Classify the singularities of \[ f(z) = {z^3+1\over z^2(z+1)} .\]
Showing a pole \(z_0\) of \(f\) is order \(n\): show that \(z_0\) is a zero of order \(n\) of \(1/f\), i.e. \(1/f = (z-z_0)^nh(z)\) with \(h\) nonvanishing in a neighborhood of \(z_0\).
Solution
Write \(f(z) = p(z)/q(z)\) and factor \(p\): a principal root is \(\omega = e^{i\pi 3}\), so \[ p(z) &= (z-\omega\zeta_3^0)(z-\omega\zeta_3^1)(z-\omega\zeta_3^2) \\ &= (z-e^{i\pi\over 3})(z-e^{3i\pi \over 3})(z-e^{5i\pi \over 3}) \\ &= (z+1)(z-\omega)(z-\bar\omega) ,\] so \(z=-1\) is a removable singularity of \(f\). Alternatively, note that \({z^3+1 \over z+1} = z^2-z+1\) and cancel the common term.
Note that \(z=0\) is a zero of order \(n=2\) of \(1/f(z)\), since \(1/f(z) = z^2h(z)\) where \(h\) is nonvanishing in a neighborhood of \(0\). Thus \(z=0\) is a pole of order \(n=2\) of \(f\). The residue is computed as \[ \Res_{z=0} f(z) &= {1\over (1-1)!} \lim_{z\to 0} \dd{}{z} (z-0)^2f(z) \\ &= \dd{}{z} {z^3+1\over z+1}\evalfrom_{z=0} \\ &= \qty{ {3z^2\over z+1} - {(z^3+1)\cdot 1\over (z+1)^2} }\evalfrom_{z=0} \\ &= -1 .\]
Alternatively, expand as a Laurent series about \(z=0\): \[ f(z) &= z^{-2}(z^3 + 1) {1\over 1+z} \\ &= (z + z^{-2})\sum_{k\geq 0}(-z)^k \\ &= (z-z^2 + z^3 - \cdots) + (z^{-2} - \inverseof{z} + 1 - z + \cdots) ,\] and read off the coefficient of \(\inverseof{z}\).