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We wish to find all complex numbers $\alpha$ such that $$f_\alpha(z)=\frac{z}{1+\alpha z^2}$$ is one to one on the unit disk. For these values of $\alpha$, find the image of the unit disk under $f_\alpha$.

One calculation shows that if $f(w)=f(z)$ and $w\neq z$ then $$ \frac{z}{1+\alpha z^2}=\frac{w}{1+\alpha w^2}\iff \alpha(zw^2-z^2w)=(w-z)\iff \alpha wz=1.$$

Is that helpful?

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  • $\begingroup$ The image of the unit disk is not defined for all $\alpha$, e.g., for $\alpha=i$ is $f(1)=\infty$. $\endgroup$ – Dietrich Burde Oct 14 '18 at 15:03
  • $\begingroup$ Isn't $$\frac{z}{1+\alpha z^\color{red}{2}}$$ $\endgroup$ – Nosrati Oct 14 '18 at 15:19
  • $\begingroup$ yes it is, sorry. $\endgroup$ – UserA Oct 14 '18 at 15:37
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Yes, since $|\alpha zw|<|\alpha|\leq1$ on the unit disk. Alternaivie way is using de Brange's theorem, $$f_\alpha(z)=\frac{z}{1+\alpha z^2}=z-\alpha z^3+\alpha^2z^5-\alpha^3z^7+\cdots$$ then $|\alpha|^n<2n+1$ for all $n$ shows $|\alpha|<\sqrt[n]{2n+1}\to1$.

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  • $\begingroup$ $|\alpha| = 1$ is possible, and that must already hold in order for $f$ to be holomorphic in the unit disk. – Bieberbach's (aka de Brange's) theorem assumes that $f$ is injective, I am not sure if that really that helps here. $\endgroup$ – Martin R Oct 14 '18 at 15:27
  • $\begingroup$ thank you....... $\endgroup$ – Nosrati Oct 14 '18 at 15:32
  • $\begingroup$ @MartinR If we want to specify the range of $\alpha$, so we should assume that $f$ is injective. $\endgroup$ – Nosrati Oct 14 '18 at 15:35
  • $\begingroup$ Bieberbach shows that if $f$ is injective then $|\alpha|<\sqrt[n]{2n+1}$ for all $n$, in the limit it follows that $|\alpha| \le 1$. But that follows already from the fact that $f$ is holomorphic in the unit disk. $\endgroup$ – Martin R Oct 14 '18 at 15:48
  • $\begingroup$ In other words: $f$ is defined only for $|\alpha| \le 1$, and for those $\alpha$ it is injective (as OP's calculation shows). No need to use Bieberbach. $\endgroup$ – Martin R Oct 14 '18 at 15:52

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