# Complex Analysis - Proving an analytic function f(z) is constant

1. Mar 18, 2012

### Daized

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

Let f(z) be an analytic function in the complex plane ℂ, and let $\phi$ be amonotonic function of a real variable.

Assume that U(x,y) = $\phi$(V(x,y)) where U(x,y) is the real part of f(z) and V(x,y) is the imaginary part of f(z). Prove that f is constant.

2. Relevant equations

The analytic function f(z) is constant if f'(z)= 0 everywhere.

The Cauchy Riemann equation...

∂u/∂x=∂v/∂y,∂v/∂x=−∂u/∂y

3. The attempt at a solution

I'm honestly a bit lost on where to start. I know that if f(z) is analytic then it is differentiable, so I thought that using the Cauchy Riemann equations for the partial derivatives might be helpful where U$_{x}$=V$_{y}$ and U$_{y}$=-V$_{x}$, but I don't know how to work with these when the function $\phi$ is there.

Just any information on what theory is best to look at would be helpful. Thank you.

Last edited: Mar 18, 2012
2. Mar 18, 2012

### sunjin09

simply substitute into the C-R equation and make use of the chain rule of taking derivatives. Also note that phi is monotonic (amonotonic=a monotonic, as I assume), so that phi'≠0

3. Mar 19, 2012

### Daized

Actually managed to figure it out just about

Assume that f(z) = u(x,y) + i v(x,y).

We know that u(x, y) = ϕ(v(x, y)).

By Cauchy-Riemann, we have
∂u/∂x = ∂v/∂y ==> ϕ'(v(x,y)) * ∂v/∂x = ∂v/∂y, and
∂u/∂y = -∂v/∂x ==> ϕ'(v(x,y)) * ∂v/∂y = -∂v/∂x.

Substituting the first equation into the second equation yields
ϕ'(v(x,y)) * [ϕ'(v(x,y)) * ∂v/∂x] = -∂v/∂x
==> [(ϕ'(v(x,y)))^2 + 1] ∂v/∂x = 0.
==> ∂v/∂x = 0, since (ϕ'(v(x,y)))^2 + 1 is nonzero.

Substituting this into the first equation yields ∂v/∂y = 0.

Since ∂v/∂x = ∂v/∂y = 0, we conclude by Cauchy-Riemann that ∂u/∂x = ∂u/∂y = 0 as well.
Hence, both u and v are constant.
Therefore, f = u + iv is also constant.

Thanks though.

4. Mar 19, 2012

### sunjin09

good work, better than what I did.