Find the number of roots on \(\abs{z} \leq 1\) of \[ f(z)=z^{6}+4 z^{2} e^{z+1}-3 .\]
Solution
- Small: \(m(z) = z^6-3\)
- Big: \(M(z) = 4z^2 e^{z+1}\), which has two such zeros
Now estimate \(m\) from above: \[ \abs{m(z)} = \abs{z^6 - 3} \leq \abs{z}^6 + 3 = 4 .\] and \(M\) from below: \[ \abs{M(z)} = \abs{4z^2 e^{z+1}} = 4e\abs{z}^4 e^{\Re(z)} = 4e e^{\Re(z)} \geq 4e e^{-1} = 4 ,\] which unfortunately isn’t quite enough. But equality occurs iff \(\Re(z) = -1\) on \(S^1\), so \(z=-1\), in which case \(\abs{m(-1)} = \abs{1-3} = 2\), so the inequality is in fact strict. So \(2 = Z_M = Z_f\).