Show that if \(f\) and \(\bar{f}\) are both holomorphic on a domain \(\Omega\), then \(f\) is constant on \(\Omega\).
Solution
Strategy: show \(f'=0\).
Write \(f = u + iv\). Since \(f\) is analytic, it satisfies CR, so \[ u_x = v_y && u_y = -v_x .\]
Similarly write \(\bar f = U + iV\) where \(U = u\) and \(V = -v\). Since \(\bar f\) is analytic, it also satisfies CR , so \[ U_x = V_y && U_y = -V_x \\ \\ \implies u_x = -v_y && u_y = v_x .\]
Add the LHS of these two equations to get \(2u_x = 0 \implies u_x = 0\). Subtract the right-hand side to get \(-2v_x = 0 \implies v_x = 0\)
Since \(f\) is analytic, it is holomorphic, so \(f'\) exists and satisfies \(f' = u_x + iv_x\). But by above, this is zero.
Since \(\Omega\) is connected and \(f'=0\) on \(\Omega\), \(f\) is constant.