QuestionMay 5, 2026

Convert into a polar coordinate: 2sqrt (3)-2i (4,(7pi )/(6)) (-4,(11pi )/(6)) (-4,(5pi )/(6)) (4,(pi )/(6))

Convert into a polar coordinate: 2sqrt (3)-2i (4,(7pi )/(6)) (-4,(11pi )/(6)) (-4,(5pi )/(6)) (4,(pi )/(6))
Convert into a polar coordinate: 2sqrt (3)-2i (4,(7pi )/(6)) (-4,(11pi )/(6)) (-4,(5pi )/(6)) (4,(pi )/(6))

Solution
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Answer

(4,\frac{11\pi}{6}) Explanation 1. Find magnitude r r = \sqrt{(2\sqrt{3})^2 + (-2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4 2. Find angle \theta \theta = \tan^{-1}\left(\frac{-2}{2\sqrt{3}}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} (lies in 4th quadrant). To express in standard polar form (0 \leq \theta < 2\pi): \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} 3. Final polar coordinate (r, \theta) = (4, \frac{11\pi}{6})

Explanation

1. Find magnitude $r$<br /> $r = \sqrt{(2\sqrt{3})^2 + (-2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4$<br />2. Find angle $\theta$<br /> $\theta = \tan^{-1}\left(\frac{-2}{2\sqrt{3}}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$ (lies in 4th quadrant). <br /> To express in standard polar form ($0 \leq \theta < 2\pi$): $\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$<br />3. Final polar coordinate<br /> $(r, \theta) = (4, \frac{11\pi}{6})$
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