We can use DeMoivre's formula which states:

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

Now plugging in our values of $\displaystyle \small n=3$ and $\displaystyle r=1$ we get the desired result.

$\displaystyle \small (\cos x+i\sin x)^n=(\cos nx+i\sin nx)$

$\displaystyle \cos 3x +i \sin 3x$

First, convert this complex number to polar form.

$\displaystyle r= \sqrt{ a^2 + b^2 } = \sqrt{1^2 +(-1)^2 }=\sqrt{2}$

$\displaystyle \sin \theta = \frac{-1}{\sqrt2}$ 

$\displaystyle \theta = \frac{5 \pi }{4} \enspace or \enspace \frac{7 \pi }{4}$

Since the point has a positive real part and a negative imaginary part, it is located in quadrant IV, so the angle is $\displaystyle \frac{7\pi}{4}$.

This gives us $\displaystyle (\sqrt{2} ( \cos \frac{7 \pi }{4} + i \sin \frac{ 7 \pi }{4} )) ^ 8$

To evaluate, use DeMoivre's Theorem:

DeMoivre's Theorem is

$\displaystyle (\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ 

We apply it to our situation to get:

$\displaystyle (\sqrt2 )^8 ( \cos \frac{ 7 \pi }{4} \cdot 8 + i \sin \frac{ 7 \pi }{4} \cdot 8 )$ simplifying

$\displaystyle 2 ^ 4 ( \cos 14 \pi + i \sin 14 \pi )$,  $\displaystyle 14 \pi$ is coterminal with $\displaystyle 0$ since it is an even multiple of $\displaystyle \pi$

$\displaystyle 2^ 4 (\cos 0 + i \sin 0 ) = 16(1 + 0i) = 16$

First convert this point to polar form:

$\displaystyle r = \sqrt{(3\sqrt3)^2 + (-3)^2 } = \sqrt{9 \cdot 3 + 9 } = \sqrt{36} = 6$

$\displaystyle \cos \theta = \frac{ 3\sqrt3}{6 } = \frac{\sqrt3}{2}$

$\displaystyle \theta = \cos ^{-1} (\frac{\sqrt3}{2} ) = \frac{ \pi }{6} \enspace \or \enspace \frac{ 11 \pi }{6}$

Since this number has a negative imaginary part and a positive real part, it is in quadrant IV, so the angle is $\displaystyle \frac{ 11 \pi }{6}$

We are evaluating $\displaystyle (6 \cos \frac{11 \pi }{6} + i \sin \frac{ 11 \pi }{6} ) ^ 6$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

$\displaystyle (\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$

We apply it to our situation to get:

$\displaystyle 6^6 (\cos \frac{ 11 \pi }{6} \cdot 6 + i \sin \frac{ 11 \pi }{6} \cdot 6)$

$\displaystyle \frac{11 \pi }{6} \cdot 6 = 11 \pi$ which is coterminal with $\displaystyle \pi$ since it is an odd multiplie

$\displaystyle 46,656(\cos \pi + i \sin \pi ) = 46,656 (-1 + 0i) = -46,656$

First, convert the complex number to polar form:

$\displaystyle r = \sqrt{ 2^2 + 2^2 } = \sqrt{4 + 4 } = \sqrt{2 \cdot 4 } = 2 \sqrt 2$

$\displaystyle \sin \theta = \frac{ 2 }{ 2 \sqrt 2 } = \frac{ 1 }{ \sqrt2 }$

$\displaystyle \theta = \sin ^{-1 } (\frac{ 1}{ \sqrt 2 }) = \frac{ \pi }{4} \enspace or \enspace \frac{ 3 \pi }{4}$

Since both the real and the imaginary parts are positive, the angle is in quadrant I, so it is $\displaystyle \frac{ \pi }{4}$

This means we're evaluating

$\displaystyle (2 \sqrt2 (\cos \frac{ \pi }{4} + i \sin \frac{ \pi }{4} ))^ 9$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

$\displaystyle (\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ 

We apply it to our situation to get.

$\displaystyle (2 \sqrt2 )^9 (\cos \frac{\pi }{4} \cdot 9 + i \sin \frac{ \pi }{4} \cdot 9 )$

First, evaluate $\displaystyle (2\sqrt2)^9$. We can split this into $\displaystyle 2^9 \cdot (\sqrt2)^9$ which is equivalent to $\displaystyle 2^9 \cdot (\sqrt2)^8 \cdot \sqrt2 = 2^9 \cdot 2^4 \cdot \sqrt 2$

[We can re-write the middle exponent since $\displaystyle \sqrt 2$ is equivalent to $\displaystyle 2^{\frac{1}{2}}$]

This comes to $\displaystyle 8,192 \sqrt 2$

Evaluating sine and cosine at $\displaystyle \frac{\pi }{4} \cdot 9 = \frac{ 9 \pi }{4}$ is equivalent to evaluating them at $\displaystyle \frac{ \pi }{4}$ since $\displaystyle \frac{ 9 \pi }{4} - 2 \pi = \frac{9 \pi }{4 } - \frac{ 8 \pi }{4} = \frac{ \pi }{4}$

This means our expression can be written as: 

$\displaystyle 8,192 \sqrt2 ( \cos \frac{\pi }{4} + i \sin \frac{ \pi }{4} ) = 8,192 \sqrt2 (\frac{1}{\sqrt2} + i \frac{ 1}{\sqrt2}) = 8,192 + 8,192i$

