Let \(f=u+iv\) be analytic on a connected open set \(D\subset\mathbb C\). Suppose there are real constants \(a,b,c\) with \[ a^2+b^2\ne0 \] and \[ au+bv=c \] throughout \(D\). Prove that \(f\) is constant on \(D\).
Solution
Define \[ g\coloneqq(a-ib)f. \]
1. The function \(g\) is analytic on \(D\) and has constant real part equalto \(c\).
Since \(a-ib\) is constant and \(f\) is analytic, \(g\) is analytic. Also \[ \begin{aligned} g &= (a-ib)(u+iv)\\ &= (au+bv)+i(av-bu). \end{aligned} \] By hypothesis, \[ \Re g=au+bv=c. \]
2. The function \(g\) is constant on \(D\).
Write \[ g=P+iQ. \] By step 1, \(P=c\), so \[ P_x=P_y=0. \] The Cauchy–Riemann equations give \[ Q_y=P_x=0, \qquad Q_x=-P_y=0. \] Thus \(Q\) is locally constant. Since \(D\) is connected, \(Q\) is constant on \(D\). Therefore \(g=P+iQ\) is constant on \(D\).
3. The function \(f\) is constant on \(D\).
The condition \[ a^2+b^2\neq0 \] implies \[ a-ib\neq0. \] Hence \[ f=\frac{g}{a-ib}. \] Step 2 therefore implies that \(f\) is constant.
4. Q.E.D.
Step 3 is the required conclusion.