## Elementary Algebra

$58+0i$
First of all, recall that $i=\sqrt{-1}$, so $i^{2}=-1$ Also, since $(a+b)(a-b)=a^{2}-b^{2}$, we obtain: $(7-3i)(7+3i)$ =$(7)^{2}-(3i)^{2}$ =$49-9i^{2}$ =$49-9(-1)$ =$49+9$ =$58$ Since we have to write the answer in the form $(a+bi)$, we write: $58=58+0i$