Answer
$-3-2i\sqrt 3$
Work Step by Step
First, we use the product theorem to multiply the absolute values and add the arguments:
$(3$ cis $135^{\circ})(2$ cis $105^{\circ})$
$=3(2)$ cis $(135^{\circ}+105^{\circ})$
$=6$ cis $(240^{\circ})$
Next, we change the expression into its equivalent form:
$=6$ cis $(240^{\circ})$
$=6 (\cos 240^{\circ}+i\sin 240^{\circ})$
Then, we use the calculator to solve the functions of $\cos$ and $\sin$ and simplify the expression:
$=6 (\cos 240^{\circ}+i\sin 240^{\circ})$
=$6 [-\frac{1}{2}+i(-\frac{\sqrt 3}{2})]$
=$-\frac{6}{2}+6i(-\frac{\sqrt 3}{2})$
=$-\frac{6}{2}-\frac{6i\sqrt 3}{2}$
=$-3-2i\sqrt 3$
