Complex polynomial properties when bounded (Liouville theorem)

[Silversonic]

Homework Statement

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

[itex] f(z) = a_1z + a_2 z^2 + a_3 z^3 + ... a_m z^m [/itex]

Homework Equations

The Attempt at a Solution

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

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

[itex] f(z) = a_0 + a_1z + a_2 z^2 + a_3 z^3 + ...[/itex]

[itex] |f(0)| = |a_0| \leq C|0| = 0 [/itex]

So [itex] a_0 = 0 [/itex]

Then take [itex] g(z) = f(z)/z [/itex]

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

[itex] |f(z)| \leq C|z|^m [/itex]

so

[itex] |f(0)| \leq C|0|^m = 0 [/itex]

So [itex] a_0 = 0 [/itex]

Again take

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

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

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

Meaning [itex] a_1 = 0 [/itex]

Keep on applying this shows that [itex] a_1 = a_2 = ... = a_{m-1} = 0 [/itex], 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
[itex] |f(z)| \geq |a||z|-|b||z|^2[/itex] for small values of z. If |z| is super small (smaller than |a|/(2|b|))
[itex] |f(z)| \geq |a||z|/2 [/itex] and no matter what our constant C is, we can pick |z| small enough so that
[tex] |a||z|/2 > C|z|^2[/tex]
for example let [itex]|z| < |a|/(2C)[/itex]