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 be a holomorphic function. A **holomorphic th root** of is a function such that for all . A **logarithm** of is a function such that for all . The main question we’ll try to figure out is: when do these exist? In particular, what if ?

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

To start us off, can we define for any complex number ?

The first obvious problem that comes up is that there for any , there are *two* numbers such that . 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 (here ) we should take the root to be

such that ; there are two choices for , differing by .

For complex numbers, we don’t have an obvious way to pick . Nonetheless, perhaps we can also get away with an arbitrary distinction: let’s see what happens if we just choose the with .

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

Here, for each , and we are constraining the 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 and ! The nearby point has been mapped very far away. This discontinuity occurs since the points on the negative real axis are at the “boundary”. For example, given , we send it to , but we have hit the boundary: in our interval , 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 . Suppose we instead insisted on the interval ; 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 -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 , 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 be holomorphic. Then we want to decide whether there is a such that

Our previous discussion when tells us we cannot hope to achieve this for ; there is a “half-ray” which causes problems. However, there are certainly functions such that a exists. As a simplest example, 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 , pick a square root of , and then try to “fudge” from there the square roots of the other points. What do I mean by fudge? Well, suppose is a point very close to , and we want to pick a square root of . While there are two choices, we also would expect to be close to . Unless we are highly unlucky, this should tells us which choice of to pick. (Stupid concrete example: if I have taken the square root of and then ask you to continue this square root to , which sign should you pick for ?)

There are two possible ways we could get unlucky in the scheme above: first, if , 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 :

We started at the point , with one of its square roots as . 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 , any ray of death must hit our red circle. The problem somehow lies with the fact that we have enclosed the very special point .

Nevertheless, we know that if we take , 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

Let . For , we already have the square root . So the burden is completing .

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

is a subset of the image of by . Since and are both punctured planes, we can identify them with .

**Ques 1**

Show that the image under is exactly once we identify .

That means that for any loop in , we need to have an *even* winding number around . This amounts to

since has no poles.

Replacing with and carrying over the discussion gives the first main result.

**Theorem 2** **(Existence of Holomorphic th Roots)**

Let be holomorphic. Then has a holomorphic th root if and only if

for every contour in .

## 4. Complex Logarithms

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

So if , then any complex number 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 since:

**Ques 3**

Show that if has any zeros then possibly can’t exist.

In fact, the map is a universal cover, since is simply connected. Thus, is *trivial*. So in addition to being zero-free, cannot have any winding number around at all. In other words:

**Theorem 4** **(Existence of Logarithms)**

Let be holomorphic. Then has a logarithm if and only if

for every contour in .

## 5. Some Special Cases

The most common special case is

**Corollary 5** **(Nonvanishing Functions from Simply Connected Domains)**

Let be continuous, where is simply connected. If for every , then has both a logarithm and holomorphic th root.

Finally, let’s return to the question of from the very beginning. What’s the best domain such that we can define ? Clearly cannot be made to work, but we can do almost as well. For note that the only zero of 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 (the point which we cannot circle around). This gives

**Theorem 6** **(Branch Cut Functions)**

There exist holomorphic functions

satisfying the obvious properties.

There are many possible choices of such functions ( choices for the th root and infinitely many for ); 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 . For , we take the such that and the imaginary part of lies in (since we can shift by integer multiples of ). Often, authors will write to emphasize this choice.

**Example 7**

Let be the complex plane minus the real interval . Then the function by has a holomorphic square root.

**Corollary 8**

A holomorphic function has a holomorphic th root for all if and only if it has a holomorphic logarithm.

I have a question on a example 7. In this case, I think U is not a simply connected domain. Then that function cannot have a holomorphic square root on U. Is it right?? Thank you.

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