 Solution

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 $R=\sqrt{2}$. On the two radial segments, $z^4$ is nonnegative real, so $|f|\ge 3$, whereas $|g|=e^{-\mathrm{Im}z}\le 1$. On the circular arc,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 satisfiesThere can be no additional zero outside the quarter-disc. Hence