1. Apr 24, 2007

### laura_a

1. The problem statement, all variables and given/known data

I have this question in my text that i'm trying to understand, I have trouble with this style of questions (more proofs than actual working with real functions)

It says

Let f be an entire function such that |f(z)| <= A|z| for all z, where A is a fixed positive number. Show that f(z)=a_1 * z where a_1 is a complex constant

2. Relevant equations

The suggestion is to use Cauchy's Inequality to show that to 2nd deriv is zero everywhere in the plane. Note also that M_R in Cauchy;s Ineq. is <= A(|z_0| + R)

Now my understanding of Cauchy's Inequality is limited but here is what I have as the formula

|f^(n) (z_0) <= (n! * M_R) / R^n (n = 1,2,3...)

3. The attempt at a solution

Now from the last section in the text I was working with Cauchy Integral Formula (I haven't gotten to residues yet) so I know what all the terms mean, but because the question is a bit airy fairy I'm not sure what the f(z) is or how to plug the given info into the equation. Any hints would be gratefully accepted :)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 24, 2007

### mjsd

hint 1: $$\frac{|f(z)|}{|z|}\leq A$$ (now look at the statement for Cauchy's estimate again.)

hint 2: eventually you would want to take your circle to be as big as possible to cover the entire complex plane... what limit would that translate to?

remark: you can extend this method to proof that if $$|f(z)|\leq A |z^n|, n\in \mathbb{Z}^+$$ then f(z) is a polynomial of degree at most n. (try it!)