Find Products of Complex Numbers in Polar Form
-$$\begin{array} { l }
z 1 = \sqrt { 3 } \left( \cos \frac { 7 \pi } { 4 } + i \sin \frac { 7 \pi } { 4 } \right) \
z 2 = \sqrt { 6 } \left( \cos \frac { 9 \pi } { 4 } + i \sin \frac { 9 \pi } { 4 } \right)
\end{array}$$
A) $3 \sqrt { 2 } ( \cos 0 + i \sin 0 )$
B) $3 \sqrt { 2 } ( \sin 0 + i \cos 0 )$
C) $3 \sqrt { 2 } \left( \cos \frac { 5 \pi } { 4 } + i \sin \frac { 5 \pi } { 4 } \right)$
D) $3 \sqrt { 2 } \left( \sin \frac { 5 \pi } { 4 } + i \cos \frac { 5 \pi } { 4 } \right)$

Question

Find Products of Complex Numbers in Polar Form
-$$\begin{array} { l } 
z 1 = \sqrt { 3 } \left( \cos \frac { 7 \pi } { 4 } + i \sin \frac { 7 \pi } { 4 } \right) \
z 2 = \sqrt { 6 } \left( \cos \frac { 9 \pi } { 4 } + i \sin \frac { 9 \pi } { 4 } \right)
\end{array}$$
A) $3 \sqrt { 2 } ( \cos 0 + i \sin 0 )$
B) $3 \sqrt { 2 } ( \sin 0 + i \cos 0 )$
C) $3 \sqrt { 2 } \left( \cos \frac { 5 \pi } { 4 } + i \sin \frac { 5 \pi } { 4 } \right)$
D) $3 \sqrt { 2 } \left( \sin \frac { 5 \pi } { 4 } + i \cos \frac { 5 \pi } { 4 } \right)$

Quizplus · Accepted Answer

To multiply complex numbers in polar form, you multiply their magnitudes and add their angles. Here, the magnitudes are $\sqrt{3}$ and $\sqrt{6}$, and their product is $\sqrt{3} \times \sqrt{6} = \sqrt{18} = 3\sqrt{2}$. The angles are $\frac{7\pi}{4}$ and $\frac{9\pi}{4}$ (or $\frac{\pi}{4}$ since $\frac{9\pi}{4} = \frac{\pi}{4} + 2\pi$), and their sum is $\frac{7\pi}{4} + \frac{\pi}{4} = 2\pi$. Thus, the product is $3\sqrt{2}(\cos(2\pi) + i\sin(2\pi))$, which simplifies to $3\sqrt{2}(\cos(0) + i\sin(0))$.

Find Products of Complex Numbers in Polar Form - A) 32(cos⁡0+isin⁡0)3 \sqrt { 2 } ( \cos 0 + i \sin 0 )32​(cos0+isin0)

Find Products of Complex Numbers in Polar Form
- A) $3 \sqrt { 2 } ( \cos 0 + i \sin 0 )$