Complex polynomial properties when bounded (Liouville theorem)

[Silversonic]

Homework Statement

Suppose f is differentiable in \mathbb{C} and |f(z)| \leq C|z|^m for some m \geq 1, C > 0 and all z \in \mathbb{C}, show that;

f(z) = a_1z + a_2 z^2 + a_3 z^3 + ... a_m z^m

Homework Equations

The Attempt at a Solution

I can't seem to show this. It does the proof when m = 1 because then;

If f(z) is differentiable, it's analytic (possible to expand into taylor series), so;

f(z) = a_0 + a_1z + a_2 z^2 + a_3 z^3 + ...

|f(0)| = |a_0| \leq C|0| = 0

So a_0 = 0

Then take g(z) = f(z)/z

|g(z)| \leq C, by Liouville's theorem this means g(z) is a constant, i.e. g(z) = a_1 so f(z) = g(z)z = a_1 zI can't seem to prove this for m > 1 though, because this means

|f(z)| \leq C|z|^m

so

|f(0)| \leq C|0|^m = 0

So a_0 = 0

Again take

g(z) = f(z)/z = a_1 + a_2 z + a_3 z^2 + ...

Then |g(z)| = |f(z)|/|z| \leq C|z|^{m-1}

So |g(0)| \leq C|0|^{m-1} = 0

Meaning a_1 = 0

Keep on applying this shows that a_1 = a_2 = ... = a_{m-1} = 0, which pretty much disproves exactly what I'm trying to prove. Any help?

 

[Office_Shredder]

I think the problem is wrong (well the statement is technically true, but Liouville gives a much stronger result). It should be that f is some constant times z^m which is what you got. For example in the m=2 case we have

f(z) = az+bz²
|f(z)| \geq |a||z|-|b||z|^2 for small values of z. If |z| is super small (smaller than |a|/(2|b|))
|f(z)| \geq |a||z|/2 and no matter what our constant C is, we can pick |z| small enough so that
|a||z|/2 > C|z|^2
for example let |z| < |a|/(2C)