How do you add #(-9+2i)+(2+4i)# in trigonometric form?

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Answer 1 · 2018-07-27T05:44:21

#color(magenta)(=> -7 + 6 i, " II Quadrant"#

#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(-9^2+ 2^2))=sqrt 85#
#r_2=sqrt(2^2+ 4^2) =sqrt 20#

#theta_1=tan^-1(2 / -9) ~~ 167.4712^@, " II quadrant"#
#theta_2=tan^-1(4/ 2)~~ 63.4349^@, " I quadrant"#

#z_1 + z_2 = sqrt 85 cos(167.4712) + sqrt 20 cos(63.4349) + i (sqrt 85 sin 167.4712 + sqrt 20 sin 63.4349)#

#=> -9 + 2 + i (2 + 4 )#

#color(magenta)(=> -7 + 6 i, " II Quadrant"#