Singularities

Write \begin{align*} P(z) &= \prod_{k\leq n} (z-a_k) \\ Q(z) &= \prod_{k\leq m}(z-b_k) \\ Q_j(z) &= \prod_{k\neq j}(z-b_k) = {Q(z) \over z-z_j} .\end{align*}

For \(b_\ell\) a simple pole, \begin{align*} {P(z) \over Q(z) } = {1\over z-b_\ell} {P(z) \over Q_\ell(z)} &= {1\over z-b_\ell}\qty{c_0 + c_1(z-b_\ell) + c_2(z-b_\ell)^2 + \cdots} \\ &= {c_0 \over z-b_\ell} + c_1 + { \mathsf{O}}(z-b_\ell) \\ & \coloneqq{P_{b_\ell}(z)} + c_1 + { \mathsf{O}}(z-b_\ell) ,\end{align*} so the principal part at \(z=z_\ell\) is given by \begin{align*} P_{z_\ell}(z) = {c_0 \over z-b_\ell} = {P(z) \over Q_\ell(z)}\Big|_{z=b_\ell} = \lim_{z\to z_\ell} {(z-b_\ell) P(z) \over Q(z)} .\end{align*}

For \(b_\ell\) a double pole, \begin{align*} {P(z) \over Q(z) } &= {1\over (z-b_\ell)^2 } {(z-z_\ell)^2P(z) \over Q(z) } \\ &= {1\over (z-b_\ell)^2}\qty{ d_0 + d_1(z-b_\ell) + d_2(z-b_\ell)^2 } \\ &= {d_0 \over (z-b_\ell)^2} + {d_1\over z-z_\ell} + d_2 + { \mathsf{O}}(z-b_\ell) \\ &\coloneqq P_{b_\ell}(z) + d_2 + { \mathsf{O}}(z-b_\ell) .\end{align*} To extract the \(d_1\) coefficient, note that \begin{align*} {(z-b_\ell)^2 P(z) \over Q(z)} &= d_0 + d_1(z-b_\ell) + \cdots \\ \implies {\frac{\partial }{\partial z}\,} {(z-b_\ell)^2 P(z) \over Q(z)} &= d_1 + 2d_2(z-b_\ell) + \cdots ,\end{align*} so \begin{align*} d_0 &= \lim_{z\to b_\ell} { (z-b_\ell)^2 P(z) \over Q(z) } \\ d_1 &= \lim_{z\to b_\ell} {\frac{\partial }{\partial z}\,} {(z-b_\ell)^2 P(z) \over Q(z) } \\ P_{b_\ell} &= {d_0 \over (z-b_\ell)^2} + {d_1\over z-b_\ell} .\end{align*}

Compute the series expansion of the RHS: \begin{align*} (z-1) f(z) &= (z-1) \sum_{n\geq 1} c_n z^k \\ &= -c_1z + \sum_{n\geq 2} (c_{n-1} - c_n) z^n \\ &\overset{z\to 1}\longrightarrow -c_1 + \sum_{n\geq 2} c_{n-1} - c_n \\ &\coloneqq\lim_{N\to\infty} -c_1 z + \sum_{n=2}^N c_{n-1} - c_n \\ &= \lim_{N\to\infty} -c_N ,\end{align*} where we’ve used that the sum is telescoping.

If \(f\) is entire, write \(f(z) = \sum_{k\geq 0}c_k z^k\) and \(g(z) \coloneqq f(1/z) = \sum_{k\geq 0}c_k z^{-k}\). If \(z=\infty\) is a pole of order \(m\) of \(f\), \(z=0\) is a pole of order \(m\) of \(g\), so \begin{align*} g(z) = \sum_{0\leq k \leq m}c_k z^{-k} \implies f(z) = \sum_{0\leq k \leq m}c_k z^k ,\end{align*} making \(f\) a polynomial of degree at most \(m\).

Write \(f(z) = P_2(z) + g(z)\) where \(P_2\) is the principal part of \(f\) at \(z=2\) and \(g\) is holomorphic at \(z=2\). Then \(g\) is an entire function with a double pole at \(\infty\), and is thus a polynomial of degree at most \(2\), so \(g(z) = c_2z^2 + c_1 z + c_0\). Since the pole of \(f\) at \(z=2\) is simple, \(P_2(z) = \sum_{k\geq -1} d_k (z-2)^k\). Combining these, we can write \begin{align*} f(z) = d_{-1}(z-2)^{-1}+ \sum_{0\leq k\leq 3} (d_k + c_k)(z-2)^k + \sum_{k\geq 3}d_k (z-2)^k .\end{align*} However, if \(d_k\neq 0\) for any \(k\geq 3\), this results in a higher order pole at \(\infty\), so \(f\) must be of the form \begin{align*} f(z) = d_{-1}(z-2)^{-1}+ \sum_{0\leq k\leq 3} (d_k + c_k)(z-2)^k .\end{align*}

Write \(f(z) = \sum_{k\geq 0} c_k z^k\) since it is entire.

