Let \(f(z)\) be analytic in a domain, and prove that \(f\) is constant if it satisfies any of the following conditions:
- \(\abs{f(z)}\) is constant.
- \(\Re(f(z))\) is constant.
- \(\arg(f(z))\) is constant.
- \(\bar{f(z)}\) is analytic.
How do you generalize (a) and (b)?
Solution: 1
Open mapping proof: use that no curve \(\gamma \subseteq \CC\) is open in \(\CC\).
If \(\abs{f} = c = r^2\) for some \(r\), then the image of \(f\) is contained in the curve \(\bd \DD_r(0)\). Since \(f\) is holomorphic on the source domain \(\Omega\), \(f\) is an open map, so if \(f\) is nonconstant the \(f(\Omega)\) is open. But \(f(\Omega) \subseteq \bd \DD_r(0)\) can not be open, so \(f\) must be constant.
The usual more direct proof: write \(\abs{f(z)} = u^2 = v^2 = r^2\). The claim is that both \(u\) and \(v\) are constant. Take partial derivatives and clear the factor of 2: \[ \del_x: \quad uu_x + vv_x &= 0\\ \del_y: \quad uu_y + vv_y &= 0 .\] Now apply CR: \(u_x= v_y, u_y=-v_x\), then \[ uu_x - vu_y &=0 \\ uu_y + vu_x &=0 .\] Multiply the first by \(u_x\) and the second by \(u_y\), then add \[ uu_x^2 - vu_y u_x &= 0 \\ uu_y^2 + vu_x u_y &=0 \\ \implies u(u_x^2 + u_y^2) &=0 .\] A similar calculation yields \(v(v_x^2 + v_y^2) = 0\), so If \(u(x,y) = v(x, y) = 0\) at any point, then \(\abs{f} = 0\) and \(f\equiv 0\), so we’re done. Otherwise, \(u,v\) do not simultaneously vanish, so we must have \[ 0 = u_x^2 + u_y^2 &\implies 0 = u_x = u_y \implies u \text{ constant }\\ 0 = v_x^2 + v_y^2 &\implies 0 = v_x = v_y \implies v \text{ constant } ,\] so \(f=u+iv\) is constant.
Solution: 2
Write \(f=u+iv\), so \(u\equiv c\) is constant. Then \(u_x = u_y = 0\), and CR yields \(v_y = u_x = 0\) and \(v_y = -u_x = 0\), so \(v\) is constant, making \(f\) constant.
Solution: 3
Slick proof: apply the open mapping theorem again, since \(\Arg(f) = \theta_0\) implies that \(\im(f) \subseteq \gamma\) for the curve \(\gamma \da \ts{t e^{i\theta_0}\st t\in \RR}\) which has no open subsets.
Note that this implies that any \(\RR\dash\)valued holomorphic function is constant.
Solution: 4
Write \(f=u+iv\) so \(\bar f = u +i\tilde v\) where \(\tilde v \da -v\). Then \(u, \tilde v\) are constant, so in particular \(\Re(f)\) is constant and by 2 \(f\) is constant.