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Modulus and complex variable

Markov

Member
Feb 1, 2012
149
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
Jan 26, 2012
19
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
Feb 1, 2012
149
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
Feb 1, 2012
149
I can't solve it yet, how to start?
 

Markov

Member
Feb 1, 2012
149
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
Feb 1, 2012
149
Perhaps on taking $g(z)=(z-1)f(z)$ or $g(z)=zf(z),$ but I still can't get it.
 

Jose27

New member
Feb 1, 2012
15
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.