Problem 1
Show that if \(f\) is analytic and non-constant on a compact domain, \(\operatorname{Re} f\) and \(\operatorname{Im} f\) assume their maxima and minima on the boundary.
Problem 3
a. Suppose \(f\) is nonconstant and analytic on \(S\) and \(f(S)=T\). Show that if \(f(z)\) is a boundary point of \(T, z\) is a boundary point of \(S\). b. Let \(f(z)=z^{2}\) on the set \(S\) which is the union of the semi-discs \(S_{1}=\\{z:|z| \leq 2 ; \operatorname{Re} z \leq 0\\}\) and \(S_{2}=\\{z:|z| \leq 1 ; \operatorname{Re} z \geq 0\\} .\) Show there are points \(z\) on the boundary of \(S\) for which \(f(z)\) is an interior point of \(f(S)\).
Problem 4
Suppose \(f\) is \(C\) -analytic in \(D(0 ; 1)\) and maps the unit circle into itself. Show then that \(f\) maps the entire disc onto itself. [Hint: Use the Maximum-Modulus Theorem to show that \(f\) maps \(D(0 ; 1)\) into itself. Then apply the previous exercise to conclude that the mapping is onto.]
Problem 5
Suppose \(f\) is entire and \(|f|=1\) on \(|z|=1\). Prove \(f(z)=C z^{n}\). [Hint: First use the maximum and minimum modulus theorem to show $$ \left.f(z)=C \prod_{i=1}^{n} \frac{z-\alpha_{i}}{1-\bar{\alpha}_{i} z} .\right] $$
Problem 6
Show that for any given rational function \(f(z)\), with poles in the unit disc, it is possible to find another rational function \(g(z)\), with no poles in the unit disc, and such that \(|f(z)|=|g(z)|\) if \(|z|=1\).
Problem 8
Suppose that \(f\) is analytic in the annulus: \(1 \leq|z| \leq 2\), that \(|f| \leq 1\) for \(|z|=1\) and that \(|f| \leq 4\) for \(|z|=2\). Prove \(|f(z)| \leq|z|^{2}\) throughout the annulus.
Problem 12
Suppose \(f\) is entire and \(|f(z)| \leq 1 /|\operatorname{Re} z|^{2}\) for all \(z\). Show that \(f \equiv 0\).
Problem 16
Suppose \(f\) is bounded and analytic in \(\operatorname{Im} z \geq 0\) and real on the real axis. Prove that \(f\) is constant.
Problem 17
Given an entire function which is real on the real axis and imaginary on the imaginary axis, prove that it is an odd function: i.e., \(f(z)=-f(-z)\).