# Isolated Singularities of Complex Square Root and Logarithmic Functions

On page 171 of Theodore Gamelin's Complex Analysis, there is an example saying the following:

"The functions $\sqrt{z}$ and $\log(z)$ do not have isolated singularities at $z=0$; they cannot be defined even continuously on any punctured disk centered at $0$."

I don't really understand why they can not be defined continuously. Suppose there is a punctured disk centered at $0$. For the square root, isn't every value on the plane defined? For either case, why are they not defined continuously?

The question is, which square root? There are two candidates for the square root of any nonzero complex number, say $f(z)$ and $-f(z)$. If you start at, say, $1$, and loop once around the origin, keeping the function continuous as you go, then when you come back to $1$ the value of $f$ that you end up with will be $-f(1)$. To get back to the same $f(1)$ it must take a jump.

Here's an animation showing this. The green dot is $z$, the red dot is $\sqrt{z}$. • Wow, haha, this is amazing.
– user41916
Dec 14, 2012 at 9:25

The log part is answered in the question Why is there no continuous log function on $\mathbb{C}\setminus\{0\}$?, so I will address only the square root.

Suppose, to reach a contradiction, that $f:\mathbb C\setminus\{0\}\to\mathbb C$ is a continuous square root function, i.e. $f(z)^2 = z$ for all $z\neq 0$. The restriction of $f$ to the unit circle $\mathbb T$ maps $\mathbb T$ to itself, and it is injective and continuous. Because $\mathbb T$ is compact and Hausdorff, this implies that $f(\mathbb T)$ is homeomorphic to $\mathbb T$. Since $f(\mathbb T)$ is compact and connected the only possibilities for $f(\mathbb T)$ are a proper closed arc or all of $\mathbb T$, and the former is ruled out because it is not homeomorphic to $\mathbb T$ (e.g., because from a proper closed arc you can remove 2 points without losing connectedness).

Hence $f(\mathbb T)=\mathbb T$, so there exist $w$ and $z$ such that $f(w)=-1$ and $f(z)=1$. This implies that $w=f(w)^2=(-1)^2=1$, and $z=f(z)^2=1^2=1$. So $z=w$, but $f(z)\neq f(w)$, which is absurd. Thus such an $f$ cannot exist.

(If instead you wanted to restrict to a small punctured disk, the same argument applies to $f$ mapping $|z|=r$ to $|z|=r^2$.)