Holomorphic function with $f(z)^2=z$

Question

Is there an holomorphic function $f:B_1(0)\setminus\{0\}\rightarrow\mathbb{C}$ with $f(z)^2=z$?

Answer 1

If $f: B(0, 1)-\{0\} \rightarrow \mathbb{C}$ is holomorphic with $f(z)^2=z$, then $$\lim_{z \rightarrow 0} f(z) = 0,$$ because $\lim_{z \rightarrow 0} f(z)^2 = 0$. Now we can extend $f$ to the whole of $B(0, 1)$, by setting $f(0)=0$, and $f$ becomes holomorphic on the whole of $B(0,1)$. This means $f$ has a Taylor series around $0$, say $$f(z) = \sum_{n=0}^{\infty} c_n z^n.$$ Now $$f(z)^2 = (\sum_{n=0}^{\infty} c_n z^n)^2 = c_0^2 +(c_0c_1+c_1c_0)z+...=z.$$ But Taylor series are unique, so $c_0=0$, in contradiction with $c_0c_1+c_1c_0=1$. So such $f$ does not exist. Let me know if I missed something!

Answer 2

There isn't even a continuous one. Suppose $f$ is a continuous one, and define $\gamma:[0,2\pi]\to\Bbb C$ by $$\gamma(t)=f\left(\frac12 e^{it}\right).$$Then $\gamma$ the index, or "winding number", of $\gamma$ about the origin would be $1/2$. Hmm.... It's clear, and follows from the fact that the exponential is a covering map, that there exist continuous functions $r:[0,2\pi]\to(0,\infty)$ and $\theta:[0,2\pi]\to\Bbb R$ with $\theta(0)=0$ and$$\gamma(t)=r(t)e^{i\theta(t)}.$$Since $f(z)^2=z$ this shows that $$e^{2i\theta(t)}=e^{it};$$now since $\theta(0)=0$ and $\theta$ is continuous it follows that that $$\theta(t)=t/2,$$hence $\gamma(2\pi)\ne\gamma(0)$, contradiction.

Answer 3

Starting as M. Van, $f$ can be extended holomorphically to the whole ball with $f(0) = 0$. Taking derivatives: $$2f(z)f'(z) = 1 \implies 2f(0)f'(0) = 1 \implies 0 = 1.$$ Contradiction.

(Is essentially the same argument without invoking explicitly the Taylor series.)