0

$\mathbb D$={${z \in \mathbb C / \mid z \mid < 1 }$}

$\mathbb S$ be the class of all functions $f$ on $\mathbb D$ such that $f$ is univalent (analytic + injective) and $f(0)=0$,$f'(0)=1$.

I want to show that class $\mathbb S$ is preserved under square-root transformation, i.e. $g(z)=\sqrt{f(z^2)} \in \mathbb S$.

I have the following logic but don't know how worth it is.

I know the proof that if $f \in \mathbb S$, then there exist an odd function $h \in \mathbb S$, such that $h^2(z)=f(z^2)$.

So, We have $g(z)=\sqrt{h^2(z)}$, then using any one of the branch, we have $g(z)=h(z)$, and as $h \in \mathbb S$, we have $g \in \mathbb S$. Is the argument ok?

ogirkar
  • 2,681
  • 14
  • 27
  • Well, not any branch. Better take the uniquely determined branch that makes $g'(0)=1$. – Hagen von Eitzen Oct 26 '19 at 12:56
  • @Hagen von Eitzen really sorry for late reply. We will get $g'(z)=\frac{h(z)}{\sqrt{h^2(z)}}h'(z)$. So in order to get $g'(0)=1$, we should choose that branch such that $\sqrt{h^2(z)}=h(z)$..Now will this argument make proof complete? – ogirkar Nov 04 '19 at 20:38

0 Answers0