Let f(z) be analytic on the set H. Let the modulus of f(z) be constant. Does f need be constant also? Explain.
Cauchy riemann equations
Hint: Prove If f and f* are both analytic on D, then f is constant.
The Attempt at a Solution
I think f need be constant.
Let f*=conjugate operator
Let f = U+iV Then f* = U-iV
Since F is analytic we can use CR equations and we get
1) Ux = Vy
2) Uy = -Vx .
Applying CR to f* gives
3) Ux = -Vy
4) Uy = Vx
1) and 3) imply Vy = -Vy and 2) and 4) imply Vx = -Vx.
But the only function that can equal its negative is zero, and thus Vx = Vy = 0, and so V = constant.
Likewise, the same argument for Ux and Uy gives Ux = Uy = 0 and so U = a constant. And both U and V constant implies that f is constant. So we have the hint proven.
Let f = u+iv. We’re given |f| = c.