show $\frac{z}{(1+z)^2}$ is injective in the unit disk

Question

So I need to show that $\frac{z}{(1+z)^2}$ is injective in the unit disk, $\mathbb{D}$.

Here is what I have thought so far:

If $\frac{z}{(1+z)^2}$ is injective, then $\frac{z}{(1+z)^2} - c$ always have 1 solution in the unit disk for any $c \in \mathbb{C}$.

So I need to find the root of $\frac{z}{(1+z)^2} - c = 0$ or $\frac{z-c(1+z)^2}{(1+z)^2} = 0$ or $c(1+z)^2 -z = cz^2-(2c-1)z +c= 0$

I tried using Rouche's theorem, but all pairs of $f$ and $g$ that I tried failed the inequality $|f(z) - g(z)| < |f(z)|$ with $|z| = 1$.

Can anyone give me a hint on how to approach this question? Thanks.

Answer 1

$$f(z)=\frac{z}{(1+z)^2}$$ is injective on the open unit disk but it is not injective on the closed unit disk.

Note that $$f(i)=f(-i) = \frac {1}{2}$$

To show $f$ is injective on the open unit disk, assume $$ \frac{z}{(1+z)^2}=\frac{w}{(1+w)^2}$$

Cross multiply and factor to get $$(z-w)(1-zw)=0$$

Thus inside the unit disk, we get $z=w$ becaluse $zw\ne 1$

Answer 2

You said

If $\frac{z}{(1+z)^2}$ is injective, then $\frac{z}{(1+z)^2} - c$ always have 1 solution in the unit disk for any $c \in \mathbb{C}$.

which is not correct (or at least, misleading). For injectivity you have to show that $f(z) = c$ has at most one solution, or equivalently, that $f(z) = f(w)$ implies $z=w$.

If $f(z) = c$ has at least one solution for all $c \in \Bbb C$ then it would be a surjective onto $\Bbb C$ (which it isn't, as shown below).

Here is a possible approach which demonstrates the injectivity and also determines the image of the unit disk under $f$:

$$ f(z) = \frac{z}{(1+z)^2} = \frac 14 - \frac 14\left( \frac{z-1}{z+1}\right)^2 $$ is the composition of three injective mappings:

• $f_1(z) = \frac{z-1}{z+1}$ is a Möbius transformation and maps the unit disk onto the left half-plane.
• $f_2(z) = z^2$ is injective in the left half-plane and maps it onto $\Bbb C \setminus (- \infty, 0] $, i.e. the entire complex plane without the negative real axis.
• $f_3(z) = \frac{1-z}{4}$.

It follows that $f = f_3 \circ f_2 \circ f_1$ is a conformal (and in particular, injective) mapping from the unit disk onto the “slit domain” $\Bbb C \setminus [\frac 14, \infty)$.

Your function $f$ is also known as “rotated Koebe function” (with $\alpha = -1$).