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
Goal: Show that \(e^z = az^n\) has exactly \(n\) solutions in the open unit disk when \(\abs a > e\), and no solutions when \(\abs a < \frac{1}{e}\).
1. On \(\abs z = 1\): \(\abs{e^z} = e^{\Re z} \leq e^{\abs z} = e\).
\(\Re z \leq \abs z = 1\) on the unit circle.
2. Case \(\abs a > e\): \(e^z - az^n\) has exactly \(n\) zeros in \(\abs z < 1\).
2.1. On \(\abs z = 1\), \(\abs{e^z} < \abs{az^n} = \abs a\).
1 and \(\abs a > e\); also \(\abs{z^n} = 1\).
2.2. \(e^z - az^n\) and \(-az^n\) have the same number of zeros in \(\abs z < 1\).
2.1.
2.3. \(-az^n\) has exactly \(n\) zeros in \(\abs z < 1\) (counting multiplicity).
Its only zero is \(z = 0\), of multiplicity \(n\) (for \(n \geq 1\); for \(n = 0\), \(-a \neq 0\) has no zeros, and “\(n = 0\) solutions” is consistent).
2.4. Conclusion.
2 and 3.
3. On \(\abs z = 1\): \(\abs{e^z} = e^{\Re z} \geq e^{-1} = \frac{1}{e}\).
\(\Re z \geq -\abs z = -1\) on the unit circle.
4. Case \(\abs a < \frac{1}{e}\): \(e^z - az^n\) has no zeros in \(\abs z < 1\).
4.1. On \(\abs z = 1\), \(\abs{az^n} = \abs a < \frac{1}{e} \leq \abs{e^z}\).
3 and the hypothesis on \(\abs a\).
4.2. \(e^z - az^n\) and \(e^z\) have the same number of zeros in \(\abs z < 1\).
4.1.
4.3. \(e^z\) has no zeros.
The exponential never vanishes.
4.4. Conclusion.
2 and 3.
5. Q.E.D.
2 gives exactly \(n\) solutions in the disk when \(\abs a > e\); 4 gives none when \(\abs a < 1/e\).