The problem asks to evaluate two complex number related expressions: b) Evaluate $(1 - \sqrt{3}i)^8$ using De Moivre's theorem. c) Find the roots of $(\sqrt{3} - i)^{\frac{1}{5}}$.

1. Problem Description

The problem asks to evaluate two complex number related expressions:

b) Evaluate $(1 - \sqrt{3}i)^8$ using De Moivre's theorem.

c) Find the roots of $(\sqrt{3} - i)^{\frac{1}{5}}$.

2. Solution Steps

b) To evaluate $(1 - \sqrt{3}i)^8$ using De Moivre's theorem, we first convert the complex number $1 - \sqrt{3}i$ to polar form.

The modulus is $r = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$.

The argument is $\theta = \arctan\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}$.

So, $1 - \sqrt{3}i = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)$.

Using De Moivre's theorem, we have

$(1 - \sqrt{3}i)^8 = \left[2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)\right]^8 = 2^8\left(\cos\left(-\frac{8\pi}{3}\right) + i\sin\left(-\frac{8\pi}{3}\right)\right)$.

Since $-\frac{8\pi}{3} = -\frac{6\pi}{3} - \frac{2\pi}{3} = -2\pi - \frac{2\pi}{3}$,

$\cos\left(-\frac{8\pi}{3}\right) = \cos\left(-\frac{2\pi}{3}\right) = -\frac{1}{2}$ and

$\sin\left(-\frac{8\pi}{3}\right) = \sin\left(-\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2}$.

Therefore, $(1 - \sqrt{3}i)^8 = 2^8\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = 256\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -128 - 128\sqrt{3}i$.

c) To find the roots of $(\sqrt{3} - i)^{\frac{1}{5}}$, we first convert the complex number $\sqrt{3} - i$ to polar form.

The modulus is $r = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.

The argument is $\theta = \arctan\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}$.

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

Then $(\sqrt{3} - i)^{\frac{1}{5}} = \left[2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right)\right]^{\frac{1}{5}} = 2^{\frac{1}{5}}\left(\cos\left(\frac{-\frac{\pi}{6} + 2k\pi}{5}\right) + i\sin\left(\frac{-\frac{\pi}{6} + 2k\pi}{5}\right)\right)$ for $k = 0, 1, 2, 3, 4$.

The roots are $2^{\frac{1}{5}}\left(\cos\left(\frac{-\pi + 12k\pi}{30}\right) + i\sin\left(\frac{-\pi + 12k\pi}{30}\right)\right)$ for $k = 0, 1, 2, 3, 4$.

The roots are $2^{\frac{1}{5}}\left(\cos\left(-\frac{\pi}{30}\right) + i\sin\left(-\frac{\pi}{30}\right)\right)$,

$2^{\frac{1}{5}}\left(\cos\left(\frac{11\pi}{30}\right) + i\sin\left(\frac{11\pi}{30}\right)\right)$,

$2^{\frac{1}{5}}\left(\cos\left(\frac{23\pi}{30}\right) + i\sin\left(\frac{23\pi}{30}\right)\right)$,

$2^{\frac{1}{5}}\left(\cos\left(\frac{35\pi}{30}\right) + i\sin\left(\frac{35\pi}{30}\right)\right) = 2^{\frac{1}{5}}\left(\cos\left(\frac{7\pi}{6}\right) + i\sin\left(\frac{7\pi}{6}\right)\right)$,

$2^{\frac{1}{5}}\left(\cos\left(\frac{47\pi}{30}\right) + i\sin\left(\frac{47\pi}{30}\right)\right)$.

3. Final Answer

b) $(1 - \sqrt{3}i)^8 = -128 - 128\sqrt{3}i$

c) The roots of $(\sqrt{3} - i)^{\frac{1}{5}}$ are $2^{\frac{1}{5}}\left(\cos\left(\frac{(12k-1)\pi}{30}\right) + i\sin\left(\frac{(12k-1)\pi}{30}\right)\right)$ for $k = 0, 1, 2, 3, 4$.

In particular, the roots are:

$2^{\frac{1}{5}}\left(\cos\left(-\frac{\pi}{30}\right) + i\sin\left(-\frac{\pi}{30}\right)\right)$,

$2^{\frac{1}{5}}\left(\cos\left(\frac{11\pi}{30}\right) + i\sin\left(\frac{11\pi}{30}\right)\right)$,

$2^{\frac{1}{5}}\left(\cos\left(\frac{23\pi}{30}\right) + i\sin\left(\frac{23\pi}{30}\right)\right)$,

$2^{\frac{1}{5}}\left(\cos\left(\frac{7\pi}{6}\right) + i\sin\left(\frac{7\pi}{6}\right)\right)$,

$2^{\frac{1}{5}}\left(\cos\left(\frac{47\pi}{30}\right) + i\sin\left(\frac{47\pi}{30}\right)\right)$.