# Modulus and complex variable

#### Markov

##### Member
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

##### New member
インテグラルキラー;490 said:
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.

#### Markov

##### Member
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.

#### Markov

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

#### Markov

##### Member
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.

#### Markov

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

#### Jose27

##### New member
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.