Example of holomorphic function with no natural root

Question

If $f$ is holomorphic over a simply connected set $\Omega\subset\Bbb C$ and $f(z)\ne0\ \forall z\in\Omega$ then it is known that for any $n\in\Bbb N^*$ there exists a holomorphic function $g$ such that $g^n=f$.

But this isn't necessarily true if one of the properties [$f\ne0\ \forall z\in\Omega$] or [$\Omega$ simply connected] isn't true.

• Why is $z\mapsto z$ over $\Bbb C$ a counter example? Namely, why is $\pm z^{1/2}=\pm e^{\log z\over 2 }$ not holomorphic?
• Is there an example where $f\ne 0\forall z\in \Omega$ but the fact that $\Omega$ isn't simply connected messes things up?

Answer 1

Let $V=\Bbb C\setminus \{0\}$. Of course there is no $f\in H(V)$ with $f(z)^2=z$. People show this by considering a function $z^{1/2}$ holomorphic near $1$, say, and then showing that if you try to continue it on a loop around the origin what you get is discontinuous.

An argument that seems cleaner and simpler to me: Recall first that if $f\in H(V)$ and $\gamma$ is any closed curve in $V$ then $$\frac1{2\pi i}\int_\gamma\frac{f'(z)}{f(z)}\,dz\in\Bbb Z,$$by the argument principle. But if $f(z)^2=z$ then $2f(z)f'(z)=1$, so $f'(z)/f(z)=1/(2 f(z)^2)=1/(2z)$and hence $$\frac1{2\pi i}\int_{|z|=1}\frac{f'(z)}{f(z)}\,dz=\frac12.$$

(Considering the argument principle, this really is the same proof: If $\gamma$ is some parametrization of the unit circle then this shows that $f\circ \gamma$ cannot be a closed curve, since the winding number about the origin would be $1/2$. In fact $1/2$ is exactly right - as $z$ traverses the unit circle, $z^{1/2}$ only goes halfway around the circle, ending at the negative of the initial point.)

About square roots in general:

Exercise. Suppose $\Omega$ is an open set in the plane and $g\in H(\Omega)$, not vanishing identically on any component of $\Omega$. There exists $f\in H(\Omega)$ with $f^2=g$ if and only if $\int_\gamma g'/g$ is an even integer for every closed curve $\gamma$ in $\Omega$.

Similarly for the non-existence of a logarithm: If $f\in H(V)$ and $\gamma$ is a closed curve in $V$ then $$\int_\gamma f'(z)\,dz=0,$$but $\int_{|z|=1}\frac1z\,dz\ne0$, hence there is no $f\in H(V)$ with $f'(z)=1/z$.

Answer 2

There are no such functions as $\pm z^{1/2}$ and $\pm e^{\frac{\log z}2}$, because

• $z^{1/2}$ can be any sequare root of $z$. So there it doesnt represent an individual complex number.
• each complex number $z$ has infinitely many logarithms. Which one do ou have in mind.

Concerning your final question,$$\begin{array}{ccc}\mathbb{C}\setminus\{0\}&\longrightarrow&\mathbb C\\z&\mapsto&z\end{array}$$has no holomorphic square root.