Let \(f(z)\) be bounded and analytic in \(\mathbb C\). Let \(a \neq b\) be any fixed complex numbers. Show that the following limit exists: \[ \lim_{R \rightarrow \infty} \int_{|z|=R} \frac{f(z)}{(z-a)(z-b)} dz .\]
Use this to show that \(f(z)\) must be a constant (Liouville’s theorem).
Solution
Apply PFD and use that \(f\) is holomorphic to apply Cauchy’s formula over a curve of radius \(R\) enclosing \(a\) and \(b\): \[ {1 \over (z-a)(z-b)} = {1 \over a-b}\qty{{1 \over z-a} - {1 \over z-b}}, \] so \[ \int_\gamma {f(z) \over (z-a)(z-b)}\dz &= {1 \over a-b}\qty{\int_\gamma {f(z) \over z-a}\dz - \int_\gamma {f(z) \over z-b}\dz} \\ &= {2\pi i \over a-b}\qty{f(a) - f(b)} .\] Since \(f\) is bounded, this number is finite and independent of \(R\), so taking \(R\to\infty\) preserves this equality. On the other hand, if \(\abs{f(z)}\leq M\), then we can estimate this integral directly as \[ I \leq \int_{\abs z = R} {M \over \abs{R-a} \cdot \abs{R-b} } = {M\cdot 2\pi R \over \abs{R-a} \cdot \abs{R-b}} \asymptotic {1\over R} \to 0 ,\] which forces \(f(a) =f(b)\). Since \(a, b\) were arbitrary, \(f\) must be constant.