Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 2 / 13A (Further Analysis)

The Liouville theorem states that every bounded entire function is constant. Let $|f(z)|\le M$ and choose distinct $a,b$. For $R>\max (|a|,|b|)$, partial fractions and the Cauchy integral formula givewith the circle traversed anticlockwise. On that circle,The displayed difference is independent of $R$, so $f(a)=f(b)$. Since $a,b$ were arbitrary, $f$ is constant. This proves the theorem with the requested two-pole contour rather than assuming it.

Set $F(z)=\exp [-(1-i)g(z)]$. It is entire, and $\mathrm{Re}[(1-i)g]=u+v$ makes $|F|=e^{-(u+v)}\le 1$. By the just-proved Liouville theorem, $F$ is constant. Differentiation gives $0=F^{\prime }=-(1-i)g^{\prime }F$. The exponential never vanishes, so $g^{\prime }=0$ everywhere and $g$ is constant. This is the entire function confined to a half-plane is constant principle.

Set $F(z)=\exp (ig(z{)}^2)$. Since $\mathrm{Im}(g^2)=2uv$, its modulus is $e^{-2uv}\le 1$. The Liouville theorem again makes $F$ constant, and $F^{\prime }=2ig{g}^{\prime }F=0$ implies $(g^2{)}^{\prime }=0$. Hence $g^2$ is a constant $c$. If $c=0$, then $g=0$. If $c\ne 0$, then $g$ never vanishes and $g{g}^{\prime }=0$ implies $g^{\prime }=0$. In either case $g$ is constant. This is also an application of the one-sided product bound for an entire function after rotating $g$ by $i$.