How do you convert $- 6 - 6 i \sqrt{3}$ $-6 - 6i sqrt 3$ to polar form?

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How do you convert $- 6 - 6 i \sqrt{3}$ $-6 - 6i sqrt 3$ to polar form?

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Answer 1 · 2016-07-14T11:51:34

$12 (sin (\frac{- 5 π}{6})) - i cos (\frac{- 5 π}{6}))$ $12(sin ((-5pi)/6)) - i cos ((-5pi)/6))$

Explanation:

In polar form, $(r sin θ + r i cos θ)$ $(rsin theta + r i cos theta)$ ... $I$ $I$
Comparing it with $- 6 - 6 i \sqrt{3}$ $-6-6isqrt3$
we get $\to - 6 = r sin θ$ $rarr -6 = r sin theta$ ... 1
$\to - 6 \sqrt{3} = r cos θ$ $rarr -6sqrt3 = r cos theta$ ... 2

Squaring and adding 1 and 2,
$r^{2} {sin}^{2} θ + r^{2} {cos}^{θ} = {(- 6)}^{2} + {(- 6 \sqrt{3})}^{2}$ $r^2 sin^2 theta + r^2 cos^theta = (-6)^2 + (-6sqrt3)^2$
$r^{2} ({sin}^{2} θ + {cos}^{2} θ) = 36 + 108$ $r^2(sin^2 theta + cos^2 theta) = 36 + 108$
$r^{2} = 144$ $r^2 = 144$
$r = 12$ $r = 12$

Dividing equation 1 by 2
$tan θ = \frac{- 6}{- 6 \sqrt{3}}$ $tan theta =( -6)/(-6sqrt3)$
$tan θ = \frac{1}{\sqrt{3}}$ $tan theta = 1/sqrt3$
$θ = \frac{π}{6}$ $theta = pi/6$

Real part of complex no = $- 6 (- x)$ $-6 (-x)$
Imaginary part of complex no = $- 6 \sqrt{3} (- y)$ $-6sqrt3 (-y)$
$:.$ the point is in 3rd quadrant.

At 3rd quadrant $alpha = theta - pi$
$alpha = (-5pi)/6$

Substituting the value in equation $I$ we get,
$12(sin ((-5pi)/6)) - i cos ((-5pi)/6))$

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How do you convert $- 6 - 6 i \sqrt{3}$ $-6 - 6i sqrt 3$ to polar form?

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