# Complex analysis prof

1. Sep 8, 2009

### squaremeplz

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

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

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

2. Relevant equations

3. The attempt at a solution

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

2. Sep 8, 2009

### gabbagabbahey

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

3. Sep 8, 2009

### squaremeplz

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

4. Sep 8, 2009

### 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 $v(x,y)$.

5. Sep 8, 2009

### 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

$$|f^n(z_o)| <= n! \frac {M}{R^n}$$

"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

$$|f^n(z)| <= n! \frac {\sqrt{|z|}}{R^n}$$

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

Last edited: Sep 8, 2009