Find all residues of the following function by writing it as a sum of principal parts at its poles: \[ f(z) = {z^3 \over z^2 + 1} .\]
Solution
Use polynomial long division to write \[ z^3 = z(z^2+1) - z \implies {z^3 \over z^2 + 1} = z - {z\over z^2 + 1} .\] Factor the latter part: \[ {z\over z^2 + 1} = {a\over z+i} + {b\over z-i} \implies a(z-i) + b(z+i) = z ,\] evaluate at \(z=i\) to get \(b=1/2\), and at \(z=-i\) to get \(a=1/2\). Thus \[ f(z) = z - {1/2 \over z+i} - {1/2 \over z-i} = P_\infty + P_{-i} + P_{i} ,\] yielding poles at \(\pm i\) with residues \[ \Res_{z=\infty} f(z) &= 1 \\ \Res_{z = i} f(z) &= -1/2 \\ \Res_{z = -i} f(z) &= -1/2 \\ .\]
Indeed, \[ f(z)=z-{1\over z}+O(z^{-3}) \qquad(z\to\infty), \] so the coefficient of \(z^{-1}\) is \(-1\) and therefore \(\Res_\infty f=-(-1)=1\). Equivalently, it is the negative of the sum of the finite residues.