How do you add #(8-9i)+(4+6i)# in trigonometric form?

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Answer 1 · 2018-06-25T07:57:38

#color(gray)(=> 12 - 3 i#

#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(8^2+-9^2))=sqrt 145#
#r_2=sqrt(4^2+ 6^2) =sqrt 52#

#theta_1=tan^-1(-9/8)~~ 311.63^@, " IV quadrant"#
#theta_2=tan^-1(6/4)~~ 56.31^@, " I quadrant"#

#z_1 + z_2 = sqrt 145 cos(311.63) + sqrt 52 cos(56.31) + i (sqrt 145 sin(311.63)+ sqrt 52 sin(56.31))#

#=> 8 + 4 + i (-9 + 6 )#
#color(gray)(=> 12 - 3 i#