How do you add #(-9+i)+(-1+3i)# in trigonometric form?

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Answer 1 · 2018-07-09T11:33:33

#color(crimson)(=> -10+ 4i#

#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+ 1^2))=sqrt 82#
#r_2=sqrt(-1^2+ 3^2) =sqrt 10#

#theta_1=tan^-1(1 / -9)~~ 173.66^@, " II quadrant"#
#theta_2=tan^-1(3/ -1)~~ 108.43^@, " II quadrant"#

#z_1 + z_2 = sqrt 82 cos(173.66) + sqrt 10 cos(108.43) + i (sqrt 82 sin 173.66 + sqrt 10 sin 108.43)#

#=> -9 - 1 + i (1 + 3 )#

#color(crimson)(=> -10+ 4i#