jmc

We have $(a-bi{)}^2=a^2-2abi+(bi{)}^2=(a^2-b^2)-2abi=8-6i$. Equating real and imaginary parts, we get $a^2-b^2=8$ and $-2ab=-6$, or $ab=3$. Since $a$ and $b$ are positive integers and $ab=3$, we know one of them is 3 and the other is 1. Since $a^2-b^2=8$, trial and error gives $a=3$, $b=1$. So $a-bi=\boxed{3-i}$.