Let \(f(z)=3z^{100}-e^z\).
How many zeros does \(f\) have in the unit disk, counting multiplicities?
Are the zeros distinct?
Solution
1. Count the zeros in the unit disk.
On \(|z|=1\), \[ |3z^{100}|=3, \] while \[ |e^z|=e^{\operatorname{Re}z}\le e<3. \] Thus \[ |-e^z|<|3z^{100}| \] on the unit circle. By Rouché’s theorem, \[ f(z)=3z^{100}-e^z \] and \(3z^{100}\) have the same number of zeros in \(|z|<1\), counted with multiplicity. The latter has a zero of multiplicity \(100\) at \(0\). Hence \[ \boxed{f\text{ has }100\text{ zeros in the unit disk, counted with multiplicity}.} \]
2. Every zero of \(f\) in the unit disk is simple.
Suppose \(z\) were a multiple zero. Then \[ f(z)=0 \] and \[ f'(z)=0. \] The first equation gives \[ e^z=3z^{100}. \] In particular \(z\ne0\). Since \[ f'(z)=300z^{99}-e^z, \] substituting the root equation yields \[ f'(z) =300z^{99}-3z^{100} =3z^{99}(100-z). \] Thus a multiple zero would have to satisfy \(z=100\), because \(z\ne0\). But \(100\) is not in the unit disk. Therefore every zero in \(|z|<1\) is simple.
Consequently the \(100\) zeros counted in step 1 are \[ \boxed{100\text{ distinct zeros}.} \]