displaystyle f(z) = g(z)^2.$

Our previous discussion when ${f = \mathrm{id}}$ tells us we cannot hope to achieve this for ${U = \mathbb C}$; there is a “half-ray” which causes problems. However, there are certainly functions ${f : \mathbb C \rightarrow \mathbb C}$ such that a ${g}$ exists. As a simplest example, ${f(z) = z^2}$ should definitely have a square root!

Now let’s see if we can fudge together a square root. Earlier, what we did was try to specify a rule to force one of the two choices at each point. This is unnecessarily strict. Perhaps we can do something like the following: start at a point in ${z_0 \in U}$, pick a square root ${w_0}$ of ${f(z_0)}$, and then try to “fudge” from there the square roots of the other points. What do I mean by fudge? Well, suppose ${z_1}$ is a point very close to ${z_0}$, and we want to pick a square root ${w_1}$ of ${f(z_1)}$. While there are two choices, we also would expect ${w_0}$ to be close to ${w_1}$. Unless we are highly unlucky, this should tells us which choice of ${w_1}$ to pick. (Stupid concrete example: if I have taken the square root ${-4.12i}$ of ${-17}$ and then ask you to continue this square root to ${-16}$, which sign should you pick for ${\pm 4i}$?)

There are two possible ways we could get unlucky in the scheme above: first, if ${w_0 = 0}$, then we’re sunk. But even if we avoid that, we have to worry that we are in a situation, where we run around a full loop in the complex plane, and then find that our continuous perturbation has left us in a different place than we started. For concreteness, consider the following situation, again with ${f = \mathrm{id}}$:

We started at the point ${z_0}$, with one of its square roots as ${w_0}$. We then wound a full red circle around the origin, only to find that at the end of it, the blue arc is at a different place where it started!

The interval construction from earlier doesn’t work either: no matter how we pick the interval for ${\alpha}$, any ray of death must hit our red circle. The problem somehow lies with the fact that we have enclosed the very special point ${0}$.

Nevertheless, we know that if we take ${f(z) = z^2}$, then we don’t run into any problems with our “make it up as you go” procedure. So, what exactly is going on?

## 3. Covering Projections

By now, if you have read the part of algebraic topology. this should all seem very strangely familiar. The “fudging” procedure exactly describes the idea of a lifting.

More precisely, recall that there is a covering projection

$\displaystyle (-)^2 : \mathbb C \setminus \{0\} \rightarrow \mathbb C \setminus \{0\}.$

Let ${V = \left\{ z \in U \mid f(z) \neq 0 \right\}}$. For ${z \in U \setminus V}$, we already have the square root ${g(z) = \sqrt{f(z)} = \sqrt 0 = 0}$. So the burden is completing ${g : V \rightarrow \mathbb C}$.

Then essentially, what we are trying to do is construct a lifting ${g}$ for the following diagram: Our map ${p}$ can be described as “winding around twice”. From algebraic topology, we now know that this lifting exists if and only if

$\displaystyle f_\ast(\pi_1(V)) \subseteq p_\ast(\pi_1(E))$

is a subset of the image of ${\pi_1(E)}$ by ${p}$. Since ${B}$ and ${E}$ are both punctured planes, we can identify them with ${S^1}$.

Ques 1

Show that the image under ${p}$ is exactly ${2\mathbb Z}$ once we identify ${\pi_1(B) = \mathbb Z}$.

That means that for any loop ${\gamma}$ in ${V}$, we need ${f \circ \gamma}$ to have an even winding number around ${0 \in B}$. This amounts to

$\displaystyle \frac{1}{2\pi} \oint_\gamma \frac{f'}{f} \; dz \in 2\mathbb Z$

since ${f}$ has no poles.

Replacing ${2}$ with ${n}$ and carrying over the discussion gives the first main result.

