How do you add #(-7+9i)# and #(-1+6i)# in trigonometric form?

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Answer 1 · 2018-07-09T11:27:26

#color(indigo)(=> -8+ 15i#

#z= a+bi= r (costheta+isintheta)#

#r=sqrt(a^2+b^2), " " theta=tan^-1(b/a)#

#r_1(cos(theta_1)+isin(theta_2))+r_2(cos(theta_2)+isin(theta_2))=r_1cos(theta_1)+r_2cos(theta_2)+i(r_1sin(theta_1)+r_2sin(theta_2))#

#r_1=sqrt(-7^2+ 9^2))=sqrt 130#
#r_2=sqrt(-1^2+ 6^2) =sqrt 37#

#theta_1=tan^-1(9 / -7)~~ 127.87^@, " II quadrant"#
#theta_2=tan^-1(6/ -1)~~ 333.43^@, " II quadrant"#

#z_1 + z_2 = sqrt 130 cos(127.87) + sqrt 37 cos(99.46) + i (sqrt 130 sin 127.87 + sqrt 37 sin 99.46)#

#=> -7 - 1 + i (9 + 6 )#

#color(indigo)(=> -8+ 15i#