How do you convert $(\sqrt{5}) - i$ $(sqrt 5)-i$ to polar form?

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How do you convert $(\sqrt{5}) - i$ $(sqrt 5)-i$ to polar form?

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Answer 1 · 2017-08-10T10:36:55

$(\sqrt{6}, - 0.42)$ $(sqrt6,-0.42)$

Explanation:

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

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

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

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

$here x = \sqrt{5} and y = - 1$ $"here "x=sqrt5" and "y=-1$

$\Rightarrow r = \sqrt{{(\sqrt{5})}^{2} + {(- 1)}^{2}} = \sqrt{6}$ $rArrr=sqrt((sqrt5)^2+(-1)^2)=sqrt6$

$\sqrt{5} - i is in fourth quadrant so we must ensure θ$ $sqrt5-i" is in fourth quadrant so we must ensure "theta$ is in the fourth quadrant.

$θ = {tan}^{- 1} (\frac{1}{\sqrt{5}}) = 0.42 \leftarrow related acute angle$ $theta=tan^-1(1/sqrt5)=0.42larrcolor(red)" related acute angle"$

$\Rightarrow θ = - 0.42 \leftarrow in fourth quadrant$ $rArrtheta=-0.42larrcolor(red)" in fourth quadrant"$

$\Rightarrow (\sqrt{5}, - 1) \to (\sqrt{6}, - 0.42)$ $rArr(sqrt5,-1)to(sqrt6,-0.42)$

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How do you convert $(\sqrt{5}) - i$ $(sqrt 5)-i$ to polar form?

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How do you convert $(\sqrt{5}) - i$ $(sqrt 5)-i$ to polar form?