Why ${(\frac{-\sqrt3}2+\frac i2)}^{12}$ in polar form is $cis(10\pi)$

Question

Use DeMoivre’s theorem to find ${(\frac{-\sqrt3}2+\frac i2)}^{12}$. Express your answer in polar form.

a. 1

b. cis (10π)

c. cis (5π)

d. cis (π)

e. None of these

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${(\frac{-\sqrt3}2+\frac i2)}^{12}$

Then, I got:

$r=1$

$\theta = 30^{\circ}$

Using DeMoivre’s theorem,

$r^{n}\ cis(\theta) = 1^{12}\ cis\ 30^{\circ} = cis\ \frac{\pi}{6}$

I got $cis\ \frac{\pi}{6}$.

Why the correct answer is $10\pi$?

Answer 1

Remember to include the negative sign on the real part of the complex number. Because it's negative, so the terminal side is in the second quadrant, the angle will be $\pi - \frac{\pi}{6}=\frac{5\pi}{6}.$

Answer 2

You have $\cos(\theta) = -\sqrt{3}/2 < 0$, so $\theta$ is in quadrants II or III, and $\sin(\theta) = 1/2 > 0$, so $\theta$ is in quadrants I or II. So $\theta$ is in quadrant II, $3\pi/6 < \theta < 6\pi/6$.

You have the correct reference angle, $\pi/6$, but the corresponding angle in quadrant II is $5\pi/6$.