How do you add $(-6+i)+(9+3i)$ in trigonometric form?

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Answer 1 · 2018-08-06T04:15:20

$color(maroon)((-6 + i) + (9 + 3i) = 3 + 4 i, " I Quadrant"$

$(-6 + i) + (9 + 3i)$

$z = x + i y$

$z = r (cos theta + i sin theta)$

$r = |z| = |sqrt(x^2 + y^2)|$

$r_1 = sqrt(-6^2 + 1^2) = sqrt37$

$theta_1 = arctan (y/x) = tan^-1 (1/-6) = 170.5377^@, " II Quadrant"$

$r_2 = sqrt(9^2 + 3^2) = sqrt90$

$theta_2 = arctan (y/x) = tan^-1 (3/9) = 18.4349^@, " I Quadrant"$

$(-6 + i) + (9 + 3i) = sqrt37(cos 170.5377 + sin170.5377) + sqrt90 (cos 18.4349 + sin 18.4349)$

$=> sqrt37 cos 170.5377 + sqrt90 cos 18.4349 + i (sqrt37 sin 170.5377 + sqrt90 sin 18.4349)$

$=> -6 + 9 + 1 i + 3 i$

$color(maroon)((-6 + i) + (9 + 3i) = 3 + 4 i, " I Quadrant"$

How do you add (-6+i)+(9+3i)(-6+i)+(9+3i) in trigonometric form?