How do you add #(-2+3i)+(6+i)# in trigonometric form?

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Answer 1 · 2018-06-25T07:00:54

#color(crimson)((-2 + 3 i) + (6 + i) = 4 + 4 i = 4(1 + 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(-2^2+3^2))=sqrt 13#
#r_2=sqrt(6^2+-1^2) =sqrt 37#

#theta_1=tan^-1(3/-2)~~ 123.69^@, " II quadrant"#
#theta_2=tan^-1(1/6)~~ 9.4623^@, " I quadrant"#

#z_1 + z_2 = sqrt13 cos(123.69) + sqrt37 cos(9.4623) + i (sqrt 13 sin(123.69)+ sqrt 37 sin(9.4623))#

#=> -2 + 6 + i (3 + 1 )#
#color(brown)(=> 4 + 4 i#

Proof:

#-2 + 3i + 6 + i#

#=> 4 + 4 i#