How do you divide #( i-3) / (7i +3)# in trigonometric form?

Question

How do you divide #( i-3)    /   (7i +3)# in trigonometric form?

Dylan Arnold · Accepted Answer

#color(crimson)(=> 0.42 ( -0.7475 + i 0.6643)# #z_1 / z_2 = (r_1 / r_2) (cos (theta_1 - theta_2) + i sin (theta_1 - theta_2))# #z_1 = i-3, z_2 = 3 + i 7# #r_1 = sqrt(1 + 3^2) = sqrt10# #theta_1 = tan ^ (-1) (1/-3) = -18.43 = 161.57^@, " II Quadrant"# #r_2 = sqrt(7^2 + 3^2) = sqrt58# #theta_2 = tan ^ (-1) (3/7) = 23.2^@# #z_1 / z_2 = sqrt(10/58) (cos (161.57- 23.2) - i sin (161.57 - 23.2))# #color(crimson)(=> 0.42 ( -0.7475 + i 0.6643)#

Aaliyah Abdullah · Answer

undefined To divide complex numbers in trigonometric form, we first express the complex numbers in polar form. Then, we divide their magnitudes and subtract their arguments.

Given $(i - 3) / (7i + 3)$, we express these complex numbers in polar form:

$i - 3 = \sqrt{1^2 + (-3)^2} \angle \tan^{-1}(\frac{-3}{1}) = \sqrt{10} \angle -71.57^\circ$

$7i + 3 = \sqrt{7^2 + 3^2} \angle \tan^{-1}(\frac{3}{7}) = \sqrt{58} \angle 22.62^\circ$

Now, divide their magnitudes and subtract their arguments:

$$\frac{\sqrt{10}}{\sqrt{58}} \angle (-71.57^\circ - 22.62^\circ)$$

$$\frac{\sqrt{10}}{\sqrt{58}} \angle -94.19^\circ$$

So, $(i - 3) / (7i + 3)$ in trigonometric form is $\frac{\sqrt{10}}{\sqrt{58}} \angle -94.19^\circ$.