Question

Given z1 = 3(cos 14° + i sin 14°) and z2 = 2(cos 121° + i sin 121°), find the product z1.z2

Answer 1

Question #9107

Given $z_1 = 3(\cos 14{}^\circ + i \sin 14{}^\circ)$ and $z_2 = 2(\cos 121{}^\circ + i \sin 121{}^\circ)$, find the product z1.z2

Solution

$$\begin{array}{l} z_1 = r_1(\cos \alpha_1 + i \sin \alpha_1) \\ z_2 = r_2(\cos \alpha_2 + i \sin \alpha_2) \\ z_1 z_2 = r_1 r_2 \left(\cos \alpha_1 \cos \alpha_2 + i \cos \alpha_1 \sin \alpha_2 + i \sin \alpha_1 \cos \alpha_2 + i^2 \sin \alpha_1 \sin \alpha_2\right) = \\ = r_1 r_2 \left((\cos \alpha_1 \cos \alpha_2 - \sin \alpha_1 \sin \alpha_2) + i (\sin \alpha_1 \cos \alpha_2 + \cos \alpha_1 \sin \alpha_2)\right) \\ = r_1 r_2 (\cos (\alpha_1 + \alpha_2) + i \sin (\alpha_1 + \alpha_2)) \\ \end{array}$$ $$z_1 \cdot z_2 = 2 \cdot 3 \left(\cos \left(14^0 + 121^0\right) + i \sin \left(14^0 + 121^0\right)\right) = 6 \left(\cos \left(135^0\right) + i \sin \left(135^0\right)\right)$$

Answer: $z_1 \cdot z_2 = 6(\cos(135^0) + i \sin(135^0))$