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

In summary, the problem is to prove that for any analytic function $f$ defined on the open ball $B(0,2)$ with $f(1)=0,$ the inequality $\left| \dfrac{f(z)}{z} \right|\le \left| \dfrac{2(z-1)}{4-z} \right|$ holds for all $z\in B(0,2).$ However, this is false as shown by the counterexample $f(z)=a(z-1)$ for small enough $a\in \mathbb{R}.$
  • #1
Markov2
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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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
I can't solve it yet, how to start?
 
  • #5
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.
 
  • #6
Perhaps on taking $g(z)=(z-1)f(z)$ or $g(z)=zf(z),$ but I still can't get it.
 
  • #7
Markov 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.

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.
 

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

What is the modulus of a complex number?

The modulus of a complex number is its distance from the origin on the complex plane. It is calculated by taking the square root of the sum of the squares of the real and imaginary parts of the number.

What is the relationship between modulus and absolute value?

The modulus of a complex number is essentially the same as the absolute value of a real number. It represents the magnitude or size of the number without considering its direction or sign.

How is the modulus used in solving equations with complex numbers?

The modulus is used to find the magnitude of a complex number, which can then be used to solve equations involving complex numbers. It is also used in polar form to simplify calculations with complex numbers.

What is a complex variable and how is it different from a real variable?

A complex variable is a number that has both a real part and an imaginary part. It is different from a real variable, which only has one part. In mathematics, complex variables are used to solve problems that involve both real and imaginary numbers.

What are some applications of modulus and complex variables?

Modulus and complex variables are used in various fields of mathematics and science, such as engineering, physics, and economics. They are also used in signal processing, control theory, and quantum mechanics.

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