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    Principal parts of $P/Q$ at simple and double poles

    E-7PAHX

    Area
    Complex Analysis
    Topics
    Poles, Polynomials, Principal Parts, Residues
    Status
    draft

    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

    Since \(P(a)\neq0\), a root of \(Q\) of order \(m\) is a pole of \(P/Q\) of order \(m\).

    1. Write \(Q(z)=(z-a)R(z)\) with \(R\) a polynomial and \(R(a)=Q'(a)\neq0\). Then \(P/R\) is holomorphic near \(a\), and \[ {P(z) \over Q(z) } = {1\over z-a} {P(z) \over R(z)} = {1\over z-a}\qty{{P(a)\over R(a)} + \bigo(z-a)} ,\] so the principal part at \(a\) is \[ \boxed{{P(a)\over Q'(a)}\,{1\over z-a}} .\]

    2. Write \(Q(z)=(z-a)^2R(z)\) with \(R(a)\neq0\). Differentiating, \(Q''(a)=2R(a)\) and \(Q'''(a)=6R'(a)\). The function \(h\coloneqq P/R\) is holomorphic near \(a\), and \[ {P(z) \over Q(z) } = {h(z)\over (z-a)^2} = {h(a) \over (z-a)^2} + {h'(a)\over z-a} + \bigo(1) .\] Here \[ h(a)={P(a)\over R(a)}={2P(a)\over Q''(a)}, \qquad h'(a)={P'(a)R(a)-P(a)R'(a)\over R(a)^2}={2P'(a)\over Q''(a)}-{2P(a)Q'''(a)\over 3Q''(a)^2} ,\] so the principal part at \(a\) is \[ \boxed{{2P(a)\over Q''(a)}\,{1\over (z-a)^2}+\qty{{2P'(a)\over Q''(a)}-{2P(a)Q'''(a)\over 3Q''(a)^2}}{1\over z-a}} .\]

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