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

[Shackleford]

Homework Statement

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

Homework Equations

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

z = x + yi

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.

 

[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.

 

[SammyS]

Homework Statement

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

Homework Equations

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

z = x + yi

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.

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

 

[Shackleford]

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

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) = u_x + iv_x

 

[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.

 

[occh]

use the cauchy-riemann conditions along with the laplace equation

 

[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.