How do you add $(-8+9i)+(4+6i)$ in trigonometric form?

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Answer 1 · 2018-06-25T09:54:29

$color(crimson)(=> -4 + 15 i$

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

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

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

$r_1=sqrt(-8^2+ 9^2))=sqrt 145$
$r_2=sqrt(4^2+ 6^2) =sqrt 52$

$theta_1=tan^-1(9 / -8)~~ 131.63^@, " II quadrant"$
$theta_2=tan^-1(6/ 4)~~ 56.31^@, " I quadrant"$

$z_1 + z_2 = sqrt 145 cos(131.63) + sqrt 52 cos(56.31) + i (sqrt 145 sin 131.63 + sqrt 52 sin 56.31)$

$=> -8 + 4 + i (9 + 6 )$

$color(crimson)(=> -4 + 15 i$

How do you add (-8+9i)+(4+6i)(-8+9i)+(4+6i) in trigonometric form?