How do you add $(9-7i)+(1+i)$ in trigonometric form?

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Answer 1 · 2018-07-09T11:45:34

$color(chocolate)(=> 10 - 6i$

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

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

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

$r_1=sqrt(9^2+ -7^2))=sqrt 130$
$r_2=sqrt(1^2+ 1^2) =sqrt 2$

$theta_1=tan^-1(-7 / 9)~~ 322.13^@, " IV quadrant"$
$theta_2=tan^-1(1/ 1)~~ 45^@, " I quadrant"$

$z_1 + z_2 = sqrt 130 cos(322.13) + sqrt 2 cos(45) + i (sqrt 130 sin 322.13 + sqrt 2 sin 45)$

$=> 9 + 1 + i (-7 + 1 )$

$color(chocolate)(=> 10 - 6i$

How do you add (9-7i)+(1+i)(9-7i)+(1+i) in trigonometric form?