How do you add $(- 7 + 9 i)$ $(-7+9i)$ and $(- 1 + 6 i)$ $(-1+6i)$ in trigonometric form?

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How do you add $(- 7 + 9 i)$ $(-7+9i)$ and $(- 1 + 6 i)$ $(-1+6i)$ in trigonometric form?

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Answer 1 · 2018-07-09T11:27:26

$\Rightarrow - 8 + 15 i$ $color(indigo)(=> -8+ 15i$

Explanation:

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

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

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

$r_{1} = \sqrt{- 7^{2} + 9^{2}}) = \sqrt{130}$ $r_1=sqrt(-7^2+ 9^2))=sqrt 130$
$r_{2} = \sqrt{- 1^{2} + 6^{2}} = \sqrt{37}$ $r_2=sqrt(-1^2+ 6^2) =sqrt 37$

$θ_{1} = {tan}^{- 1} (\frac{9}{- 7}) \approx {127.87}^{\circ}, II quadrant$ $theta_1=tan^-1(9 / -7)~~ 127.87^@, " II quadrant"$
$θ_{2} = {tan}^{- 1} (\frac{6}{- 1}) \approx {333.43}^{\circ}, II quadrant$ $theta_2=tan^-1(6/ -1)~~ 333.43^@, " II quadrant"$

$z_{1} + z_{2} = \sqrt{130} cos (127.87) + \sqrt{37} cos (99.46) + i (\sqrt{130} sin 127.87 + \sqrt{37} sin 99.46)$ $z_1 + z_2 = sqrt 130 cos(127.87) + sqrt 37 cos(99.46) + i (sqrt 130 sin 127.87 + sqrt 37 sin 99.46)$

$\Rightarrow - 7 - 1 + i (9 + 6)$ $=> -7 - 1 + i (9 + 6 )$

$\Rightarrow - 8 + 15 i$ $color(indigo)(=> -8+ 15i$

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How do you add $(- 7 + 9 i)$ $(-7+9i)$ and $(- 1 + 6 i)$ $(-1+6i)$ in trigonometric form?

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Explanation:

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Math

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... and beyond

How do you add (−7+9i)(-7+9i) and (−1+6i)(-1+6i) in trigonometric form?

1 Answer

Explanation:

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How do you add $(- 7 + 9 i)$ $(-7+9i)$ and $(- 1 + 6 i)$ $(-1+6i)$ in trigonometric form?