# 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?

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The log part is answered in math.stackexchange.com/questions/104710/…. – Jonas Meyer Dec 14 '12 at 9:00

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 $\sqrt{z}$.

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Wow, haha, this is amazing. – Kiyoshi Dec 14 '12 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 $|z|=r^2$.)

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