## Trigonometry (10th Edition)

$6i$
First, we use the product theorem to multiply the absolute values and add the arguments: $(\sqrt{6} cis 120^{\circ})(\sqrt{6} cis (-30^{\circ})) \\=\sqrt{6}\sqrt{6} cis (120^{\circ}-30^{\circ}) \\=6 cis (90^{\circ})$ Next, we change the expression into its equivalent form: $=6 cis (90^{\circ}) \\6(\cos90^{\circ}+i\sin90^{\circ})$ Since we know that $\cos90^{\circ}=0$ and $\sin90^{\circ}=1$, we substitute these values in the expression and simplify: $6(\cos90^{\circ}+i\sin90^{\circ}) \\=6(i) \\=6i$