Square roots of a complex number

System. $a^2-b^2=3$, $2ab=4$, so $ab=2$.

Also. $|w|^2=|z|=\sqrt{9+16}=5$, so $a^2+b^2=5$.

Solve. $a^2=(5+3)/2=4$, so $a=\pm2$. $b^2=(5-3)/2=1$, so $b=\pm1$. With $ab=2>0$, $a$ and $b$ have the same sign: $(a,b)=(2,1)$ or $(-2,-1)$.

Conclusion. The square roots are $\pm(2+i)$.

Check. $(2+i)^2=4+4i+i^2=3+4i$. Correct.