# If f(z) holomorphic show that f(z) is constant

1. Mar 7, 2015

### Shackleford

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

If f(z) = u + iv is a holomorphic function on a domain D satisfying |f(z)| = constant, show that f(z) is constant.

2. Relevant equations

f(z) = u(z) + y(z)i = u(x,y) + y(x,y)i

z = x + yi

3. The attempt at a solution

|f(z)| = √u2+v2) = √u(x,y)2+v(x,y)2)

I know this is a simple problem, but I'm just not seeing the connection. I'm probably missing some obvious relation/equation to consider.

2. Mar 7, 2015

### WWGD

Maybe you can just quote Liouville? Maybe extending into an entire function with
constant nodulus.

If not, use Maximum Modulus http://en.wikipedia.org/wiki/Maximum_modulus_principle

Maybe let $c=u^2+v^2$ , then $0= 2uu_x+2vv_x$ and $0 =2uu_y+ 2vv_y$ and use Cauchy -Riemann.

Last edited: Mar 7, 2015
3. Mar 8, 2015

### SammyS

Staff Emeritus
It seems to me that you might want to start out by using the definition of holomorphic .

You also have a couple of typos in your post.

I assume you mean: f(z) = u(z) + v(z)i = u(x,y) + v(x,y)i

4. Mar 8, 2015

### Shackleford

Shoot. Yes. Sorry.

If f is complex differentiable at any point a ∈ D, f is called a holomorphic function on D.

Then of course f is holomorphic ⇔ Cauchy-Riemann Equations hold.

If f is a holomorphic function, then f'(z) = ux + ivx

Last edited: Mar 8, 2015
5. Mar 8, 2015

### Shackleford

WWGD, I had that idea last night. I haven't been able to divine a different "proof," so that's likely what I'll turn in.

6. Mar 8, 2015

### occh

use the cauchy-riemann conditions along with the laplace equation

7. Mar 9, 2015

### WWGD

Or, like occh suggested, use the fact that a harmonic function cannot be bounded. Then, if |f| is bounded, both the Real, Complex part must be bounded.

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