= Solution
\b[Rouché's theorem.] Suppose $f,g$ are <holomorphic functions> on a neighbourhood of the closure of a bounded domain enclosed by a simple, positively oriented, piecewise smooth curve. If $|g(z)|<|f(z)|$ on the boundary, then $f$ and $f+g$ have the same number of zeros inside, counted with <multiplicity>.
Apply <Rouche theorem> with $f(z)=z^4+3$ and $g(z)=e^{iz}$ on the first-quadrant quarter-disc of radius $R=\sqrt2$. On the two radial segments, $z^4$ is nonnegative real, so $|f|\geq3$, whereas $|g|=e^{-\operatorname{Im}z}\leq1$. On the circular arc,
$$
|f(z)|\geq|z|^4-3=1.
$$
At every arc point with positive imaginary part, $|g|<1$, giving the strict inequality; at the remaining endpoint $z=R$, $|f|=7>1=|g|$. Thus the strict boundary inequality holds everywhere, including the origin.
The four zeros of $f$ have modulus $3^{1/4}$ and arguments $\pi/4,3\pi/4,5\pi/4,7\pi/4$. Exactly one lies inside the chosen domain. <Rouche theorem> gives exactly one zero of $f+g$ there, counted with <multiplicity>, so that zero is simple. Finally any zero in the open first quadrant satisfies
$$
|z|^4=|3+e^{iz}|\leq3+e^{-\operatorname{Im}z}<4.
$$
There can be no additional zero outside the quarter-disc. Hence
$$
\boxed{\text{Exactly one zero lies in the first quadrant, and }|z_0|<\sqrt2.}
$$
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