Define \[ f(z) &= {\pi^2 \over \sin^2 \qty{\pi z} } \\ g(z) &= \sum_{n\in \ZZ} {1\over (z-n)^2} .\]
Show that \(f\) and \(g\) have the same singularities in \(\CC\).
Show that \(f\) and \(g\) have the same singular parts at each of their singularities.
Show that \(f, g\) each have period one and approach zero uniformly on \(0\leq x \leq 1\) as \(\abs{y}\to \infty\).
Conclude that \(f = g\).
Solution
Part 1: This is clear: \(\sin^2(\pi z) = 0 \iff z = k\) for \(k\in \ZZ\), and this is a pole of order 2 for \(f\). Every \(k\in \ZZ\) is visibly an order 2 pole of \(g\).
Part 2: By periodicity, it suffices to consider the singularity at \(z_0 = 0\). Expanding \(\sin(\pi z) = \pi z - {1\over 3!}(\pi z)^3 + {1\over 5!} (\pi z)^5 + \cdots\) and considering \(\sin(\pi z)^2\) shows that \(z=0\) is a pole of order 2. So \(z^2f(z)\) has a removable singularity at \(z=0\), and can be expanded: \[ z^2f(z) &= \qty{\pi z\over \sin(\pi z)}^2 \\ &= (\pi z)^2 \qty{ \inverseof{(\pi z)} + {1\over 3!}(\pi z) + {7\over 360} (\pi z^3) + \cdots}^2 \\ &= (\pi z)^2 \qty{ (\pi z)^{-2} + \bigo(1) } \\ &= 1 + \bigo(z^2) \\ \implies f(z) &= z^{-2} + \bigo(1) ,\] so the singular part of \(f\) at \(z=0\) is \(z^{-2}\). This coincides with the \({1\over z^2}\) term in \(g\). The remaining principal parts at \(z=k\) are \({1\over (z-k)^2},\) using the fact that \(f(z+1) = f(z)\), so \(f(k) = f(0)\) and the Laurent expansions are gotten by substituting \(z-k\) in for \(z\) everywhere.
Part 3: Periodicity is clear for \(f\). For \(g\), \[ g(z+1) = \sum_{k\in \ZZ} ((z-1)-k)^{-2} = \sum_{k'\in \ZZ} (z-k)^{-2} ,\] where \(k' \da k+1\), and the equality is true since both sums run over all of \(\ZZ\).
For convergence: take \(z=it\), then for \(f\) \[ f(it) \sim \csc^2(i\pi t) &\sim \qty{ e^{i\pi (it) } - e^{-i\pi (it)}}^{-2} \\ &= \qty{e^{-\pi t} - e^{\pi t}}^{-2} \\ &\leq {1\over e^{-\pi t} + e^{\pi t} } \\ &\sim e^{-\pi t} \\ &\to 0 ,\] using the reverse triangle inequality and that the \(e^{-\pi t}\) term in the denominator is negligible for large \(t\).
For \(g\), \[ g(it) &\sim t^{-2} + \sum_{k\geq 1} \inverseof{(t^2 + k^2)} \\ &\leq t^{-2} + \sum_{1\leq k \leq N}\inverseof{(t^2 + k^2)} + \sum_{k\geq N}\inverseof{(t^2 + k^2)} \\ &\leq t^{-2} + \sum_{1\leq k \leq N}\inverseof{(t\cdot k^2)} + \sum_{k\geq N}\inverseof{(k^2)} \\ &\leq t^{-2} + \inverseof{t} \sum_{1\leq k \leq N}\inverseof{(k^2)} + \sum_{k\geq N}\inverseof{(k^2)} \\ &\convergesto{N\to\infty\implies t\to\infty} 0 ,\] where given \(N\) we can pick \(t\) large enough so that \(t^2 + k^2 \geq tk^2\) for all \(k\leq N\). These converge to zero as \(N\to\infty\) since \(\sum k^{-2} < \infty\), making the last term the tail of a convergent sum.
Part 4: Since \(f,g\) uniformly converge to zero on the strip \(0<\Re(x) < 1\), they are bounded on this strip. Since this is a fundamental domain for their periods, they are bounded on \(\CC\). Write \(h\da f-g\), then \(h\) is entire since \(f,g\) have the same singular parts, and bounded since \(\abs{h}\leq \abs{f} + \abs{g}\). By Liouville, \(h\) is constant with \(\lim_{t\to\infty} h(it) = 0\), so \(h\equiv 0\) and \(f\equiv g\).