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
Multiple versions of Rouches theorem!
Set \(h(z) = 3z^2\) and \(g(z) = z^3 + bz + b^2\).
Then on \(\abs{z} = 1\), \[\begin{align*} \abs{g(z)} \leq 1 + b + b^2 < 3 = 3\abs{z}^2 = \abs{3z^2} = \abs{h} ,\end{align*}\] so \(g, h\) have the same number of roots in \(\abs{z} \leq_? 1\).
But \(h\) evidently has two roots in this region.