Complex Analysis Proof Using Cauchy-Riemann?

Question

Another sample problem for an exam I'm going to take:

Use the Cauchy-Riemann equations to show that an analytic function satisfying |f(z)| = 1 for all z must be a constant. You may assume that the domain is connected.

I know this must be true by Liouville's Theorem, but I can't imagine how to show it as a consequence of the C-R equations.

Thanks!

Stephen, thanks so much. That makes perfect sense.

As for the others, I appreciate the time you took, although Stephen is correct that a conjugate function need not be analytic.

Answer 1

Charlie M's approach is correct, but flawed. F(z) doesn't have to analytic, since it is the conjugate. For example, f(z) = z, while the conjugate is not analytic. However, since |f(z)| = 1,
1 / f(z) = F(z) = u - i * v is analytic, and the rest of the proof carries through.

Steve

Answer 2

show that f-conjugate is also analytic. then get that all the C-R partials are equal to 0. then f is constant.

Answer 3

For typing let F(z) = conjugate of f(z)
Consider |f(z)I ^2 = f(z) * F(z) = 1
then if f(z) is analytic on D then so also is F(z)
Now F(z) = u -iv where f(z) = u+iv

C-R says for f(z) du / dx = dv / dy and dv / dx = - du /dy
C-R says for F(z) du / dx = - dv / dy and dv / dx = du /dy

and so du / dx = dv / dy = dv / dx = du /dy = 0

Thus f(z) is locally constant on D and constant if D is (as you say ) connected

Hope this helps