Solution
= Solution
Rearrange to $z^4=1+i\sqrt3=2e^{i\pi/3}$. A fourth root has modulus $2^{1/4}$ and an argument one quarter of $\pi/3+2\pi k$. Therefore all four solutions are
$$
\boxed{z_k=2^{1/4}e^{i(\pi/12+k\pi/2)},\qquad k=0,1,2,3.}
$$
Their arguments differ by $\pi/2$, so they are distinct. Raising any of them to the fourth power returns $1+i\sqrt3$.