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

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Answer 1 · 2018-06-25T08:29:57

$color(cyan)(=> -2 - 10 i$

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

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

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

$r_1=sqrt(-1^2+-8^2))=sqrt 65$
$r_2=sqrt(-1^2+-2^2) =sqrt 5$

$theta_1=tan^-1(-8/-1)~~ 262.87^@, " III quadrant"$
$theta_2=tan^-1(-2/-1)~~ 243.43^@, " III quadrant"$

$z_1 + z_2 = sqrt 65 cos(262.87) + sqrt 5 cos(243.43) + i (sqrt 65 sin(262.87)+ sqrt 5 sin(243.43))$

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

$color(cyan)(=> -2 - 10 i$

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