5.14
5.14 Suppose \(f\) is entire and there exists \(M >0\) such that \(\mid f(z) \mid \geq M\) for all \(x \in \mathbb{C}\). Prove that \(f\) is constant.
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Suppose \(f\) is entire with bounded real portion, writing \(f(z)=u(z)+iv(z)\), there exists \(M > 0\) such that \(\mid u(z) \mid \leq M\) for all \(z \in \mathbb{C}\). Now consider \(exp\) \(f(z)=exp(u(z)+iv(z))=e^{u(z)+uv(z)}\). Since \(\mid u(z) \mid \leq M\), we know that \(\mid e^{u(z)} \mid \leq e^M\) for all \(z \in \mathbb{C}\) so \(\mid e^{u(z)} \mid\) is bounded. Note that \(\mid exp\) \(f(z) \mid = \mid e^{u(z)} \mid\) and since both \(exp\) and \(f(z)\) are entire, then \(exp\) \(f(z)\) is entire. Then \(exp\) \(f(z)\) is a bounded entire function per Corollary 5.14, \(exp\) \(f(z)\) is constant. Let \(exp\) \(f(z)=x+iy\) for some \(x,y \in \mathbb{R}\). Then we have \(\sqrt{x^2+y^2}=\mid exp\) \(f(z) \mid =\mid e^{u(z)} \mid = e^{u(z)} \mid\) and thus \(u(z)=ln\sqrt{x^2+y^2}\). So \(u(z)\) is constant. Then by Theorem 2.2, we have \(0=f'(z)=\frac{\partial f}{\partial x}(z)\). So \(f'(z)=u'(x)=0\) for all \(z \in \mathbb{C}\). Therefore \(f\) is constant by Theorem 2.20.