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    $z^3+3z^2+bz+b^2$ has exactly two roots in the unit disk when $|b|<1$

    P-SCDUI

    Area
    Complex Analysis
    Topics
    Polynomials, Rouché, Zeros
    Status
    draft

    Assume that \(\abs b < 1\) and show that the following polynomial has exactly two roots (counting multiplicity) in \(\abs{z} < 1\): \[ f(z) \definedas z^3 + 3z^2 + bz + b^2 .\]

    Solution

    Big: \(M(z) = 3z^2\). Small: \(m(z) = z^3+bz + b^2\). Then on \(\abs{z} = 1\): \[ \abs{m(z)} \leq \abs{z}^3 + b\abs{z} + b^2 = 1 + b + b^2 < 3 = \abs{M(z)} ,\] and \(M(z)\) has exact two roots in \(\DD\).

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