Show an analytical function of constant absolute value is constant?

Answer 1

This is based on exercise 10 on p 165.4 of "Introduction to Complex Analysis" by Nevanlinna and Paatero, Chelsea Publishing, NY.

With w(z) == u(x,y) + i v(x,y), ( |w(z)| = constant ) ==>

1) uu + vv = const. Using notation ux == du/dx, uy == du/dy, etc., 1) ==>

2) 0 = u ux + v vx and 0 = u uy + v vy.

Using the Cauchy-Riemann equations, ux = vy and uy = -vx to eliminate derivatives of v, 2) becomes

3) 0 = u ux - v uy and 0 = u uy + v ux. Exercise 10 asserts that from this one can show that

4) 0 = (uu + vv)(ux ux + uy uy). I have not obtained that formula, but using 3) one can form the combination

5) 0 = (u ux - v uy)(u ux + v uy) + (u uy + v ux)(u uy - v ux) which simlifies to

6) 0 = (uu - vv)(ux ux + uy uy).

If w(z) is not a constant, then from the continuity of derivatives of analytic functions there must exist some domain throughout which ux or uy is non-zero. 6) indicates that in that domain either w(z) must vanish or its argument must be constant (45 or 225 degrees) so that in that domain w(z) must vanish or its modulus and argument must both be constant making w(z) constant. But if w(z) is constant in a domain, then by the uniqueness theorem (Nevanlinna and Paatero page 139) it is constant altogether.