Prove: Show $\frac{|f'(z)|}{1-|f(z)|^2}<\frac{1}{1-|z|^2}$ for $f(z)=z^n$ and any $n \in N$

Question

Prove that if $f:\mathbb{D} \to \mathbb{D}$ (where $\mathbb{D}$ is the unit disk) is given by $f(z)=z^2$, the for all $z \in \mathbb{D}$, we have

$$\frac{|f'(z)|}{1-|f(z)|^2}<\frac{1}{1-|z|^2}$$

Proof:

Let $t=|z|$, then $0 \leq t <1$. We have

$$0<(1-t)^2$$ $$0<t^2-2t+1$$ $$2t<t^2+1=\frac{(1+t^2)(1-t^2)}{1-t^2}$$

and so $$\frac{2t}{1-(t^2)^2}<\frac{1}{1-t^2}$$

Challenge

This is where I get stuck

Prove this holds for $f(z)=z^n$ and any $n \in N$.

My first instinct is to use the binomial theorem but I wasn't sure how well it applies here. . . Perhaps induction?

Extra Challange

Prove this for any non-injective analytic function $f: \mathbb{D} \to \mathbb{D}$. What happens if $f$ is injective (think mobius)?

If $f$ is non-injective then for at least $z_0,z_1 \in \mathbb{D}$ with $z_0 \neq z_1$ we have $f(z_0)=f(z_1)$. Since $f$ is analytic we have $f=u+iv$ and

$$u_x=v_y \text{ and } u_y=-v_x$$

We know the derivative isn't non-zero everywhere on $\mathbb{D}$ since our function is non-injective but I am unsure of where to take this?

Thanks for your help!

Answer 1

$\dfrac{|f'(z)|}{1-|f(z)|^2}<\dfrac{1}{1-|z|^2} \iff \dfrac{nt^{n-1}}{1-t^{2n}} < \dfrac{1}{1-t^2} \iff nt^{n-1} < 1+t^2+t^4+t^6+... t^{2(n-1)}$

it is AM-GM.