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    Liouville's theorem via $\lim_{R\to\infty}\int_{|z|=R}\frac{f(z)}{(z-a)(z-b)}\,dz$ for bounded entire $f$

    P-FL5SY

    Area
    Complex Analysis
    Topics
    Cauchy Integral Formula, Entire Functions, Liouville's Theorem, Residues
    Status
    draft

    Let \(f(z)\) be bounded and analytic in \(\CC\). Let \(a\neq b\) be any fixed complex numbers. Show that the following limit exists: \[ \lim_{R\to \infty} \int_{\abs z = R} {f(z) \over (z-a)(z-b)} \,dz .\]

    Use this to show that \(f(z)\) must be constant.

    Solution

    Let \(|f(z)|\le M\) on \(\mathbb C\). For \(R>2\max\{|a|,|b|\}\), \[ \left| \int_{|z|=R}\frac{f(z)}{(z-a)(z-b)}\,dz \right| \le 2\pi R\frac{M}{(R-|a|)(R-|b|)}. \] The right-hand side tends to \(0\) as \(R\to\infty\), so the requested limit exists and equals \(0\).

    For such \(R\), the residue theorem gives \[ \int_{|z|=R}\frac{f(z)}{(z-a)(z-b)}\,dz =2\pi i\left( \frac{f(a)}{a-b}+\frac{f(b)}{b-a} \right) =2\pi i\frac{f(a)-f(b)}{a-b}. \] The left-hand side tends to \(0\), while the right-hand side is independent of \(R\). Hence \[ f(a)=f(b). \] Because \(a\ne b\) were arbitrary, \(f\) is constant.

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