How do you convert $- 6 + j 3$ $-6 + j3$ to polar form?

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How do you convert $- 6 + j 3$ $-6 + j3$ to polar form?

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Answer 1 · 2018-06-24T06:08:53

$Polar form is (6.7082, {153.43}^{\circ}$ $color(orange)("Polar form is "(6.7082, 153.43^@$

Explanation:

$z = - 6 + j 3, also = r, θ$ $z= -6 + j 3, " also " = r, theta$

$r = \sqrt{{(- 6)}^{2} + 3^{2}} = \sqrt{45} = 6.7082$ $r = sqrt((-6)^2 + 3^2) = sqrt45 = 6.7082$

$θ = {tan}^{- 1} (\frac{3}{- 6}) = {tan}^{- 1} (- 0.5) = - {26.57}^{\circ} = {153.43}^{\circ}, point lies in II quadrant$ $theta = tan ^ (-1) (3 / -6) = tan ^ (-1) (-0.5) = -26.57^@ = 153.43^@, " point lies in II quadrant"$

$Polar form is (6.7082, {153.43}^{\circ}$ $color(orange)("Polar form is "(6.7082, 153.43^@$

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How do you convert $- 6 + j 3$ $-6 + j3$ to polar form?

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