How do you add $(1+7i)+(-4-2i)$ in trigonometric form?

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Answer 1 · 2018-06-25T06:52:44

$color(brown)(=> -3 + 5i$

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

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

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

$r_1=sqrt(1^2+7^2))=sqrt50$
$r_2=sqrt(-4^2+-2^2) =sqrt20$

$theta_1=tan^-1(7/1)~~81.87^@, " I quadrant"$
$theta_2=tan^-1(-2/-4)~~206.57^@, " III quadrant"$

$z_1 + z_2 = sqrt50 cos(81.87) + sqrt20 cos(206.57) + i (sqrt50 sin(81.87)+ sqrt20 sin(206.57))$

$=> 1 -4 + i (7- 2 )$
$color(brown)(=> -3 + 5i$

Proof:

$1 + 7i - 4 - 2i$

$=> -3 + 5i$

How do you add (1+7i)+(-4-2i)(1+7i)+(-4-2i) in trigonometric form?