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Thread: Conformal Mapping

  1. #1
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    Conformal Mapping

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

    Not really important but if anyone could help me out, I would appreciate it, thanks!
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  2. #2
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    Quote Originally Posted by shadow_2145 View Post
    (1) Let $\displaystyle f$ be a one-to-one analytic function on a domain $\displaystyle D$ and suppose that there is an analytic function $\displaystyle h$ on $\displaystyle D$ with $\displaystyle h^2 = f$. Show that $\displaystyle h$ 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 $\displaystyle x,y\in D$ are such that $\displaystyle h(x)=h(y)$ then, squaring both sides, $\displaystyle h(x)^2=h(y)^2$, i.e. $\displaystyle f(x)=f(y)$, and hence $\displaystyle x=y$ because $\displaystyle f$ is one-to-one. This shows that $\displaystyle h$ is one-to-one as well.

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