How many zeroes does $f(z)=z^4+e^z$ have in the unit disc?
ADDED: can you calculate them?
Here the same question is asked about the disk of radius $2$. It can be solved easily by Rouché's theorem since when $|z|=2$, if $z=x+iy$ then $|e^z|=e^x\leq e^2$ so $$|f(z)-z^4|=|e^z|\leq e^2\leq 9<16=|z^4|$$ therefore there are $4$ roots in that disc (up to multiplicity).
Now when $|z|=1$ we don't have this inequality, nor can we use the same trick by subtracting $e^z$ instead, since when $|z|=1$, $1/e\leq |e^z|\leq e$, one side is less than $1$ and the other greater.
This can be used to show that there are no roots in the right half of the disc, where by going around a curve approximating the boundary of the right half of the disc, so always $x>0$, we get $e^x>1\geq |z|$. A direct calculation also shows no roots are on the $y$ axis.
Any ideas on how to deal with the left half?