Branches of $\sqrt{z}$ when removing the negative real axis

Question

In Complex Analysis by Ahlfors, he defines the complex square root $w(z)=\sqrt{z}$ on the complement of the negative axis $\Re(z)\leqslant 0$ to be the square root with positive real part. He then shows that this is continuous as follows: $$|z-a|=|w^2(z)-w^2(a)|=|w(z)-w(a)||w(z)+w(a)| \geqslant |w(z)-w(a)|\cdot\Re(w(a))$$ which implies that given $\epsilon>0$ we can see that letting $\delta=\epsilon \cdot \Re(w(a))$ we have $$|z-a|<\delta \implies |w(z)-w(a)|\leqslant\frac{|z-a|}{\Re(w(a))}<\epsilon.$$ I can see that he used the fact that we cut out the negative real axis when dividing by $\Re(w(a))$ as the square roots of negative reals are purely complex. But, why did he decide the square root to be the one with positive real value? I believe that the same computation shows that if we define \begin{cases} 0\leqslant \text{Arg}(w)\leqslant \pi/2 & \text{if } 0\leqslant \theta \leqslant \pi/2\\ \pi\leqslant \text{Arg}(w)\leqslant 3\pi/2 & \text{if } \pi/2\leqslant \theta<\pi\\ \pi/2\leqslant \text{Arg}(w)\leqslant\pi & \text{if } \pi<\theta\leqslant 3\pi/2\\ 3\pi/2\leqslant \text{Arg}(w)\leqslant 2\pi & \text{if } 3\pi/2\leqslant\theta\leqslant2\pi \end{cases} then we can replace $\Re(w(a))$ with $|\Re(w(a))|$ and then $w$ is continuous. But we would have 16 choices for a continuous square root on $\mathbb{C}-\mathbb{R}^{\leqslant0}$ which seems untrue. Is there an error in my understanding here?

Answer 1

Let $D$ denote the complex plane with the non-positive real axis removed. There are precisely two branches of square root on $D$, each the negative of the other. (A branch of square root in $D$ is a holomorphic function $f$ in $D$ satisfying $f(z)^{2} = z$ for all $z$ in $D$.) The diagram shows the real parts of the two branches, whose graphs are mirror images across the $z$-plane:

One branch, the so-called principal branch (gray), is positive on the positive reals, and extends the real square root (yellow) to $D$. The other branch (green mesh) is negative on the positive reals. That's about as close as one can get to answering "Why did [Ahlfors] decide the square root to be the one with positive real value?"

As for what appears to be wrong with the "sixteen branches" argument, in the "piecewise" formulas of the original question there are discontinuities in $\operatorname{Arg}(w)$ on the ray where $\theta = \pi/2$ (the stated value jumps from $\pi/2$ to $\pi$), and similarly discontinuities where $\theta = 3\pi/2$ and where $\theta = 0$ (a.k.a., $2\pi$). If we could independently select branches on each quarter plane, there would be $2^{4} = 16$ branches of square root. Continuity, however, reduces our freedom to just two choices.