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

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Answer 1 · 2018-06-25T09:54:29

#color(crimson)(=> -4 + 15 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)~~ 131.63^@, " II quadrant"#
#theta_2=tan^-1(6/ 4)~~ 56.31^@, " I quadrant"#

#z_1 + z_2 = sqrt 145 cos(131.63) + sqrt 52 cos(56.31) + i (sqrt 145 sin 131.63 + sqrt 52 sin 56.31)#

#=> -8 + 4 + i (9 + 6 )#

#color(crimson)(=> -4 + 15 i#