How do you add $(-9+2i)+(2+4i)$ in trigonometric form?

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Answer 1 · 2018-07-27T05:44:21

$color(magenta)(=> -7 + 6 i, " II Quadrant"$

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

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

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

$r_1=sqrt(-9^2+ 2^2))=sqrt 85$
$r_2=sqrt(2^2+ 4^2) =sqrt 20$

$theta_1=tan^-1(2 / -9) ~~ 167.4712^@, " II quadrant"$
$theta_2=tan^-1(4/ 2)~~ 63.4349^@, " I quadrant"$

$z_1 + z_2 = sqrt 85 cos(167.4712) + sqrt 20 cos(63.4349) + i (sqrt 85 sin 167.4712 + sqrt 20 sin 63.4349)$

$=> -9 + 2 + i (2 + 4 )$

$color(magenta)(=> -7 + 6 i, " II Quadrant"$

How do you add (-9+2i)+(2+4i)(-9+2i)+(2+4i) in trigonometric form?