We have for example the problem: Show that all the roots of $f(z) = z^5 + 3z + 1$ are in the disk $|z| < 2.$
The solution to this is not so hard as we can let $g(z)=-z^5$, this has $5$ roots in the disk $|z|<2,$ then for some $z\in \Gamma$, $$|f(z)+g(z)|\le 3|z|+1=7<|g(z)|=32$$
Clearly by Rouché's Theorem, $f$ has $5$ roots inside $|z|=2.$
Then come the problem I am stuck in, unlike above which I solved.
Problem: Prove that $e^z+z^3$ has no root in $\{z:|z|<3/4\}$ and has three roots in $\{z:|z|<2\}$.
My attempt: Following the above method that I used, if we let some $f(z)=-z^3,$ then we have $g(z)=e^z+z^3$ which gives us in $|z|=2:$
$$|g(z)+f(z)|=|e^z|= e^x\le e^{2}<|f(z)|=2^3$$
Then by Rouché's theorem, $f$ and $g$ has the same number of roots, and since $f(z)$ has $3$ roots (counting multiplicity) therefore $g(z)$ must have $3$ roots inside the disk $|z|=2.$
Where I am stuck: I am stuck in trying to show the case for when there are no roots inside the disk $|z|=3/4.$ I am assuming we work with $e^z$ since $e^z$ has no roots anywhere as its never $0.$
I would appreciate the help.