Question

For an entire function \(f\), which one of the following statements is false?

• A. \(f\) is constant, if the range of \(f\) is contained in a straight line.
• B. \(f\) is constant, if \(f\) has uncountably many zeros.
• C. \(f\) is constant, if \(f\) is bounded on \(\{z\in\mathbb{C}:\operatorname{Re}(z)\le0\}\).
• D. \(f\) is constant, if the real part of \(f\) is bounded.

Hint:

Try \(f(z)=e^z\) as a candidate that is bounded on the left half-plane but not constant.

Answer 1

The Correct Option is C

Step 1: Check option (A). If the range of entire \(f\) lies on a line \(\{a+tb:t\in\mathbb{R}\}\), then \(g(z)=\dfrac{f(z)-a}{b}\) is entire and real-valued everywhere. For an analytic function \(u+iv\) with \(v\equiv0\), the Cauchy-Riemann equations force \(u_x=u_y=0\), so \(u\) (and hence \(f\)) is constant. This statement is true.

Step 2: Check option (B). The zero set of a non-identically-zero entire function consists of isolated points, hence is countable. So if \(f\) has uncountably many zeros, \(f\) must be identically zero, which is (trivially) constant. This statement is true.

Step 3: Check option (C). Consider \(f(z)=e^{z}\), which is entire. For \(\operatorname{Re}(z)\le0\), \(|f(z)|=e^{\operatorname{Re}(z)}\le e^0=1\), so \(f\) is bounded on this half-plane, yet \(f\) is clearly non-constant. This is a direct counterexample, so statement (C) is false.

Step 4: Check option (D). If \(\operatorname{Re}(f)\) is bounded, say \(\operatorname{Re}(f(z))\le M\) for all \(z\), then \(h(z)=e^{f(z)-M}\) is entire with \(|h(z)|=e^{\operatorname{Re}(f(z))-M}\le1\), so \(h\) is a bounded entire function. By Liouville's theorem \(h\) is constant, and since \(h'(z)=f'(z)h(z)\) with \(h\) never zero, \(f'\equiv0\), so \(f\) is constant. This statement is true.

Step 5: Only option (C) is false.

\[\boxed{\text{Option (C) is false}}\]