# For what complex values of $\alpha$ is $f_\alpha$ one to one?

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 $$\frac{z}{1+\alpha z^2}=\frac{w}{1+\alpha w^2}\iff \alpha(zw^2-z^2w)=(w-z)\iff \alpha wz=1.$$

• The image of the unit disk is not defined for all $\alpha$, e.g., for $\alpha=i$ is $f(1)=\infty$. – Dietrich Burde Oct 14 '18 at 15:03
• Isn't $$\frac{z}{1+\alpha z^\color{red}{2}}$$ – Nosrati Oct 14 '18 at 15:19
• yes it is, sorry. – UserA Oct 14 '18 at 15:37

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 $$|\alpha|<\sqrt[n]{2n+1}\to1$$.
• $|\alpha| = 1$ 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. – Martin R Oct 14 '18 at 15:27
• @MartinR If we want to specify the range of $\alpha$, so we should assume that $f$ is injective. – Nosrati Oct 14 '18 at 15:35
• 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. – Martin R Oct 14 '18 at 15:48
• 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. – Martin R Oct 14 '18 at 15:52