Analyzing Complex Functions with a Given Inequality - Homework Problem

[squaremeplz]

Homework Statement

suppose that f(z) is an analytic function on all of C, and suppose that, for all z in C, we have

|f(z)| <= sqrt{|z|}

Homework Equations

The Attempt at a Solution

I'm unsure of how to start the proof. any help is greatly appreciated.

 

[gabbagabbahey]

What are you supposed to prove? Your problem statement only lists the premises, and not the conclusion that you are meant to deduce.

 

[squaremeplz]

sorry, we are supposed to prove that f(z) must be a constant

 

[gabbagabbahey]

There may be an easier way to do this, but I would start by writing z=x+iy and f(z)=u(x,y)+iv(x,y) and then seeing what restrictions the condition |f(z)|\leq\sqrt{|z|} places on u(x,y) and v(x,y).

 

[squaremeplz]

my prof told us to look at the liouville theorem to do this proof

i looked in my book and found this paragraph

let f(z) be analytic inside and on a circle C of radius R centered about z_0. If |f(z)| <= M

|f^n(z_o)| &lt;= n! \frac {M}{R^n}

"This innocuous looking theorem actually places rather severe restrictions on the behaviour of analytic functions. Suppose, for instance, that f(z) is analytic and bounded by some number M over the whole plane. Then the conditions of the theorem hold for any z_o and for any R. Taking n = 1 in the above equations and letting R -> infinity, we counclude that f' vanishes everywhere; i.e., f must be constant"

So, how can I apply this theorem over the whole complex plane instead of a circle centered at z_o?

could I say that sqrt(|z|) = M for all z in C


|f^n(z)| &lt;= n! \frac {\sqrt{|z|}}{R^n}

and thus have the proof done by the argument in the above paragraph?