Proving that a map with the holomorphic branch of the square root is a conformal map

Question

Let $\Omega = \mathbb{C} \setminus \{z=iy \in \mathbb{C} \mid y \in \mathbb{R}, |y| \geq 1 \} $. Also let $R : \Omega \rightarrow \mathbb{C}:z\mapsto (z^2+1)^{1/2}$ be the holomorphic branch of the square root that is real and positive for $z \in \mathbb{R}$. We're tasked to show that \begin{equation}\tag{1} z \mapsto \frac{R(z) - R(x)}{z-x} \end{equation} is a conformal map from $\Omega$ to the unit disk for every $x \in \mathbb{R}$. (A conformal map is a bijective holomorphic map)

Without using this I managed to find a conformal map from $\Omega$ to the unit disk. I got \begin{equation} z \mapsto \frac{-i(-1+\sqrt{z^2+1})}{z} \end{equation} after some calculations (which is essentially the map in $(1)$ for $x=0$ and rotated over an angle $\pi / 2$).

I fail however how to prove that $(1)$ is a conformal map. Hints all the way to full solutions are welcome!

Answer 1

Let $f(z) = \dfrac{R(z)-R(x)}{z-x} = \dfrac{\sqrt{z^2+1}-\sqrt{x^2+1}}{z-x}$. By multiplying the numerator and denominator by $\sqrt{z^2+1}+\sqrt{x^2+1}$ gives $f(z) = \dfrac{z^2-x^2}{(z-x)(\sqrt{z^2+1}+\sqrt{x^2+1})} = \dfrac{z+x}{\sqrt{z^2+1}+\sqrt{x^2+1}}$. Now, a map is conformal iff it is holomorphic and its derivative is everywhere non-zero. It is easy to see that this function is holomorphic in the way it has just been written. We compute $f'(z) = \dfrac{\sqrt{z^2+1}+\sqrt{x^2+1} - (z+x)(\frac z{\sqrt{z^2+1}})}{(\sqrt{z^2+1}+\sqrt{x^2+1})^2}$. Remember that $z^2+1\neq 0$ for $z\in \Omega$.

For this to be nonzero, we show there are no roots to the numerator. Set it equal to 0. After some algebra, we get $\sqrt{x^2+1}\sqrt{z^2+1}=xz-1$. Square both sides*, giving $(z+x)^2 = 0$ after some algebra. Thus the derivative is (supposedly) zero when $z=-x$. However, you can check that this is not true. *This erroneous solution appeared when we squared both sides. Thus, the derivative is everywhere non-zero.

Answer 2

Let $f_x(z)=\frac{R(z) - R(x)}{z-x}$

I will give a sketch of a proof assuming we know as in the OP that $f_0(z)$ is a conformal map onto the unit disc by showing that the map $g_x(w)=f_x \circ f_0^{-1}(w)$ is a conformal automorphism of the unit disc or equivalently that it is a Blaschke product with precisely one zero.

First some preliminary stuff:

We have that on the principal branch of the square root that $\Re \sqrt {z} >0, z \in \mathbb C -\mathbb R_{-}$, so in particular $\Re R(z)>0$ for all $z \in \Omega$

We have consequently that $R(z)=R(-z)$ so $f_{x}(z)=-f_{-x}(-z)$ hence it is enough to prove the result for $x>0$

Also clearly $f_x$ has only one zero at $z=-x$ since $R^2(z)=z^2+1$ hence $R(z)$ is non real unless $z \in \mathbb R$ or $z=iy, |y|<1$ and for the latter we have $R(iy)<1$ while $R(x)>1, x \in \mathbb R$, so $R(z)=R(x)$ implies $z = \pm x$ and clearly for $x>0$ we have $f_x(x)=R'(x)>0$ so only $z=-x$ is a root

By the above, it follows that it is enough to show that $f_x$ maps $\Omega$ into the unit disc in such a way that $|f_x(z)| \to 1, z \to \partial \Omega$ (here we understand of course $\partial \Omega$ in the extended plane so including the point at infinity) as this immediately implies that $g_x(w)=f_x \circ f_0^{-1}(w)$ is a map from the unit disc to itself that satisfies $|g_x(w)| \to 1, |w| \to 1^-$, hence by Fatou's theorem, $g_x$ is a finite Blaschke product and has only one zero so we are done!

Now it is clear that $|f_x(z)| \to 1, |z| \to \infty$ while if $z \to \pm i$ it is also clear that $f_x(z) \to \frac{-R(x)}{\pm i-x}$ and $|\pm i-x|=\sqrt {x^2+1}=R(x)$ so we are good here too.

If now say $z \to iy, y>1$ from the right (ie with $\Re z >0$) then $\Im{z^2+1} \to 0$ through the upper half-plane, so looking at how we take the square root above the negative axis we have that $\sqrt {z^2+1} \to i\sqrt {y^2-1}$ (while of course if we go from the left $\sqrt {z^2+1} \to -i\sqrt {y^2-1}$ and reverse that for $z \to -iy, y>1$); in any case $R(z)-R(x) \to \pm i\sqrt {y^2-1}-\sqrt {x^2+1}$ and of course this means $|R(z)-R(x)| \to \sqrt {x^2+y^2}$ as does $|z-x|$, hence $|f_x(z)| \to 1$ in this case too.

So it remains to prove $|f_x(z)| <1, z \in \Omega$ and while one can give an argument that it follows from the above boundary behavior, we can also do it directly

First it is clear that since $R(-z)=R(z), x \ge 0$ and for $\Re z \ge 0$ we have $|z+x| \ge |-z+x|$ it is enough to prove that $|R(z)+R(x)| > |z+x|$ for $\Re z \ge 0$.

But now $\Re R(z) > 0, \Re z \ge 0, \Re R(x)>0$ so $R(x)+R(z) \ne 0$ and hence using that $R^2(z)-R^2(x)=z^2-x^2$ we get that $f_x(z)=\frac{z+x}{R(x)+R(z)}$ (this of course showing again that $f_x$ has a unique zero at $-x$) and then $|z^2+1| +1 > |z|^2$ (since $z \in \Omega$ we have strict inequality in the triangle inequality here showing that the proof applies for $x=0$ too), $\Re (R(z)-z)=\Re \frac{1}{ R(z)-z}=\Re (R(z)+z) > 0$,

so by squaring we get $|R(z)+R(x)|^2 =|z^2+1|+1+x^2+2R(x)\Re R(z)>|z|^2+x^2+2x\Re z=|z+x|^2$ so finally we are done with the problem assuming again that we know that $f_0$ is a conformal map from $\Omega$ onto the unit disc.

Note also that it is enough to prove that $f_0$ (or any $f_x$ for that matter) is injective for the above considerations to show that it is a conformal map from $\Omega$ onto the unit disc and since $f_0(z)=\frac{\sqrt {z^2+1}-1}{z}$ that is a direct though fairly tedious exercise in setting $f(z)=f(w)$ and simplifying, squaring, simplifying, squaring again etc and finally arriving at $w^2+z^2=2zw$ or $zw=0$ so at either $z=w$ or $z=w=0$ respectively...) hence the proof here can be made self contained.