Let \(n \in \NN\). Prove that the equation \(\exp(z) = a z^n\) has \(n\) solutions in the open unit disk \(D\) if \(\abs{a} > e\) and none if \(\abs{a} < \frac{1}{e}\).
Solution
Set \[ F(z)=e^z-az^n. \]
1. On the unit circle,\[ e^{-1} \leq \abs{e^z} \leq e. \]
If \(\abs{z}=1\), then \[ -1\leq\operatorname{Re}z\leq1. \] Since \[ \abs{e^z}=e^{\operatorname{Re}z}, \] the displayed bounds follow.
2. If \(\abs{a}>e\), then on \(\abs{z}=1\),\[ \abs{e^z} < \abs{-az^n}. \]
By step 1, \[ \abs{e^z}\leq e. \] On the unit circle, \[ \abs{-az^n}=\abs{a}>e. \]
3. If \(\abs{a}>e\), then the equation\[ e^z=az^n \]has exactly \(n\) solutions in the open unit disk, counting multiplicity.
Write \[ F(z)=-az^n+e^z. \] By step 2 and Rouché’s theorem, \(F\) and \(-az^n\) have the same number of zeros in \(\abs{z}<1\). The latter has exactly \(n\) zeros there, all at \(z=0\) counted with multiplicity.
4. If \(\abs{a}<e^{-1}\), then on \(\abs{z}=1\),\[ \abs{az^n} < \abs{e^z}. \]
On the unit circle, \[ \abs{az^n}=\abs{a}<e^{-1}. \] Step 1 gives \[ \abs{e^z}\geq e^{-1}. \] Thus the inequality is strict.
5. If \(\abs{a}<e^{-1}\), then the equation\[ e^z=az^n \]has no solutions in the open unit disk.
By step 4 and Rouché’s theorem, \(F=e^z-az^n\) and \(e^z\) have the same number of zeros in \(\abs{z}<1\). The exponential function has no zeros anywhere in \(\CC\), so this number is zero.
6. Therefore the equation has exactly \(n\) solutions in the unit diskwhen \(\abs{a}>e\), and none when \(\abs{a}<e^{-1}\).
This combines steps 3 and 5.
7. Q.E.D.
Step 6 is the required conclusion.