## Precalculus (6th Edition)

Simplify the given expression to obtain: $=\sqrt{36(-1)} \\=6\sqrt{-1} \\=6i$ RECALL: The complex number $a+bi$ is: (i) a pure imaginary number when $a=0$ and $b\ne0$; (ii) a real number when $b=0$; (iii) a nonreal complex number when $b\ne0$. The given expression is equivalent to $6i$. $6i$ has $a=0$ and $b\ne0$. Thus, it is a: pure imaginary number; nonreal complex number; complex number