QuestionJuly 15, 2025

(b) 6sqrt (3)-6i The polar form is square Give an exact answer.

(b) 6sqrt (3)-6i The polar form is square Give an exact answer.
(b) 6sqrt (3)-6i The polar form is square Give an exact answer.

Solution
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Answer

12(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})) Explanation 1. Calculate the magnitude Use the formula r = \sqrt{a^2 + b^2} where a = 6\sqrt{3} and b = -6. So, r = \sqrt{(6\sqrt{3})^2 + (-6)^2} = \sqrt{108 + 36} = \sqrt{144} = 12. 2. Calculate the angle Use the formula \theta = \tan^{-1}\left(\frac{b}{a}\right) where b = -6 and a = 6\sqrt{3}. So, \theta = \tan^{-1}\left(\frac{-6}{6\sqrt{3}}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}. 3. Write in polar form The polar form is r(\cos \theta + i\sin \theta). Substitute r = 12 and \theta = -\frac{\pi}{6}.

Explanation

1. Calculate the magnitude<br /> Use the formula $r = \sqrt{a^2 + b^2}$ where $a = 6\sqrt{3}$ and $b = -6$. So, $r = \sqrt{(6\sqrt{3})^2 + (-6)^2} = \sqrt{108 + 36} = \sqrt{144} = 12$.<br />2. Calculate the angle<br /> Use the formula $\theta = \tan^{-1}\left(\frac{b}{a}\right)$ where $b = -6$ and $a = 6\sqrt{3}$. So, $\theta = \tan^{-1}\left(\frac{-6}{6\sqrt{3}}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$.<br />3. Write in polar form<br /> The polar form is $r(\cos \theta + i\sin \theta)$. Substitute $r = 12$ and $\theta = -\frac{\pi}{6}$.
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