# Use De Moivre’s theorem to find (√3 + i) 3 .

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De Moivre's theorem states that for any complex number $$z = r(\cos \theta + i \sin \theta)$$, where $$r$$ is the modulus of $$z$$ and $$\theta$$ is the argument of $$z$$, the $$n$$-th power of $$z$$ is given by:

$z^n = r^n (\cos (n\theta) + i \sin (n\theta))$

For the complex number $$\sqrt{3} + i$$, we can express it in polar form:

$\sqrt{3} + i = 2\left(\cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right)\right)$

Now, raise $$\sqrt{3} + i$$ to the power of 3 using De Moivre's theorem:

$(\sqrt{3} + i)^3 = 2^3 \left(\cos\left(3 \cdot \frac{\pi}{6}\right) + i \sin\left(3 \cdot \frac{\pi}{6}\right)\right)$

Simplify the expression:

$(\sqrt{3} + i)^3 = 8 \left(\cos\left(\frac{\pi}{2}\right) + i \sin\left(\frac{\pi}{2}\right)\right)$

Since $$\cos\left(\frac{\pi}{2}\right) = 0$$ and $$\sin\left(\frac{\pi}{2}\right) = 1$$, the expression becomes:

$(\sqrt{3} + i)^3 = 8 \cdot (0 + i) = 8i$

Therefore, $$(\sqrt{3} + i)^3 = 8i$$.