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    Holomorphic $f$ with constant $\Re f$, $\Im f$, or $\abs{f}$ is constant

    E-SS1.EX-13

    Area
    Complex Analysis
    Topics
    Cauchy-Riemann, Holomorphic Functions, Open Mapping Theorem
    Status
    draft

    If \(f\) is holomorphic on \(\Omega\) and any of the following hold, then \(f\) is constant:

    1. \(\Re(f)\) is constant.

    2. \(\Im(f)\) is constant.

    3. \(\abs{f}\) is constant.

    Solution

    Here \(\Omega\) is a region, that is, open and connected. Write \(f=u+iv\) with \(u,v\) real.

    1. (1) If \(u\) is constant on \(\Omega\), then \(f\) is constant.

    At each point \(f'=\partial f/\partial x=u_x+iv_x\), and the Cauchy–Riemann equation \(v_x=-u_y\) gives \(f'=u_x-iu_y=0\). So \(f'=0\) on the connected set \(\Omega\), and \(f\) is constant.

    2. (2) If \(v\) is constant on \(\Omega\), then \(f\) is constant.

    The function \(-if\) is holomorphic with real part \(v\), so step 1 makes \(-if\), and hence \(f\), constant.

    3. (3) If \(\abs{f} = c\) on \(\Omega\), then \(f\) is constant.

    3.1. If \(c=0\), then \(f=0\).

    \(\abs{f(z)} = 0\) if and only if \(f(z) = 0\).

    3.2. If \(c>0\), then \(\bar f\) is holomorphic on \(\Omega\).

    Since \(\abs{f}=c>0\), \(f\) has no zeros on \(\Omega\), and \(f\bar{f} = \abs{f}^2 = c^2\) gives \(\bar{f}=c^2/f\), a quotient of holomorphic functions with nonvanishing denominator.

    3.3. Q.E.D.

    In the case \(c>0\), both \(f\) and \(\bar f\) are holomorphic, so \(\partial f/\partial\bar z=0\) and \(\overline{f'}=\overline{\partial f/\partial z}=\partial\bar f/\partial\bar z=0\). Hence \(f'=0\) on the connected set \(\Omega\), and \(f\) is constant. The case \(c=0\) is step 1.

    4. Q.E.D.

    Steps 1, 2 and 3 are the three cases.