Question

Show that analytic function with constant amplitude is constant.

Answer 1

Let f(z)=u(x,y)+iv(x,y) be an analytic function. Since the argument of f is constant, this implies that |f|2=(u2+v2)= constant.

Using this, we can take partial derivatives on both sides with respect to x,y to get the following equations:

∂(u2+v2)∂x=∂(u2+v2)∂y=0

Thus, we have: u∂u∂x+v∂v∂y=u∂u∂y+v∂v∂y=0

Now, we can use the Cauchy's Riemann conditions to get the following:

u∂v∂y+v∂v∂x=−u∂v∂x+v∂v∂y=0

Using some algebraic manipulation on the equations, such as multiplying the first equation by u, second by v and adding the equations, we get:

(u2+v2)∂v∂y=0

Now, if (u2+v2)=0, then f=0 throughout the domain, since |f|= constant. Otherwise, if (u2+v2)≠0, then ∂v∂y=0, and similarly ∂v∂x=0. Now using Cauchy Riemann conditions, we get ∂u∂y=0=∂u∂x so, we have f′(z)=0, which implies that f(z) is constant. Now, since f(z) is entire, we can use the Liouvile's theorem to conclude that f is constant.