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 reads

\[ (\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta \]

(1) First, we obtain the magnitude, which is called modulus squared. It multiplies by its conjugate:

\[ \left| \frac{1+\sqrt{3}i}{2} \right| = \sqrt{\left(\frac{1}{2}\right)^2+ \left(\frac{\sqrt{3}}{2}\right)^2} = 1 \]

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 first separate the numerator and denominator as follows:

\[ \frac{(\sqrt{3}+i)^6}{(1+i)^6} \]

For the numerator, the magnitude and the 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

\[ |1+i| = \sqrt{2}   \]

\[    \cos\theta_2 = \frac{1}{\sqrt{2}}, \sin\theta_2 = \frac{1}{\sqrt{2}} \rightarrow \theta_2 = 45^o   \]

Namely, \(1+i = \sqrt{2}(\cos 45^o + i\sin 45^o)\).

We can have the denominator as

\[ \frac{\sqrt{3}+i}{1+i}=\frac{2(\cos 30^o + i\sin 30^o)}{\sqrt{2}(\cos 45^o + i\sin 45^o)} \]

\[    = \sqrt{2}[\cos(-15^o)+i\sin(-15^o)]   \]

Finally, use De Moivre’s theorem:

\[ (\sqrt{2}[\cos(-15^o)+i\sin(-15^o)])^6 = 2^3[\cos(-6\times 15^o) + i\sin(-6\times 15^o)]  \]

\[               = 8[\cos(-90^o) + i\sin(-90^o)]  \]

\[               =-8i \]