Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.3 The Product and Quotient Theorems - 8.3 Exercises - Page 369: 3

$$[3\cos60^o+i\sin60^o][2\cos90^o+i\sin90^o]=-3\sqrt{3}+3i$$

$$A=[3\cos60^o+i\sin60^o][2\cos90^o+i\sin90^o]$$ The Product Theorem states that $$[r_1\cos\theta_1+i\sin\theta_1][r_2\cos\theta_2+i\sin\theta_2]=r_1r_2[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)]$$ Therefore, $$A=3\times2[\cos(60^o+90^o)+i\sin(60^o+90^o)]$$ $$A=6[\cos150^o+i\sin150^o]$$ $$A=6(-\frac{\sqrt{3}}{2}+\frac{1}{2}i)$$ $$A=-3\sqrt{3}+3i$$