How do you add $(2-5i)$ and $(-2-2i)$ in trigonometric form?

Question

1479 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-09T11:40:21

$color(red)(=> - 7i$

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

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

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

$r_1=sqrt(2^2+ -5^2))=sqrt 29$
$r_2=sqrt(-2^2+ -2^2) =sqrt 8$

$theta_1=tan^-1(-5 / 2)~~ 291.8^@, " IV quadrant"$
$theta_2=tan^-1(-2/ -2)~~ 225^@, " III quadrant"$

$z_1 + z_2 = sqrt 29 cos(291.8) + sqrt 8 cos(225) + i (sqrt 29 sin 291.8 + sqrt 8 sin 225)$

$=> 2 - 2 + i (-5 - 2 )$

$color(red)(=> - 7i$

How do you add (2-5i)(2-5i) and (-2-2i)(-2-2i) in trigonometric form?