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

Question

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

Quizplus · Accepted Answer

To find the quotient of two complex numbers in polar form, we divide their magnitudes and subtract their angles. For $z_1 = 4i$, its magnitude is 4 and angle is $\frac{\pi}{2}$. For $z_2 = -6 + 6i$, its magnitude is $6\sqrt{2}$ and angle is $\frac{3\pi}{4}$. The quotient's magnitude is $\frac{4}{6\sqrt{2}} = \frac{\sqrt{2}}{3}$ and the angle is $\frac{\pi}{2} - \frac{3\pi}{4} = -\frac{\pi}{4}$, which corresponds to $\frac{7\pi}{4}$ in the standard interval $[0, 2\pi)$.

Find Quotients of Complex Numbers in Polar Form - z1=4iz2=−6+6i\begin{array} { l } \mathrm { z } _ { 1 } = 4 \mathrm { i } \\\mathrm { z } _ { 2 } = - 6 + 6 \mathrm { i }\end{array}z1​=4iz2​=−6+6i​

Find Quotients of Complex Numbers in Polar Form
- $\begin{array} { l } \mathrm { z } _ { 1 } = 4 \mathrm { i } \\\mathrm { z } _ { 2 } = - 6 + 6 \mathrm { i }\end{array}$