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

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

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Answer 1 · 2018-07-09T11:45:34

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

Explanation:

$z = a + b i = r (cos θ + i sin θ)$ $z= a+bi= r (costheta+isintheta)$

$r = \sqrt{a^{2} + b^{2}}, θ = {tan}^{- 1} (\frac{b}{a})$ $r=sqrt(a^2+b^2), " " theta=tan^-1(b/a)$

$r_{1} (cos (θ_{1}) + i sin (θ_{2})) + r_{2} (cos (θ_{2}) + i sin (θ_{2})) = r_{1} cos (θ_{1}) + r_{2} cos (θ_{2}) + i (r_{1} sin (θ_{1}) + r_{2} sin (θ_{2}))$ $r_1(cos(theta_1)+isin(theta_2))+r_2(cos(theta_2)+isin(theta_2))=r_1cos(theta_1)+r_2cos(theta_2)+i(r_1sin(theta_1)+r_2sin(theta_2))$

$r_{1} = \sqrt{9^{2} + - 7^{2}}) = \sqrt{130}$ $r_1=sqrt(9^2+ -7^2))=sqrt 130$
$r_{2} = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$ $r_2=sqrt(1^2+ 1^2) =sqrt 2$

$θ_{1} = {tan}^{- 1} (- \frac{7}{9}) \approx {322.13}^{\circ}, IV quadrant$ $theta_1=tan^-1(-7 / 9)~~ 322.13^@, " IV quadrant"$
$θ_{2} = {tan}^{- 1} (\frac{1}{1}) \approx 45^{\circ}, I quadrant$ $theta_2=tan^-1(1/ 1)~~ 45^@, " I quadrant"$

$z_{1} + z_{2} = \sqrt{130} cos (322.13) + \sqrt{2} cos (45) + i (\sqrt{130} sin 322.13 + \sqrt{2} sin 45)$ $z_1 + z_2 = sqrt 130 cos(322.13) + sqrt 2 cos(45) + i (sqrt 130 sin 322.13 + sqrt 2 sin 45)$

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

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

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

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