How do you add $(-1+8i)+(4-2i)$ in trigonometric form?

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Answer 1 · 2018-06-25T10:58:11

$color(indigo)(=> 3+ 6 i$

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

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

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

$r_1=sqrt(-1^2+ 8^2))=sqrt 65$
$r_2=sqrt(-2^2+ 4^2) =sqrt 20$

$theta_1=tan^-1(8 / -1)~~ 97.13^@, " II quadrant"$
$theta_2=tan^-1(-2/ 4)~~ 333.43^@, " IV quadrant"$

$z_1 + z_2 = sqrt 65 cos(97.13) + sqrt 20 cos(333.43) + i (sqrt 65 sin 97.13 + sqrt 20 sin 333.43)$

$=> -1 + 4 + i (8 - 2 )$

$color(indigo)(=> 3+ 6 i$

How do you add (-1+8i)+(4-2i)(-1+8i)+(4-2i) in trigonometric form?