Conformal Mapping

**shadow_2145** · November 6th 2008, 03:29 PM

(1) Let $f$ be a one-to-one analytic function on a domain $D$ and suppose that there is an analytic function $h$ on $D$ with $h^2 = f$ . Show that $h$ is also one-to-one.

Not really important but if anyone could help me out, I would appreciate it, thanks!

**Laurent** · November 6th 2008, 11:20 PM

Originally Posted by shadow_2145

(1) Let $f$ be a one-to-one analytic function on a domain $D$ and suppose that there is an analytic function $h$ on $D$ with $h^2 = f$ . Show that $h$ is also one-to-one.

Not really important but if anyone could help me out, I would appreciate it, thanks!

I may be wrong, but it doesn't depend on analysticity:
If $x,y\in D$ are such that $h(x)=h(y)$ then, squaring both sides, $h(x)^2=h(y)^2$ , i.e. $f(x)=f(y)$ , and hence $x=y$ because $f$ is one-to-one. This shows that $h$ is one-to-one as well.

A related question would be to show that $h$ is locally one-to-one if $f$ is locally one-to-one. This is easy if you know that $f$ is locally one-to-one if and only if, for $x\in D$ , $f'(x)\neq 0$ . Derivating $h(x)^2=f(x)$ , we get that $2h(x)h'(x)\neq 0$ hence $h'(x)\neq 0$ and $h$ is thus locally one-to-one as well.

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