Let \(n\in \ZZ^{\geq 0}\) and show that the equation \[ e^z = az^n \] has \(n\) solutions in the open unit disc if \(\abs{a} > e\), and no solutions if \(\abs{a} < {1\over e}\).
Solution
Note that \(\abs{e^z} = e^{\Re(z)}\), which is maximizes on \(S^1\) at \(z=1 \in \RR\) and minimized at \(z=-1\). Write \(f(z) = e^z-az^n\), so solution correspond to zeros of \(f\).
Case 1: suppose \(\abs{a} > e\). Big: \(M(z) = az^n\). Small: \(m(z) = e^z\). On \(\abs{z} = 1\), \[ \abs{m(z)} = \abs{e^z} = e^\Re(z) \leq e^1 < \abs{a} = \abs{az^n} = \abs{M(z)} ,\] so \(f\) has \(\size Z_M = n\) zeros.
Case 2: suppose \(\abs{a} < 1/e\). Big: \(M(z) = e^z\). Small: \(m(z) = az^n\). On \(\abs{z} = 1\), \[ \abs{m(z)} = \abs{az^n} = \abs{a} < \inverseof{e} \leq e^{\Re(z)} = \abs{e^z} = \abs{M(z)} ,\] and \(M\) has no zeros in \(\DD\) (and in fact none in \(\CC\)), so neither does \(f\).