## College Algebra (10th Edition)

Recall, $i=\sqrt{-1}$, so $i^{2}=-1$. Thus, we multiply by the conjugate to obtain: $z*\overline{z}$ $=(3-4i)(\overline{3-4i})$ $=(3-4i)(3+4i)$ $=9+12i-12i-16i^{2}$ $=9-16*-1$ $=9+16$ $=25$