How do you add $(8+4i)$ and $(3-3i)$ in trigonometric form?

Question

1670 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-25T09:59:49

$color(crimson)(=> 11+ i$

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

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

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

$r_1=sqrt(4^2+ 8^2))=sqrt 80$
$r_2=sqrt(3^2+ -3^2) =sqrt 18$

$theta_1=tan^-1(4 / 8)~~ 26.57^@, " I quadrant"$
$theta_2=tan^-1(-3/ 3)~~ 315^@, " IV quadrant"$

$z_1 + z_2 = sqrt 80 cos(26.57) + sqrt 18 cos(315) + i (sqrt 80 sin 26.57 + sqrt 18 sin 315)$

$=> 8 + 3 + i (4 - 3 )$

$color(crimson)(=> 11+ i$

How do you add (8+4i)(8+4i) and (3-3i)(3-3i) in trigonometric form?