Answer
$-3i$
Work Step by Step
First, we use the product theorem to multiply the absolute values and add the arguments:
$(\sqrt{3} cis 45^{\circ})(\sqrt{3} cis 225^{\circ})
\\=\sqrt{3} \sqrt{3} cis (45^{\circ}+225^{\circ})
\\=3 cis (270^{\circ})$
Next, we change the expression into its equivalent form:
$=3 cis (270^{\circ})
\\=3(\cos270^{\circ}+i\sin270^{\circ})$
Since we know that $\cos270^{\circ} =0$ and $\sin270^{\circ}=-1$, we substitute these values in the expression and simplify:
$3(\cos270^{\circ}+i\sin270^{\circ})
\\=3(-i)
\\=-3i$