Zeroes of $z^4+e^z$ in the unit disk 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
$$|f(z)-z^4|=|e^z|\leq e^2\leq 9&lt;16=|z^4|$$ 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 $x&gt;0$,  we get $e^x&gt;1\geq |z|$. A direct calculation also shows no roots are on the $y$ axis.

Any ideas on how to deal with the left half?
This is not a solution, just a graphical representation of the modulus $|f(z)|$ (with colors tending to red when values tend to zero). This shows that there are two roots inside the unit disk with conjugate (approximate) values $-0.67003127469869344 \ \pm 0.5161249765067623 i$ (the two others are outside the unit disk).


Another way to be convinced graphically of the fact that there are 2 zeros inside the unit circle is by plotting $\frac{\text{arg }f(e^{it})}{2 \pi}$ (unwrapped) (using the argument principle) .
This is not a solution, just a graphical representation of the modulus $|f(z)|$ (with colors tending to red when values tend to zero). This shows that there are two roots inside the unit disk with conjugate (approximate) values $-0.67003127469869344 \ \pm 0.5161249765067623 i$ (the two others are outside the unit disk).


Another way to be convinced graphically of the fact that there are 2 zeros inside the unit circle is by plotting $\frac{\text{arg }f(e^{it})}{2 \pi}$ (unwrapped) (using the argument principle) .