Vtgyjfy

Let $|a|>3$, $ngeq1$, $ninmathbb{N}$,
then the function $$f(z)=e^z-az^n$$
has exactly $n$ roots in the disc ${zmid|z|<1}$, 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!

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 $tin [0,2pi)$.
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.

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