How do you express 1 – 3i in polar form?

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How do you express 1 – 3i in polar form?

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Answer 1 · 2016-01-23T20:21:37

$(\sqrt{10}, - 1.25)$ $(sqrt10 , - 1.25 )$

Explanation:

Using the formulae that links Cartesian to Polar coordinates.

$∙ r^{2} = x^{2} + y^{2}$ $• r^2 = x^2 + y^2$

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

[ 1 - 3i is in 4th quadrant and care must be taken to ensure

that $θ$ $theta$ is in the 4th quadrant ]

(Here x = 1 and y = - 3 )

$r^{2} = 1^{2} + {(- 3)}^{2} = 1 + 9 = 10$ $r^2 = 1^2 + (-3)^2 = 1 + 9 = 10$

$\Rightarrow r = \sqrt{10}$ $rArr r = sqrt10$

and $θ = {tan}^{- 1} (- 3) = - 1.25 radians$ $theta = tan^-1 (-3) = - 1.25 color(black)("radians" )$

$θ = - 1.25 is in the 4th quadrant$ $theta = - 1.25 color(black)(" is in the 4th quadrant")$

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How do you express 1 – 3i in polar form?

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