What is the trigonometric form of $(- 2 + 9 i)$ $(-2+9i)$ ?

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What is the trigonometric form of $(- 2 + 9 i)$ $(-2+9i)$ ?

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Answer 1 · 2016-01-01T04:10:09

$\sqrt{85} (cos ({tan}^{- 1} (- \frac{9}{2})) + i sin ({tan}^{- 1} (- \frac{9}{2})))$ $sqrt(85)(cos(tan^-1(-9/2))+isin(tan^-1(-9/2)))$

Explanation:

$(- 2 + 9 i)$ $(-2+9i)$

$r cos (θ) = - 2$ $rcos(theta)=-2$
$r sin (t h e a) = 9$ $rsin(thea) = 9$

Squaring both and adding

$r^{2} {cos}^{2} (θ) = 4$ $r^2cos^2(theta) = 4$
$r^{2} {sin}^{2} (θ) = 81$ $r^2sin^2(theta) = 81$

$r^{2} {cos}^{2} (θ) + r^{2} {sin}^{2} (θ) = 4 + 81$ $r^2cos^2(theta)+r^2sin^2(theta) = 4+81$
$r^{2} ({cos}^{2} (θ) + {sin}^{2} (θ)) = 85$ $r^2(cos^2(theta)+sin^2(theta))=85$
$r^{2} = 85$ $r^2=85$
$r = \sqrt{85}$ $r=sqrt(85)$

$r \frac{sin (θ)}{r} cos (θ) = \frac{9}{- 2}$ $rsin(theta)/rcos(theta) = 9/-2$
$tan (θ) = - \frac{9}{2}$ $tan(theta) = -9/2$

$θ = {tan}^{- 1} (- \frac{9}{2})$ $theta = tan^-1(-9/2)$

The complex number in trigonometric form is

$\sqrt{85} (cos ({tan}^{- 1} (- \frac{9}{2})) + i sin ({tan}^{- 1} (- \frac{9}{2})))$ $sqrt(85)(cos(tan^-1(-9/2))+isin(tan^-1(-9/2)))$

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What is the trigonometric form of $(- 2 + 9 i)$ $(-2+9i)$ ?

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Science

Math

Humanities

... and beyond

What is the trigonometric form of (−2+9i) (-2+9i) ?

1 Answer

Explanation:

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Impact of this question

What is the trigonometric form of $(- 2 + 9 i)$ $(-2+9i)$ ?