Proving Cauchy's Inequality for Analytic Functions with Distance Constraints"

[WannaBe22]

Homework Statement

Let $$f(z)=\sum_{n=0}^{\infty} a_n z^n$$ be analytic at {z: |z|<R} and satisfies:
$$|f(z)| \leq M$$ for every |z|<R.
Let's define: d=the distance between the origin and the closest zero of f(z).

Prove: $$d \geq \frac{R|a_0|}{M+|a_0|}$$.

Hope you'll be able to help me


Thanks !

Homework Equations

The Attempt at a Solution

I've tried using Cauchy's Inequality... But it doesn't give anything new for $$a_0$$.
I've also tried isolating $$a_0$$ from this inequality, but it gives me nothing...


Hope someone will be able to help me

Thanks in advance

 

[mighty2000]

Thank you for reaching out for help with your problem. I am happy to assist you in finding a solution.

First, let's consider the Cauchy's inequality you mentioned. It states that for a complex function f(z) analytic in a disk of radius R centered at z_0, we have:

|f^{(n)}(z_0)| \leq \frac{n!M}{R^n}

where M is the maximum of |f(z)| in the disk. Applying this to our function f(z), we have:

|a_0| \leq \frac{0!M}{R^0} = M

This doesn't seem to give us any new information, as you mentioned. However, let's take a closer look at the inequality:

|f(z)| \leq M

This means that the maximum of |f(z)| in the disk is M. This also implies that there is a point z_1 in the disk such that |f(z_1)| = M. Now, let's consider the distance between the origin and this point z_1. By the triangle inequality, we have:

|z_1| \geq |z_1 - 0| = |z_1| - |0| = |z_1|

But we know that |f(z_1)| = M, so we can rewrite this as:

|z_1| \geq M

Now, let's go back to our definition of d, the distance between the origin and the closest zero of f(z):

d = the distance between the origin and the closest zero of f(z)

We can rewrite this as:

d = |z_1|

Therefore, we have:

d \geq M

Now, let's go back to our original inequality:

|f(z)| \leq M

We can rewrite this as:

|a_0 + a_1z + a_2z^2 + ...| \leq M

Using the triangle inequality, we have:

|a_0| + |a_1z| + |a_2z^2| + ... \leq M

Since we know that |a_0| \leq M, we can rewrite this as:

|a_1z| + |a_2z^2| + ... \leq 0

But since z is in the