How do you convert $8 - 15 i$ $8 - 15i$ to polar form?

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How do you convert $8 - 15 i$ $8 - 15i$ to polar form?

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Answer 1 · 2017-05-06T09:42:41

$(17, - 1.08)$ $(17,-1.08)$

Explanation:

$to convert from cartesian to polar form$ $"to convert from "color(blue)"cartesian to polar form"$

$that is (x, y) \to (r, θ) where$ $"that is " (x,y)to(r,theta)" where"$

$∙ r = \sqrt{x^{2} + y^{2}}$ $• r=sqrt(x^2+y^2)$

$∙ θ = {tan}^{- 1} (\frac{y}{x}); - π < θ \leq π$ $• theta=tan^-1(y/x); -pi < theta <= pi$

$here (x, y) = (8, - 15)$ $"here " (x,y)=(8,-15)$

$\Rightarrow r = \sqrt{8^{2} + {(- 15)}^{2}} = 17$ $rArrr=sqrt(8^2+(-15)^2)=17$

$now 8 - 15 i is in the fourth quadrant so we must ensure$ $"now " 8-15i" is in the fourth quadrant so we must ensure"$
$that θ is in the fourth quadrant$ $"that "theta" is in the fourth quadrant"$

$θ = {tan}^{- 1} (- \frac{15}{8}) = - 1.08 \leftarrow in fourth quadrant$ $theta=tan^-1(-15/8)=-1.08larrcolor(red)" in fourth quadrant"$

$\Rightarrow 8 - 15 i \to (17, - 1.08)$ $rArr8-15ito(17,-1.08)$

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How do you convert $8 - 15 i$ $8 - 15i$ to polar form?

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