jmc

We have $(a+bi{)}^2=a^2+2abi+(bi{)}^2=(a^2-b^2)+2abi=3+4i$. Equating real and imaginary parts, we get $a^2-b^2=3$ and $2ab=4$. The second equation implies $ab=2$. Since $a$ and $b$ are positive integers and $ab=2$, we know one of them is 2 and the other is 1. Since $a^2-b^2=3$, we have $a=2$, $b=1$. So $a+bi=\boxed{2+i}$.