Find Products of Complex Numbers in Polar Form
-$$\begin{array} { l }
\mathrm { z } 1 = 2 + 2 \mathrm { i } \
\mathrm { z } 2 = \sqrt { 3 } - \mathrm { i }
\end{array}$$
A) $4 \sqrt { 2 } \left( \cos \frac { \pi } { 12 } + i \sin \frac { \pi } { 12 } \right)$
B) $4 \sqrt { 2 } \left( \cos \frac { 23 \pi } { 12 } + i \sin \frac { 23 \pi } { 12 } \right)$
C) $4 \left( \cos \frac { \pi } { 12 } + i \sin \frac { \pi } { 12 } \right)$
D) $4 \left( \cos \frac { 23 \pi } { 12 } + i \sin \frac { 23 \pi } { 12 } \right)$

Question

Find Products of Complex Numbers in Polar Form
-$$\begin{array} { l } 
\mathrm { z } 1 = 2 + 2 \mathrm { i } \
\mathrm { z } 2 = \sqrt { 3 } - \mathrm { i }
\end{array}$$
A) $4 \sqrt { 2 } \left( \cos \frac { \pi } { 12 } + i \sin \frac { \pi } { 12 } \right)$
B) $4 \sqrt { 2 } \left( \cos \frac { 23 \pi } { 12 } + i \sin \frac { 23 \pi } { 12 } \right)$
C) $4 \left( \cos \frac { \pi } { 12 } + i \sin \frac { \pi } { 12 } \right)$
D) $4 \left( \cos \frac { 23 \pi } { 12 } + i \sin \frac { 23 \pi } { 12 } \right)$

Quizplus · Accepted Answer

To find the product of two complex numbers in polar form, we first convert each number to polar form and then multiply their magnitudes and add their angles. For $z1 = 2 + 2i$, the magnitude is $\sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$ and the angle is $\tan^{-1}(2/2) = \tan^{-1}(1) = \pi/4$. For $z2 = \sqrt{3} - i$, the magnitude is $\sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{4} = 2$ and the angle is $\tan^{-1}(-1/\sqrt{3}) = -\pi/6$. Multiplying the magnitudes gives $2\sqrt{2} \times 2 = 4\sqrt{2}$, and adding the angles gives $\pi/4 - \pi/6 = \pi/12$. Therefore, the product in polar form is $4\sqrt{2}(\cos(\pi/12) + i\sin(\pi/12))$.

Find Products of Complex Numbers in Polar Form - z1=2+2iz2=3−i\begin{array} { l } \mathrm { z } 1 = 2 + 2 \mathrm { i } \\\mathrm { z } 2 = \sqrt { 3 } - \mathrm { i }\end{array}z1=2+2iz2=3​−i​

Find Products of Complex Numbers in Polar Form
- $\begin{array} { l } \mathrm { z } 1 = 2 + 2 \mathrm { i } \\\mathrm { z } 2 = \sqrt { 3 } - \mathrm { i }\end{array}$