Final answer:
Show that an analytic function f on a connected open set that satisfies Re(f(z)) = (Im(f(z)))^2 is constant, we apply the Cauchy-Riemann equations to prove that the derivatives of the real and imaginary parts of f reduce to zero.
Step-by-step explanation:
The problem asks us to show that if f:Ω → C is analytic and Re(f(z)) = (Im(f(z)))^2 for every z in Ω, where Ω is a nonempty open connected subset of C, then f must be constant on Ω. We approach this by considering the Cauchy-Riemann equations, which are necessary conditions for a function to be analytic. Let f(z) = u(x, y) + i*v(x, y), where u and v are the real and imaginary parts of f, respectively, and z = x + iy.
Given that u = v^2, we can use the Cauchy-Riemann equations u_x = v_y and u_y = -v_x to show that the partial derivatives of u and v must be zero, leading to the conclusion that f is constant. As an example, we differentiate u = v^2 with respect to x to get u_x = 2v*v_x and substitute the Cauchy-Riemann conditions to find that 2v*v_y = -2v*v_x, implying that v (and thus u) is constant on Ω.