Theorem 2 (Existence of Holomorphic ${n}$th Roots)

Let ${f : U \rightarrow \mathbb C}$ be holomorphic. Then ${f}$ has a holomorphic ${n}$th root if and only if

$\displaystyle \frac{1}{2\pi i}\oint_\gamma \frac{f'}{f} \; dz \in n\mathbb Z$

for every contour ${\gamma}$ in ${U}$.

## 4. Complex Logarithms

The multivalued nature of the complex logarithm comes from the fact that

$\displaystyle \exp(z+2\pi i) = \exp(z).$

So if ${e^w = z}$, then any complex number ${w + 2\pi i k}$ is also a solution.

We can handle this in the same way as before: it amounts to a lifting of the following diagram. There is no longer a need to work with a separate ${V}$ since:

Ques 3

Show that if ${f}$ has any zeros then ${g}$ possibly can’t exist.

In fact, the map ${\exp : \mathbb C \rightarrow \mathbb C\setminus\{0\}}$ is a universal cover, since ${\mathbb C}$ is simply connected. Thus, ${p(\pi_1(\mathbb C))}$ is trivial. So in addition to being zero-free, ${f}$ cannot have any winding number around ${0 \in B}$ at all. In other words:

Theorem 4 (Existence of Logarithms)

Let ${f : U \rightarrow \mathbb C}$ be holomorphic. Then ${f}$ has a logarithm if and only if

$\displaystyle \frac{1}{2\pi i}\oint_\gamma \frac{f'}{f} \; dz = 0$

for every contour ${\gamma}$ in ${U}$.

## 5. Some Special Cases

The most common special case is

Corollary 5 (Nonvanishing Functions from Simply Connected Domains)

Let ${f : \Omega \rightarrow \mathbb C}$ be continuous, where ${\Omega}$ is simply connected. If ${f(z) \neq 0}$ for every ${z \in \Omega}$, then ${f}$ has both a logarithm and holomorphic ${n}$th root.

Finally, let’s return to the question of ${f = \mathrm{id}}$ from the very beginning. What’s the best domain ${U}$ such that we can define ${\sqrt{-} : U \rightarrow \mathbb C}$? Clearly ${U = \mathbb C}$ cannot be made to work, but we can do almost as well. For note that the only zero of ${f = \mathrm{id}}$ is at the origin. Thus if we want to make a logarithm exist, all we have to do is make an incision in the complex plane that renders it impossible to make a loop around the origin. The usual choice is to delete negative half of the real axis, our very first ray of death; we call this a branch cut, with branch point at ${0 \in \mathbb C}$ (the point which we cannot circle around). This gives

Theorem 6 (Branch Cut Functions)

There exist holomorphic functions

\displaystyle \begin{aligned} \log &: \mathbb C \setminus (-\infty, 0] \rightarrow \mathbb C \\ \sqrt[n]{-} &: \mathbb C \setminus (-\infty, 0] \rightarrow \mathbb C \end{aligned}

satisfying the obvious properties.

There are many possible choices of such functions (${n}$ choices for the ${n}$th root and infinitely many for ${\log}$); a choice of such a function is called a branch. So this is what is meant by a “branch” of a logarithm.

The principal branch is the “canonical” branch, analogous to the way we arbitrarily pick the positive branch to define ${\sqrt{-} : \mathbb R_{\ge 0} \rightarrow \mathbb R_{\ge 0}}$. For ${\log}$, we take the ${w}$ such that ${e^w = z}$ and the imaginary part of ${w}$ lies in ${(-\pi, \pi]}$ (since we can shift by integer multiples of ${2\pi i}$). Often, authors will write ${\text{Log } z}$ to emphasize this choice.

Example 7

Let ${U}$ be the complex plane minus the real interval ${[0,1]}$. Then the function ${U \rightarrow \mathbb C}$ by ${z \mapsto z(z-1)}$ has a holomorphic square root.

Corollary 8

A holomorphic function ${f : U \rightarrow \mathbb C}$ has a holomorphic ${n}$th root for all ${n \ge 1}$ if and only if it has a holomorphic logarithm.