Is the function $f:\mathbb C \to \mathbb C$ defined as $f(z)=\sqrt z$ uniformly continuous ?

Question

Is the function $f:\mathbb C \to \mathbb C$ defined as $f(z)=\sqrt z$ uniformly continuous ? Is it uniformly continuous on some non-compact restriction of the complex plane (other than $(0,\infty)$ ) ?

Answer 1

So the problem is that you can't define $\sqrt{z}$ to be even continuous on $\mathbb C$.

The reason is that, as you go along the circle $z=e^{i\theta}$ as $\theta\in[0,2\pi)$, if you want $\sqrt{z}$ to remain continuous (and $\sqrt{1}=1$) you'd need $\sqrt{z}=e^{i\theta/2}$. But then as $\theta\to 2\pi$, we have $z=e^{i\theta}\to 1$ and $\sqrt{z}=e^{i\theta/2}\to e^{i\pi}=-1$.

There are a number of ways around this sort of problem in complex analysis.

• We can define $\sqrt{z}$ as a multi-valued function with two possible values. As you can see, in the above argument, we got $\sqrt{1}=\pm 1$, for example.
• We can define $\sqrt{z}$ so that it remains undefined (or defined and discontinuous) on some line (or more generally a curve) starting at zero and going to infinity. The most common way to do this is to leave $\sqrt{z}$ undefined on the negative real line $(-\infty,0)$, which is simplest. (This always strikes me as funny because when we first learn about complex numbers, we are told $i=\sqrt{-1}$ and then we are told that $\sqrt{z}$ is left undefined at $z=-1$. But those are essentially two different usages of the "square root" symbol. In particular, $i$ is not actually defined as the square root of $-1$, that is just a motivating idea.)
• We can define the notion of a branch of a function. This is a deeper concept, both related to the concept of the cut and the multi-valued view of the function.

Answer 2

An additional answer to clarify some minor details. Here's the Riemann surface of the square root map:

The map contains two branches. The two branches intersect at the negative-axis $(-\infty,0]$.

On this surface, you can preserve the notion of continuity using the convention of counterclockwise continuity. In other words, if you want to cross between branches, then you cross them in the order you stitched them together using a suitable branch cut.

In order to establish this convention however, you need to consider a cut that separates the two branches, which is the joining curve of the two sheafs. This way, the upper sheaf is smoothly connected to the lower sheaf, forming a single surface.

Precisely because the map $z^{1/2}$ is two-valued on $(0,+\infty)$ (i.e., $\sqrt{x}=\pm\sqrt{|x|}$), one convention is to use the positive real axis as the branch cut of this map (as in the second figure).

Then, if you constrain your $x,y$ in each branch separately, you can use the regular argument to prove that each branch is uniformly continuous on its respective domain.

If you allow however a domain such as the one shown on the pic above, either crossing your cut or crossing after encircling, you can see that there are going to be problems. In this case, you can choose your $x$ and $y$ to be very close on the complex plane, say $x=x_0+\epsilon i$, $y=x_0-\epsilon i$, with $x_0\in\mathbb{R}^+$ and some $\epsilon\gt 0$ and then uniform continuity will fail, because although $|x-y|=|x_0+\epsilon i-x_0+\epsilon i|\le 2\epsilon$, you use different branches of the map, so for any $x_0\in\mathbb{R}^+$, with $\sqrt{|x_0|}\gt\epsilon$:

$$|\sqrt{x}-\sqrt{y}|=|\sqrt{x_0+\epsilon i}-(-\sqrt{x_0-\epsilon i})|\gt |2\sqrt{x_0}|\gt 2\epsilon$$

The net effect of the above is that you are prevented from choosing a domain that wraps around the origin or crossing the branch, so continuity is preserved by the counterclockwise continuity convention and hence you can say the map $z^{1/2}$ is uniformly continuous on $\mathbb{C}$, which is qualified as saying that each branch of the map (i.e. the principal and the secondary branch) is uniformly continuous on its respective domain $\mathbb{C}\setminus (0,+\infty)$.