Perform the operation and leave the result in trigonometric form. $$\frac { 16 \left( \cos 98 ^ { \circ } + i \sin 98 ^ { \circ } \right) } { 2 \left( \cos 128 ^ { \circ } + i \sin 128 ^ { \circ } \right) }$$
A) $8 \left( \cos 330 ^ { \circ } + i \sin 330 ^ { \circ } \right)$
B) $8 \left( \cos 98 ^ { \circ } - i \sin 98 ^ { \circ } \right)$
C) $8 \left( \cos 128 ^ { \circ } + i \sin 128 ^ { \circ } \right)$
D) $8 \left( \cos 330 ^ { \circ } - i \sin 330 ^ { \circ } \right)$
E) $8 \left( \cos 98 ^ { \circ } + i \sin 98 ^ { \circ } \right)$

Question

Perform the operation and leave the result in trigonometric form.​ $$\frac { 16 \left( \cos 98 ^ { \circ } + i \sin 98 ^ { \circ } \right) } { 2 \left( \cos 128 ^ { \circ } + i \sin 128 ^ { \circ } \right) }$$ ​
A) $8 \left( \cos 330 ^ { \circ } + i \sin 330 ^ { \circ } \right)$
B)​ $8 \left( \cos 98 ^ { \circ } - i \sin 98 ^ { \circ } \right)$
C)​ $8 \left( \cos 128 ^ { \circ } + i \sin 128 ^ { \circ } \right)$
D)​ $8 \left( \cos 330 ^ { \circ } - i \sin 330 ^ { \circ } \right)$
E)​ $8 \left( \cos 98 ^ { \circ } + i \sin 98 ^ { \circ } \right)$

Quizplus · Accepted Answer

To divide two complex numbers in trigonometric form, you divide their magnitudes and subtract their angles. Here, the magnitude of the numerator is 16, and the magnitude of the denominator is 2, so their quotient is 8. The angle of the numerator is 98 degrees, and the angle of the denominator is 128 degrees, so the difference is 98 - 128 = -30 degrees. Since a negative angle is measured clockwise, -30 degrees is equivalent to 330 degrees in the standard counter-clockwise measurement, resulting in $8(\cos 330^\circ + i\sin 330^\circ)$.