Show that there is an entire function \(h\) such that \[ {\pi^2\over \sin^2(\pi z)} = \sum_{k\in \ZZ} {1\over (z-k)^2} + h(z) .\]
- The principal part of \(f\) at \(z=z_0\) is gotten by expanding \(f(z) = \sum_{k\in \ZZ} c_k z^k\) and taking \(\sum_{k\leq 1} c_k z^k\).
- Common trick: to control a singularity, subtract off a function with the same principal part at that point.
Solution
Write
- \(f(z) = {\pi^2 \over \sin^2(\pi z)}\)
- \(g(z) = \sum_{k\in \ZZ} {1\over (z-k)^2}\)
Write the above equation as \(f(z) = g(z) - h(z)\) and consider \(h(z) \da f(z) - g(z)\). Then \(h\) is meromorphic with singularities precisely on the set \(\ZZ\), and are thus isolated. By the classification of isolated singularities, these can be removable, poles, or essential. If they are removable, then \(h\) is entire.
Consider the singularity at \(z_0 = 0\). This is a pole of \(f\), and a computation shows it is order 2: \[ f(z) &= \qty{\pi \over \sin(\pi z)}^2 \\ &= \qty{\pi \over \pi z - {1\over 3!}(\pi z)^3 + \bigo(z^5) }^2 \\ &= \qty{\pi \over \pi z (1 - {1\over 3!}(\pi z)^2 + \bigo(z^4))}^2 \\ &= {1\over z^2} \qty{1 \over 1 - {1\over 3!}(\pi z)^2 + \bigo(z^4)}^2 \\ &= {1\over z^2}\qty{1 + \bigo(z^2)}^2 \\ &= {1\over z^2} + \bigo(z^2) ,\] so \(z=0\) is a zero of order 2 of \(1/f\). This expansion also shows that the principal part of \(f\) at \(z=0\) is \({1\over z^2}\), which is precisely that of \(g\) at \(z=0\), i.e. \({1\over (z-0)^2} = 1/z^2\), Since \(f-g\) subtracts off this part, \(z=0\) becomes a removable singularity for \(h\) since \(\lim_{z\to 0} \qty{ f(z) - {1\over z^2}} = 1<\infty\).
Now note that \(f\) is periodic, and since the period is 1, a similar argument shows that the remaining singularities on \(\ZZ\smz\) are all removable for \(f-g\). By Riemann’s removable singularity theorem, \(f-g\) extends over these singularities, yielding an entire function \(h\) that restricts to \(f-g\) on \(\CC\sm\ZZ\).