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

Question

1649 views around the world

You can reuse this answer
Creative Commons License

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"#