Showing that, for $|a|>3$ and $n\geq1$, the function $f(z)=e^z-az^n$ has exactly $n$ roots (all simple and different) in the open unit disc

Question

Let $|a|>3$, $n\geq1$, $n\in\mathbb{N}$, then the function $$f(z)=e^z-az^n$$ has exactly $n$ roots in the disc $\{z\mid|z|<1\}$, and that they are all simple.

Hint: look at $f(z)-f'(z)$

I am not sure how do I even approach proving existence of roots, rather then proving maybe that the function is not vanishing (and then it has logarithm, and root, but how do I show existence of n roots?)

I thought also of maybe using somehow Rouche's Theorem, since:

$$f(z)-f'(z)=az^{n-1}\cdot(n-z)$$

and from there, that showing that $f$ has $n$ roots in the disk.

Thank you!

Answer 1

Proving existence of roots by using Rouché's Theorem is a good idea. Let $g(z)=az^n$ then for $|z|=1$, $$|g(z)|=|a|>3>|e^z|=e^{\cos(t)}=|f(z)-g(z)|$$ where $t\in [0,2\pi)$. What may we conclude?

Moreover if $|w|<1$ and $f(w)=e^w-aw^n=0$ then $w$ is NOT simple iff $$0=f'(w)=e^w-anw^{n-1}=aw^{n-1}(w-n)$$ which holds iff $w=0$ which is not a root because $f(w)=1$. So $w$ is simple.

Answer 2

The fast way is with Rouche's Theorem.

Alternatively, set $f_t(z)=az^n-te^z$.

Observe that, $f_t(z)\ne 0$, for $|z|=1$, and $t\in [0,1]$, and that the $$ g(t)=\frac{1}{2\pi i}\int_{|z|=1}\frac{f_t'(z)\,dz}{f_t(z)}=\text{number of roots of $f$ is $\{|z|<1\}$} \in \mathbb Z $$ Meanwhile, $g$ is continuous on $[0,1]$, and hence constant. Thus $$ \text{number of roots of $f$ is $\{|z|<1\}$}=\frac{1}{2\pi i}\int_{|z|=1}\frac{f'(z)\,dz}{f(z)}=g(1) \\=g(0)=\frac{1}{2\pi i}\int_{|z|=1}\frac{nz^{n-1}\,dz}{z^n}=n. $$