How do you add $(-2-3i)+(6-9i)$ in trigonometric form?

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Answer 1 · 2018-06-25T09:37:15

$color(blue)(=> 4 - 12 i$

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

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

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

$r_1=sqrt(-2^2+ -3^2))=sqrt 13$
$r_2=sqrt(6^2+ -9^2) =sqrt 117$

$theta_1=tan^-1(-3/-2)~~ 236.31^@, " III quadrant"$
$theta_2=tan^-1(-9/ 6)~~ 303.69^@, " IV quadrant"$

$z_1 + z_2 = sqrt 13 cos(236.31) + sqrt 117 cos(303.69) + i (sqrt 13 sin 236.31 + sqrt 117 sin 303.69)$

$=> -2+ 6 + i (-3 - 9 )$

$color(blue)(=> 4 - 12 i$

How do you add (-2-3i)+(6-9i)(-2-3i)+(6-9i) in trigonometric form?