Solve the problem. Leave your answer in polar form.
-$$\begin{array} { l }
z = 8 \left( \cos \frac { \pi } { 6 } + i \sin \frac { \pi } { 6 } \right) \
w = 3 \left( \cos \frac { \pi } { 2 } + i \sin \frac { \pi } { 2 } \right)
\end{array}$$
Find zw.
A) $24 \left( \cos \frac { 2 \pi } { 3 } + i \sin \frac { 2 \pi } { 3 } \right)$
B) $24 \left( \cos \frac { \pi } { 3 } + i \sin \frac { \pi } { 3 } \right)$
C) $12 \left( \cos \frac { \pi } { 3 } + i \sin \frac { \pi } { 3 } \right)$
D) $12 \left( \cos \frac { 2 \pi } { 3 } + i \sin \frac { 2 \pi } { 3 } \right)$

Question

Solve the problem. Leave your answer in polar form.
-$$\begin{array} { l } 
z = 8 \left( \cos \frac { \pi } { 6 } + i \sin \frac { \pi } { 6 } \right) \
w = 3 \left( \cos \frac { \pi } { 2 } + i \sin \frac { \pi } { 2 } \right)
\end{array}$$
Find zw.
A) $24 \left( \cos \frac { 2 \pi } { 3 } + i \sin \frac { 2 \pi } { 3 } \right)$
B) $24 \left( \cos \frac { \pi } { 3 } + i \sin \frac { \pi } { 3 } \right)$
C) $12 \left( \cos \frac { \pi } { 3 } + i \sin \frac { \pi } { 3 } \right)$
D) $12 \left( \cos \frac { 2 \pi } { 3 } + i \sin \frac { 2 \pi } { 3 } \right)$

Quizplus · Accepted Answer

To find the product of two complex numbers in polar form, you multiply their magnitudes and add their angles. Here, $|z| = 8$, $|w| = 3$, $\theta_z = \frac{\pi}{6}$, and $\theta_w = \frac{\pi}{2}$. Thus, $|zw| = 8 \times 3 = 24$ and $\theta_{zw} = \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3}$. Therefore, $zw = 24 \left( \cos \frac { 2 \pi } { 3 } + i \sin \frac { 2 \pi } { 3 } \right)$.

Solve the Problem Find Zw A) 24(cos⁡2π3+isin⁡2π3)24 \left( \cos \frac { 2 \pi } { 3 } + i \sin \frac { 2 \pi } { 3 } \right)24(cos32π​+isin32π​)

Solve the Problem Find Zw
A) $24 \left( \cos \frac { 2 \pi } { 3 } + i \sin \frac { 2 \pi } { 3 } \right)$