Yet another Rouché's theorem example for $f(z) = e^z-3z^{2019}$ - solution verification

Question

Use Rouché's Theorem to conclude that $$ f(z) = e^z-3z^{2019} $$ has exactly 2019 zeroes in $D(0,1)$.

My main question is, since we're working for $|z| = 1$ can I conclude that $|e^z| \sim e^1 = e \sim 2.71...$?

Because if that is true then taking $f(z) = -3z^{2019}$ and $g(z) = e^z -3z^{2019}$ we could utilise Rouché's Theorem as follows: $$ |f - g| = |-e^z| = 2.71... < 3 = |-3z^{2019}| = |f| $$ and since $-3z^{2019}$ has 2019 zeroes, then so does $g(z) = e^z -3z^{2019}$.

Is that correct? If not, how can one solve this example?

Answer 1

I don't know what is it that you mean when you write $|e^z|\sim e^1$. If $|z|=1$, then$$|e^z|=e^{\operatorname{Re}z}\leqslant e^1=e<3=|3z^{2\,019}|.$$That's all you need.