Precalculus: Mathematics for Calculus, 7th Edition

$\sqrt -3\times\sqrt -12=-6$
Since $i=\sqrt -1$ we can write $\sqrt -3\times\sqrt -12=i\sqrt 3\times i\sqrt 12$. We can write $i\sqrt 12=2i\sqrt 3$ hence $i\sqrt 3\times i\sqrt 12=i\sqrt 3\times2i\sqrt 3=2i^{2}(\sqrt 3\times\sqrt 3)=2i^{2}\times3=6i^2$ however we know that $i^{2}=-1$ hence $6i^{2}=-6$. Hence the final answer written in the form $a+bi$ where $a$ is the real part and $b$ is the imaginary part is $-6$ as $a=-6$ and $b=0$.