How do you divide $(-4+5i)/(6+i)$ in trigonometric form?

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Answer 1 · 2018-06-15T04:00:43

$color(purple)(=> 1.05 (-0.482 + i 0.876)$

$z_1 / z_2 = r_1/ r_2 (cos (theta_1 + thet_2) + i sin (theta_1 + theta_2)$

$z_1 = (-4 + i5), z_2 = (6 + i)$

$=> r_1 = sqrt(4^2 + 5^2) = sqrt41 = 6.4$

$theta_1 = tan^(-1) (5/-4)= -51.34^@ = 128.26^@, " II Quadrant"$

$=> r_2 = sqrt(1^2 + 6^2) = sqrt37 = 6.08$

$theta_2 = tan^(-1) (1/6)= 9.46^@$

$z_1 / z_2 = sqrt41/sqrt37 (cos (128.26 - 9.46) + i sin(128.26 - 9.46)0$

$color(purple)(=> 1.05 (-0.482 + i 0.876)$

How do you divide (-4+5i)/(6+i) (-4+5i)/(6+i) in trigonometric form?