How do you convert $- 7 - 7 \sqrt{3} i$ $-7-7sqrt3i$ to polar form?

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How do you convert $- 7 - 7 \sqrt{3} i$ $-7-7sqrt3i$ to polar form?

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Answer 1 · 2017-09-13T18:45:35

$(14, \frac{4 π}{3})$ $(14,(4pi)/3)$

Explanation:

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

$that is x + y i \to (x, y) \to (r, θ) using$ $"that is "x+yito(x,y)to(r,theta)" using"$

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

$∙ x θ = {tan}^{- 1} (\frac{y}{x})$ $•color(white)(x)theta=tan^-1(y/x)$

$here x = - 7 and y = - 7 \sqrt{3}$ $"here "x=-7" and "y=-7sqrt3$

$\Rightarrow r = \sqrt{{(- 7)}^{2} + {(- 7 \sqrt{3})}^{2}}$ $rArrr=sqrt((-7)^2+(-7sqrt3)^2)$

$\Rightarrow r = \sqrt{49 + 147} = \sqrt{196} = 14$ $color(white)(rArrr)=sqrt(49+147)=sqrt196=14$

$since - 7 - 7 \sqrt{3} i is in the third quadrant we must$ $"since "-7-7sqrt3i" is in the third quadrant we must"$
$ensure that θ is in the third quadrant$ $"ensure that "theta" is in the third quadrant"$

$θ = {tan}^{- 1} (\sqrt{3}) = \frac{π}{3} \leftarrow related acute angle$ $theta=tan^-1(sqrt3)=pi/3larrcolor(blue)"related acute angle"$

$\Rightarrow θ = (π + \frac{π}{3}) = \frac{4 π}{3} \leftarrow in third quadrant$ $rArrtheta=(pi+pi/3)=(4pi)/3larrcolor(blue)" in third quadrant"$

$\Rightarrow (- 7, - 7 \sqrt{3}) \to (14, \frac{4 π}{3})$ $rArr(-7,-7sqrt3)to(14,(4pi)/3)$

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How do you convert $- 7 - 7 \sqrt{3} i$ $-7-7sqrt3i$ to polar form?

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