- Find all the poles of the function
\[ f ( z ) = \frac { e ^ { \pi z } } { ( z ^ { 2 } + 1 ) ^ { 2 } } . \]
Determine the residue of f at each pole.
Solution
The only poles are \(i\) and \(-i\), both of order two, with \[ \boxed{\operatorname{Res}_{i}f=\frac{\pi+i}{4},\qquad \operatorname{Res}_{-i}f=\frac{\pi-i}{4}.} \]
1. Factoring the denominator identifies all poles and their orders.
The denominator is \((z-i)^2(z+i)^2\), and the numerator \(e^{\pi z}\) is entire and never zero. Thus the quotient is holomorphic away from \(\pm i\), and at either point the denominator has order two while the numerator is nonzero. Both are genuine double poles; there are no other finite poles.
2. Differentiate the holomorphic factors to compute the residues.
For a double pole at \(a\), the residue of \(H(z)/(z-a)^2\) is \(H'(a)\), by the Taylor expansion of the holomorphic factor \(H\) [@SS03]. At \(i\), this gives \[ \begin{aligned} \operatorname{Res}_{i}f &=\left.\frac{d}{dz}\frac{e^{\pi z}}{(z+i)^2}\right|_{z=i}\\ &=e^{\pi i}\left(\frac{\pi}{(2i)^2}-\frac{2}{(2i)^3}\right) =\frac{\pi+i}{4}. \end{aligned} \] At \(-i\), the same calculation with the other factor yields \[ \begin{aligned} \operatorname{Res}_{-i}f &=\left.\frac{d}{dz}\frac{e^{\pi z}}{(z-i)^2}\right|_{z=-i}\\ &=e^{-\pi i}\left(\frac{\pi}{(-2i)^2}-\frac{2}{(-2i)^3}\right) =\frac{\pi-i}{4}. \end{aligned} \] Here \(e^{\pi i}=e^{-\pi i}=-1\), \((2i)^3=-8i\) and \((-2i)^3=8i\). These are the residues at all the poles.