Let \(P, Q\) be polynomials with no common zeros. Assume \(a\) is a root of \(Q\). Find the principal part of \(P/Q\) at \(z=a\) in terms of \(P\) and \(Q\) if \(a\) is (1) a simple root, and (2) a double root.
Solution
If \(a\) is a simple zero of \(Q\), write \(Q(z)=(z-a)q(z)\) with \(q(a)=Q'(a)\ne0\). Then \[ \frac{P(z)}{Q(z)}=\frac{P(z)/q(z)}{z-a}, \] so the principal part is \[ \boxed{\frac{P(a)}{Q'(a)}\frac1{z-a}}. \]
Now suppose \(a\) is a double zero. Write \[ Q(z)=(z-a)^2q(z), \qquad q(a)=\frac{Q''(a)}2\ne0. \] Set \(h=P/q\), which is holomorphic near \(a\). Then \[ \frac{P(z)}{Q(z)}=\frac{h(z)}{(z-a)^2} =\frac{h(a)}{(z-a)^2}+\frac{h'(a)}{z-a}+\text{holomorphic terms}. \] Since \[ q'(a)=\frac{Q'''(a)}6, \] we have \[ h(a)=\frac{2P(a)}{Q''(a)} \] and \[ h'(a) =\frac{2P'(a)}{Q''(a)} -\frac{2P(a)Q'''(a)}{3Q''(a)^2}. \] Thus the principal part is \[ \boxed{ \frac{2P(a)}{Q''(a)}\frac1{(z-a)^2} +\left( \frac{2P'(a)}{Q''(a)} -\frac{2P(a)Q'''(a)}{3Q''(a)^2} \right)\frac1{z-a}.} \]