How do you add $(1+6i)+(-3+i)$ in trigonometric form?

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Answer 1 · 2018-06-25T09:47:25

$color(blue)(=> -2 + 7 i$

$z= a+bi= r (costheta+isintheta)$

$r=sqrt(a^2+b^2), " " theta=tan^-1(b/a)$

$r_1(cos(theta_1)+isin(theta_2))+r_2(cos(theta_2)+isin(theta_2))=r_1cos(theta_1)+r_2cos(theta_2)+i(r_1sin(theta_1)+r_2sin(theta_2))$

$r_1=sqrt(1^2+ 6^2))=sqrt 37$
$r_2=sqrt(-3^2+ 1^2) =sqrt 10$

$theta_1=tan^-1(6/1)~~ 80.54^@, " I quadrant"$
$theta_2=tan^-1(1/ -3)~~ 161.57^@, " II quadrant"$

$z_1 + z_2 = sqrt 37 cos(80.54) + sqrt 10 cos(161.57) + i (sqrt 37 sin 80.54 + sqrt 10 sin 161.57)$

$=> 1- 3 + i (6 + 1 )$

$color(blue)(=> -2 + 7 i$

How do you add (1+6i)+(-3+i)(1+6i)+(-3+i) in trigonometric form?