• If \(f\) has finitely many zeros, \(f\) is nonconstant and entire, and thus unbounded by Liouville. If \(f\) is nonconstant, \(z=\infty\) can not be removable, since this would force \(f\) to be constant. So \(z=\infty\) can be a pole or an essential singularity. Both possibilities can occur: if \(f\) is a polynomial, it is entire with finitely many zeros and a pole at \(z=\infty\). Taking \(f(z)= e^z\) has no zeros and an essential singularity at \(z=\infty\).

• If \(f\) has infinitely many zeros, if \(f\) is nonconstant then infinitely many \(c_k\) are nonzero – otherwise \(f\) is a polynomial and can only have finitely many zeros. Then \(g(z) \coloneqq f(1/z) = \sum_{k\geq 0}{c_k\over z^k}\) has infinitely many nonzero terms, making \(z=0\) an essential singularity for \(g\) and \(z=\infty\) essential for \(f\).

Part 1: This is clear: \(\sin^2(\pi z) = 0 \iff z = k\) for \(k\in {\mathbf{Z}}\), and this is a pole of order 2 for \(f\). Every \(k\in {\mathbf{Z}}\) 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: \begin{align*} z^2f(z) &= \qty{\pi z\over \sin(\pi z)}^2 \\ &= (\pi z)^2 \qty{ (\pi z) ^{-1}+ {1\over 3!}(\pi z) + {7\over 360} (\pi z^3) + \cdots}^2 \\ &= (\pi z)^2 \qty{ (\pi z)^{-2} + { \mathsf{O}}(1) } \\ &= 1 + { \mathsf{O}}(z^2) \\ \implies f(z) &= z^{-2} + { \mathsf{O}}(1) ,\end{align*} 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 \(g\), \begin{align*} g(z+1) = \sum_{k\in {\mathbf{Z}}} ((z-1)-k)^{-2} = \sum_{k'\in {\mathbf{Z}}} (z-k)^{-2} ,\end{align*} where \(k' \coloneqq k+1\), and the equality is true since both sums run over all of \({\mathbf{Z}}\).

For convergence: take \(z=it\), then for \(f\) \begin{align*} 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 ,\end{align*} using the reverse triangle inequality and that the \(e^{-\pi t}\) term in the denominator is negligible for large \(t\).

For \(g\), \begin{align*} g(it) &\sim t^{-2} + \sum_{k\geq 1} (t^2 + k^2)^{-1}\\ &\leq t^{-2} + \sum_{1\leq k \leq N}(t^2 + k^2)^{-1}+ \sum_{k\geq N}(t^2 + k^2)^{-1}\\ &\leq t^{-2} + \sum_{1\leq k \leq N}(t\cdot k^2)^{-1}+ \sum_{k\geq N}(k^2)^{-1}\\ &\leq t^{-2} + t^{-1}\sum_{1\leq k \leq N}(k^2)^{-1}+ \sum_{k\geq N}(k^2)^{-1}\\ &\overset{N\to\infty\implies t\to\infty}\longrightarrow 0 ,\end{align*} 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 \({\mathbf{C}}\). Write \(h\coloneqq f-g\), then \(h\) is entire since \(f,g\) have the same singular parts, and bounded since \({\left\lvert {h} \right\rvert}\leq {\left\lvert {f} \right\rvert} + {\left\lvert {g} \right\rvert}\). By Liouville, \(h\) is constant with \(\lim_{t\to\infty} h(it) = 0\), so \(h\equiv 0\) and \(f\equiv g\).

Note that \(z=0\) can not be a removable singularity, since then \(f\) would extend to a holomorphic function over \(z=0\), and by continuity \(0 = \lim f(z_n) = f(\lim z_n) = f(0)\). By the identity principle, this would force \(f\equiv 0\), contradicting that \(f\) is nonconstant.

It can not be a pole, because then \(f(z_n)\to \infty\), but \({\left\lvert {f(z_n)} \right\rvert} = 0 < {\varepsilon}\) for any \({\varepsilon}\) infinitely many times.