How to define the complex square root $ \sqrt{z} $?

Question

We need to define the complex square root $ \sqrt{z} $ on a small open $ U \subset \mathbb{C} $, for example a disc. Let put : $ \mathcal{F} (U) = \{\ f: U \to \mathbb {C} \ / \ f \ \text{is continuous} \ \forall z \in U \: \ (f (z))^2 = z \ \} $

Questions:

• Why is : $ 0 \in U \ \ \Longrightarrow \ \ \mathcal{F} (U) = \emptyset $ ?
• So either: $ U \subset \mathbb {C}^* $ and connected non-empty: If: $ U = U_1 \bigcup U_2 $, open $ U_1 $ and $ U_2 $ are disjoint, why the application: $ \mathcal{F } (U) \to \mathcal{F} (U_1) \times \mathcal{F} (U_2) $ to which $ f $ combines $ (f_{|U_{1} } , f_{|U_{2}}) $ is a bijection?

Thank you in advance.

Answer 1

$\mathcal{F}(U)$ is empty if $U$ contains $0$ because we can compose a square root function $f$ with the phase function $\phi:U\setminus\{0\}\to S^1$ and should still get something continuous. But taking a square root divides phase by $2$, so along a loop around zero $\phi\circ f$ runs only from $0$ to $\pi$ or $\pi$ to $2\pi$, and so cannot be continuous. The second statement should be obvious if you understand what it means to pick a branch of square root. There are two choices of $f$ on each of $U_1$ and $U_2$ if the $U_i$ are connected, and we can take them independently.

Answer 2

There can be no continuous function $z\mapsto f(z)$ with $f^2(z)=z$ for all $z$ in a neighborhood $U$ of $0$.

For given $z=re^{i\phi}\ne0$ the equation $w^2=z$ has exactly two solutions $w_1$, $w_2$ by the fundamental theorem of algebra. Using polar coordinates these can be written explicitly as $\pm \sqrt{r}e^{i\phi/2}$. It is then obvious that there are exactly two continuous square roots of $z$ in the complex plane minus the negative real axis ($=:H_r$), namely $$g_\pm(re^{i\phi})=\pm \sqrt{r}e^{i\phi/2}\qquad(-\pi<\phi<\pi)\ .$$ Similarly, there are exactly two continuous square roots of $z$ in the complex plane minus the positive real axis ($=:H_l$), namely $$h_\pm(-re^{i\psi})=\pm i \sqrt{r}e^{i\psi/2}\qquad(-\pi<\psi<\pi)\ .$$ Assume now for simplicity that $f(z)$ is a continuous square root of $z$ in a full neighborhood of $0$ containing the unit circle. Then $f(i)\in\{e^{i\pi/4},-e^{i\pi/4}\}$. If $f(i)=e^{i\pi/4}$ then $f(z)\equiv g_+(z)$ in $H_r$, and $f(z)\equiv h_+(z)$ in $H_l$ (check this!).

But $-i\in H_r\cap H_l$, and $g_+(-i)=e^{-i\pi/4}\ne e^{3i\pi/4}=h_+(-i)$.

Answer 3

If you've learned complex analysis in a more elementary fashion, then this is all about branches and branch cuts and stuff.

If you pick a branch of the square root at one point in an open set and insist on continuity, then you've chosen a branch everywhere connected to that point.

So if you have disconnected regions, you can pick a branch independently on each one.

And if a region encloses $0$, then you can't pick a branch at all, because looping around the origin would switch to the opposite branch. i.e. a square root function has to have a branch cut, and if your region encloses $0$, the branch cut has to pass through it.