Begin by converting the complex number to polar form:

$\displaystyle 3+4i=5(\cos53.13^\circ+i\sin53.13^\circ)$

Next, put this in its generalized form, using k which is any integer, including zero:

$\displaystyle 5[\cos(53.13^\circ+k360^\circ)+i\sin(53.13^\circ+k360^\circ)]$

Using De Moivre's theorem, a fifth root of $\displaystyle 3+4i$ is given by:

$\displaystyle \left [5[\cos(53.13^\circ+k360^\circ)+i\sin(53.13^\circ+k360^\circ)] \right ]^\frac{1}{5}$

$\displaystyle =5^\frac{1}{5}\left (\cos\frac{53.13^\circ+k360^\circ}{5}+i\sin\frac{53.13^\circ+k360^\circ}{5} \right )$

$\displaystyle =\sqrt[5]{5}\left [\cos(10.63^\circ+k72^\circ)+i\sin(10.63^\circ+k72^\circ) \right ]$

Assigning the values $\displaystyle k=0,1,2,3,4$ will allow us to find the following roots. In general, use the values $\displaystyle k=0, 1, 2, ... , n-1$.

$\displaystyle k=0: \sqrt[5]{5}\left (\cos10.63^\circ+i\sin10.63^\circ \right )=R_1$

$\displaystyle k=1: \sqrt[5]{5}\left (\cos82.63^\circ+i\sin82.63^\circ \right )=R_2$

$\displaystyle k=2: \sqrt[5]{5}\left (\cos154.63^\circ+i\sin154.63^\circ \right )=R_3$

$\displaystyle k=3: \sqrt[5]{5}\left (\cos226.63^\circ+i\sin226.63^\circ \right )=R_4$

$\displaystyle k=4: \sqrt[5]{5}\left (\cos298.63^\circ+i\sin298.63^\circ \right )=R_5$

These are the fifth roots of $\displaystyle 3+4i$.

Begin by converting the complex number to polar form:

$\displaystyle 1=\cos0^\circ+i\sin0^\circ$

Next, put this in its generalized form, using k which is any integer, including zero:

$\displaystyle 1=\cos(0^\circ+k360^\circ)+i\sin(0^\circ+k360^\circ)$

Using De Moivre's theorem, a fifth root of 1 is given by:

$\displaystyle [\cos(k360^\circ)+i\sin(k360^\circ)] ^\frac{1}{3}$

$\displaystyle =1^\frac{1}{3}\cdot \left [\cos\left (\frac{k360^\circ}{3} \right )+i\sin\left (\frac{k360^\circ}{3} \right ) \right ]$

$\displaystyle =\cos k 120^\circ+i\sin k120^\circ$

Assigning the values $\displaystyle k=0, 1, 2$ will allow us to find the following roots. In general, use the values $\displaystyle k=0, 1, 2, ... , n-1$.

$\displaystyle k=0: \cos0^\circ+i\sin0^\circ=1=R_1$

$\displaystyle k=1: \cos120^\circ+i\sin120^\circ=-\frac{1}{2}+i\frac{1}{2}\sqrt{3}=R_2$

$\displaystyle k=2: \cos240^\circ+i\sin240^\circ=-\frac{1}{2}-i\frac{1}{2}\sqrt{3}=R_3$

These are the cube roots of 1.

Begin by converting the complex number to polar form:

$\displaystyle -8-8i\sqrt{3}=16(\cos240^\circ+i\sin240^\circ)$

Next, put this in its generalized form, using k which is any integer, including zero:

$\displaystyle 16[\cos(240^\circ+k360^\circ)+i\sin(240^\circ+k360^\circ)]$

Using De Moivre's theorem, a fifth root of $\displaystyle -8-8i\sqrt{3}$ is given by:

$\displaystyle \left [16\cos(240^\circ+k360^\circ)+i\sin(240^\circ+k360^\circ) \right ]^\frac{1}{4}$

$\displaystyle =16^\frac{1}{4}\left [\cos\left (\frac{240^\circ}{4}+\frac{k360^\circ}{4} \right )+i\sin\left (\frac{240^\circ}{4}+\frac{k360^\circ}{4} \right ) \right ]$

$\displaystyle =2\left [\cos\left (60^\circ+k90^\circ \right )+i\sin\left (60^\circ+k90^\circ \right ) \right ]$

Assigning the values $\displaystyle k=0,1,2,3$ will allow us to find the following roots. In general, use the values $\displaystyle k=0, 1, 2, ... , n-1$.

$\displaystyle \\k=0: 2\left (\cos60^\circ+i\sin60^\circ \right )=2\left ( \frac{1}{2}+i\frac{1}{2}\sqrt{3} \right )=1+i\sqrt{3}=R_1\\ k=1: 2\left (\cos150^\circ+i\sin150^\circ \right )=2\left ( -\frac{1}{2}\sqrt{3}+\frac{1}{2}i \right )=-\sqrt{3}+i=R_2\\ k=2: 2\left (\cos240^\circ+i\sin240^\circ \right )=2\left ( -\frac{1}{2}-i\frac{1}{2}\sqrt{3} \right )=-1-i\sqrt{3}=R_3\\ k=3: 2\left (\cos330^\circ+i\sin330^\circ \right )=2\left ( \frac{1}{2}\sqrt{3}-\frac{1}{2}i \right )=\sqrt{3}-i=R_4$

These are the fifth roots of $\displaystyle -8-8i\sqrt{3}$.