Take \(\begin{array} { r } { f ( z ) = \frac { \pi ^ { 2 } } { \sin ^ { 2 } \pi z } } \end{array}\) and \(\begin{array} { r } { g ( z ) = \sum _ { n = - \infty } ^ { \infty } \frac { 1 } { ( z - n ) ^ { 2 } } } \end{array}\)
Show these functions have the same singularities in \(\mathbb { C }\)
Show that \(f\) and \(g\) have the same singular parts at each of their singularities.
Note that \(f\) and \(g\) each have period one and that both approach zero uniformly on \(0 \leq x \leq 1\) as \(| y | \to \infty\) .
Conclude that \(f = g\)
Solution
Write \[ f(z)=\frac{\pi^2}{\sin^2(\pi z)} \] and \[ g(z)=\sum_{n\in\ZZ}\frac1{(z-n)^2}. \]
1. The series defining \(g\) converges locally uniformly on\(\CC\sm\ZZ\) and therefore defines a holomorphic function there.
Let \(K\subseteq\CC\sm\ZZ\) be compact and put \[ M=\max_{z\in K}\abs{z}. \] For every integer \(n\) with \(\abs{n}>2M\) and every \(z\in K\), \[ \abs{z-n} \geq \abs{n}-\abs{z} > \frac{\abs{n}}{2}. \] Hence \[ \abs{\frac1{(z-n)^2}} \leq \frac4{n^2}. \] The numerical series \[ \sum_{n\in\ZZ\sm\{0\}}\frac1{n^2} \] converges. The Weierstrass M-test therefore gives uniform convergence of the tail on \(K\). Adding the finitely many remaining holomorphic terms proves local uniform convergence on \(\CC\sm\ZZ\), hence holomorphy there.
2. The singularities of \(g\) are exactly the integers, and each integer\(k\) is a double pole with singular part\[ \frac1{(z-k)^2}. \]
Fix \(k\in\ZZ\) and choose \(0<r<1/2\). On the disk \[ \abs{z-k}<r, \] write \[ g(z) = \frac1{(z-k)^2} + \sum_{n\in\ZZ\sm\{k\}} \frac1{(z-n)^2}. \] The second series converges locally uniformly on this disk by the argument of step 1 and is holomorphic there, since none of its denominators vanishes. Thus the first term is the complete principal part at \(k\), so \(k\) is a double pole. Step 1 shows there are no other singularities.
3. The singularities of \(f\) are exactly the integers, and at every\(k\in\ZZ\) its singular part is also\[ \frac1{(z-k)^2}. \]
The zeros of \(\sin(\pi z)\) are exactly the integers and are simple, so \(f\) has no singularities away from \(\ZZ\) and has a double pole at each integer.
For \(z=k+w\), \[ \sin(\pi z) = (-1)^k\sin(\pi w) = (-1)^k\pi w \left( 1-\frac{\pi^2w^2}{6}+O(w^4) \right). \] Therefore \[ \begin{aligned} f(k+w) &= \frac1{w^2} \left( 1-\frac{\pi^2w^2}{6}+O(w^4) \right)^{-2}\\ &= \frac1{w^2} +\frac{\pi^2}{3} +O(w^2). \end{aligned} \] Thus the entire principal part is \(w^{-2}=(z-k)^{-2}\).
4. Parts (a) and (b) hold: \(f\) and \(g\) have the same singularities andthe same singular parts at each one.
Steps 2 and 3 show that both functions have exactly the integers as double poles and that the principal part at every integer \(k\) is \((z-k)^{-2}\).
5. Both \(f\) and \(g\) have period \(1\).
Since \[ \sin(\pi(z+1))=-\sin(\pi z), \] one has \(f(z+1)=f(z)\).
For \(g\), absolute convergence away from the integers allows reindexing: \[ \begin{aligned} g(z+1) &= \sum_{n\in\ZZ}\frac1{(z+1-n)^2}\\ &= \sum_{m\in\ZZ}\frac1{(z-m)^2}\\ &= g(z), \end{aligned} \] where \(m=n-1\).
6. Uniformly for \(0\leq x\leq1\),\[ f(x+iy)\longrightarrow0 \]as \(\abs{y}\to\infty\).
For real \(x,y\), \[ \abs{\sin(\pi(x+iy))}^2 = \sin^2(\pi x)+\sinh^2(\pi y). \] Hence \[ \abs{f(x+iy)} \leq \frac{\pi^2}{\sinh^2(\pi y)}. \] The right-hand side is independent of \(x\) and tends to \(0\) as \(\abs{y}\to\infty\).
7. Uniformly for \(0\leq x\leq1\),\[ g(x+iy)\longrightarrow0 \]as \(\abs{y}\to\infty\).
For \(z=x+iy\) with \(0\leq x\leq1\) and \(y\neq0\), \[ \abs{g(z)} \leq \sum_{n\in\ZZ} \frac1{(x-n)^2+y^2}. \] The terms \(n=0,1\) contribute at most \(2/y^2\). For \(n\geq2\), \(\abs{x-n}\geq n-1\), and for \(n\leq-1\), \(\abs{x-n}\geq\abs{n}\). Therefore \[ \abs{g(x+iy)} \leq \frac2{y^2} + 2\sum_{m=1}^{\infty}\frac1{m^2+y^2}. \] Since \(t\mapsto(t^2+y^2)^{-1}\) is decreasing on \([0,\infty)\), \[ \sum_{m=1}^{\infty}\frac1{m^2+y^2} \leq \int_0^{\infty}\frac{dt}{t^2+y^2} = \frac{\pi}{2\abs{y}}. \] Consequently \[ \abs{g(x+iy)} \leq \frac2{y^2} + \frac{\pi}{\abs{y}}, \] uniformly in \(x\in[0,1]\). This bound tends to \(0\) as \(\abs{y}\to\infty\).
8. The difference\[ h=f-g \]extends to an entire, \(1\)-periodic function.
By step 4, at each integer the principal parts of \(f\) and \(g\) cancel. Thus every singularity of \(h\) at an integer is removable. After filling in those removable singularities, \(h\) is entire. Step 5 gives \(h(z+1)=h(z)\).
9. The entire function \(h\) is bounded on \(\CC\).
By steps 6 and 7, \[ h(x+iy)\longrightarrow0 \] uniformly for \(0\leq x\leq1\) as \(\abs{y}\to\infty\). Choose \(Y>0\) so that \[ \abs{h(x+iy)}\leq1 \] whenever \(0\leq x\leq1\) and \(\abs{y}\geq Y\).
On the compact rectangle \[ 0\leq x\leq1, \qquad \abs{y}\leq Y, \] the entire function \(h\) is bounded. Hence \(h\) is bounded on the whole vertical strip \(0\leq\operatorname{Re}z\leq1\). By \(1\)-periodicity, every point of \(\CC\) translates by an integer into this strip, so the same bound holds on all of \(\CC\).
10. One has\[ \boxed{ \frac{\pi^2}{\sin^2(\pi z)} = \sum_{n\in\ZZ}\frac1{(z-n)^2}. } \]
By step 9 and Liouville’s theorem, \(h\) is constant. Steps 6 and 7 show that \(h(z)\to0\) as \(\abs{\operatorname{Im}z}\to\infty\) in a fundamental strip. A constant function with this limit is zero. Therefore \(h\equiv0\), which is the displayed identity.
11. Q.E.D.
Steps <1>4–<1>10 establish parts (a)–(d).