How do you add $(-5+5i)+(7+i)$ in trigonometric form?

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Answer 1 · 2018-06-25T09:30:16

$color(cyan)(=> 2 + 6 i$

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

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

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

$r_1=sqrt(-5^2+ 5^2))=sqrt 50$
$r_2=sqrt(7^2+ 1^2) =sqrt 50$

$theta_1=tan^-1(5/-5)~~ 135^@, " II quadrant"$
$theta_2=tan^-1(1/ 7)~~ 8.13^@, " I quadrant"$

$z_1 + z_2 = sqrt 50 cos(135) + sqrt 50 cos(8.13) + i (sqrt 50 sin 135 + sqrt 50 sin 8.13)$

$=> -5+ 7 + i (5 + 1 )$

$color(cyan)(=> 2 + 6 i$

How do you add (-5+5i)+(7+i)(-5+5i)+(7+i) in trigonometric form?