Show that \(\alpha z e^z = 1\) where \(\abs{\alpha} > e\) has exactly one solution in \(\DD\).
Solution
Set \(f(z) = \alpha z\) and \(g(z) = e^{-z}\).
Estimate at \(\abs{z} =1\) we have \(\abs{g} =\abs{e^{-z}} = e^{-\Re(z)} \leq e^1 < \abs{\alpha} = \abs{f(z)}\)
\(f\) has one zero at \(z_0 = 0\), thus so does \(f+g\).