# Branch of $\sqrt{1-z^2}$

Show that a branch of $\sqrt{1-z^2}$ can be defined in any region $\Omega$ where the points $1,-1$ are in the same component of its complement.

This is a question in Ahlfors' Complex Analysis (P.148 Q5) that I came across while trying to self-study the book. I tried to tackle the problem by considering $\Omega=\mathbb{C} \backslash [-1,1]$ first, and tried the approach as in Section 4.4 Corollary 2, namely find a branch of the corresponding log first. For this $\Omega$, the image of $1-z^2$ is $\mathbb{C} \backslash [0,1]$, so a branch of $log(1-z^2)$ cannot be defined; evidently one needs to construct the branch of $\sqrt{1-z^2}$ directly. Here is where I ran out of ideas... Any help is appreciated!

• On the complement of the closed unit disk, you can take $\pm iz \sqrt{1 - \frac{1}{z^2}}$ with the principal branch of $\sqrt{1-w}$ for $\lvert w\rvert < 1$. Analytic continuation then extends it to $\mathbb{C}\setminus [-1,1]$. Dec 1, 2015 at 15:54
• Yes, that would work for the case when $\Omega=\mathbb{C} \backslash [-1,1]$... but how about for general $\Omega$? Dec 1, 2015 at 15:59

Let $\gamma$ be a closed curve in $\Omega$. Then $$\int_\gamma \bigl( \frac{1}{z-1} - \frac{1}{z+1} \bigr) \, dz = 2 \pi i \bigl( n(\gamma, 1) - n(\gamma , -1) \bigr)$$ where $n$ denotes the winding number. $1$ and $-1$ are in the same component of $\mathbb C \setminus \Omega$ and therefore also in the same component of $\mathbb C \setminus |\gamma|$. Since the winding number is constant on each component of $\mathbb C \setminus |\gamma|$, the integral is zero.

This holds for all closed curves in $\Omega$. It follows that the integrand has a holomorphic antiderivative $$F : \Omega \to \mathbb C, \quad F'(z) = \frac{1}{z-1} - \frac{1}{z+1} \, .$$ Now define $$f : \Omega \to \mathbb C, \quad f(z) = (z+1)e^{F(z)/2} \, ,$$ this is (almost) the holomorphic branch of $\sqrt{1-z^2}$.

The function $$g : \Omega \to \mathbb C, \quad g(z) = \frac{f(z)^2}{z^2 -1} = \frac{z+1}{z -1} e^{F(z)}$$ satisfies $$\frac{g'(z)}{g(z)} = \frac{1}{z+1} - \frac{1}{z-1} + F'(z) = 0$$ It follows that $g$ is constant and therefore $$f(z)^2 = C (1-z^2)$$ for some constant $C$. Then $f(z)/\sqrt{C}$ is a holomorphic branch of $\sqrt{1-z^2}$ in $\Omega$.

The idea behind those calculations was to show that $F(z) = \log \frac{z-1}{z+1}$ has a holomorphic branch in $\Omega$, then $$e^{F(z)} = \frac{z-1}{z+1} = \frac{z^2-1}{(z+1)^2}$$ which leads to the "candidate" $f(z) = (z+1)e^{F(z)/2}$.

• Very neat answer, thank you very much! Dec 2, 2015 at 4:00

A branch of $$\sqrt{1-z^2}$$ is a lift of $$f(z)=1-z^2$$ with respect to the covering map $$p:\mathbb{C}^\times\to\mathbb{C}^\times,\,p(z)=z^2$$. Fix a basepoint $$z_0\in\Omega$$, and let $$\tilde{w_0}\in \mathbb{C}^\times$$ be a point such that $$p(\tilde{w_0})=f(z_0)$$. By the theory of covering spaces, $$f$$ has a lift with respect to $$p$$ if and only if $$f_\ast(\pi_1(\Omega,z_0))\subset p_\ast(\pi _1(\mathbb{C}^\times,\tilde{w_0}))$$. Now we may identify $$\pi_1(\mathbb{C}^\times,{f(z_0)})$$ with $$\mathbb{Z}$$ via the integration $$\gamma\mapsto \frac{1}{2\pi i}\int_\gamma \frac{1}{z}dz,$$ and under this identification $$\operatorname{im}p_\ast$$ corresponds to $$2\mathbb{Z}$$. THerefore, we have $$\operatorname{im}f_\ast\subset\operatorname{im}p_\ast$$ if and only if $$\frac{1}{2\pi i} \int_{f\ast\gamma}\frac{1}{z}dz\in 2\mathbb{Z}$$ for (every piecewise $$C^1$$) curve. On the other hand, $$\frac{1}{2\pi i} \int_{f\ast\gamma}\frac{1}{z}dz=\frac{1}{2\pi i}\int_{\gamma}\frac{2z}{z^2-1}dz=\frac{1}{2\pi i}\int_\gamma\left(\frac{1}{1-z}+\frac{1}{1+z}\right)dz$$ is the sum of the winding number of $$\gamma$$ around $$1$$ and $$-1$$. By our hypothesis on $$\Omega$$, these winding numbers are equal, and we are done.

• This answer is very good... Never imagined my topology course will he usefull in such marvelous ways...this proof also helped me to biuldany insights Apr 11, 2023 at 5:46