De Moivre's theorem and application

Question:
By using De Moivre's theorem, simplify the following expressions:
(1) $\left(\frac{1+\sqrt{3}i}{2}\right)^{10}$
(2) $\left(\frac{\sqrt{3}+i}{1+i}\right)^{6}$

Answer:
De Moivre's theorem shows
\[ (\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta \]
(1) First, we obtain the magnitude.
\[ \left| \frac{1+\sqrt{3}i}{2} \right| = \sqrt{\left(\frac{1}{2}\right)^2+ \left(\frac{\sqrt{3}}{2}\right)^2} = 1 \]
The argument $\theta$ ranges from 0 to 360 degrees. Comparing with $\cos\theta + i\sin\theta$, we can have
\[ \cos\theta = \frac{1}{2}, \sin\theta = \frac{\sqrt{3}}{2} \]
Therefore, $\theta = 60^o$. Use the theorem.
\[ (\cos 60^o + i\sin 60^o)^{10} = \cos(10\times60^o)+i\sin(10\times60^o)   \\
                     = \cos 600^o + i\sin 600^o   \\
                     = \cos 240^o + i\sin 240^o    \\
                     = -\frac{1}{2}-\frac{\sqrt{3}}{2}i  \]

(2)
Let us separate the numerator and denominator.
\[ \frac{(\sqrt{3}+i)^6}{(1+i)^6} \]
For the numerator, the magnitude and argument are
\[ |\sqrt{3}+i| = 2   \\
    \cos\theta_1 = \frac{\sqrt{3}}{2}, \sin\theta_1 = \frac{1}{2} \rightarrow \theta_1 = 30^o   \]
Thus, $\sqrt{3}+i = 2(\cos 30^o + i\sin 30^o)$.
For the denominator, we have
\[ |1+i| = \sqrt{2}   \\
    \cos\theta_2 = \frac{1}{\sqrt{2}}, \sin\theta_2 = \frac{1}{\sqrt{2}} \rightarrow \theta_2 = 45^o   \]
Thus, $1+i = \sqrt{2}(\cos 45^o + i\sin 45^o)$.
We have
\[ \frac{\sqrt{3}+i}{1+i}=\frac{2(\cos 30^o + i\sin 30^o)}{\sqrt{2}(\cos 45^o + i\sin 45^o)} \\
    = \sqrt{2}[\cos(-15^o)+i\sin(-15^o)]   \]
Use the theorem.
\[ (\sqrt{2}[\cos(-15^o)+i\sin(-15^o)])^6 = 2^3[\cos(-6\times 15^o) + i\sin(-6\times 15^o)]  \\
               = 8[\cos(-90^o) + i\sin(-90^o)]  \\
               =-8i \]

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