Can Analytic Functions in the Unit Disk Meet Specific Modulus Conditions?

[Markov2]

Let $f:B(0,2)\to B(0,2)$ be an analytic function so that $f(1)=0.$ Prove that for all $z\in B(0,2)$ is $\left| \dfrac{f(z)}{z} \right|\le \left| \dfrac{2(z-1)}{4-z} \right|.$

What's the trick here? I don't see it.

 

[AlexYoucis]

Let $f:B(0,2)\to B(0,2)$ be an analytic function so that $f(1)=0.$ Prove that for all $z\in B(0,2)$ is $\left| \dfrac{f(z)}{z} \right|\le \left| \dfrac{2(z-1)}{4-z} \right|.$

What's the trick here? I don't see it.

You have to have some starting ideas, even failed ones. Give us some of them.

 

[Markov2]

I don't know how to use the data $f(1)=0,$ I thought on defining a function, but, I don't get a thing.

 

[Markov2]

I can't solve it yet, how to start?

 

[Markov2]

Does anyone have any idea? I think a function needs to be defined by using the initial condition, but I can't think of it.

 

[Markov2]

Perhaps on taking $g(z)=(z-1)f(z)$ or $g(z)=zf(z),$ but I still can't get it.

 

[Jose27]

Let $f:B(0,2)\to B(0,2)$ be an analytic function so that $f(1)=0.$ Prove that for all $z\in B(0,2)$ is $\left| \dfrac{f(z)}{z} \right|\le \left| \dfrac{2(z-1)}{4-z} \right|.$

What's the trick here? I don't see it.

This is false as written: Take \(f(z)=a(z-1)\) for \(a\in \mathbb{R}\) small enough. This is because your inequality implies that \(f(0)=0\) which need not happen.