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 \(\displaystyle -8-8i\sqrt{3}\).

Given the polar coordinates \(\displaystyle (r,\theta )\), the  \(\displaystyle x\)-coordinate is  \(\displaystyle x= r \cos \theta\). We can find this coordinate by substituting \(\displaystyle r = 2.1, \theta = \frac{7\pi }{3}\):

\(\displaystyle x = r \cos \theta = 2.1 \cdot \cos \frac{7\pi }{3} = 2.1 \cdot 0.5 = 1.05\)

Likewise, given the polar coordinates \(\displaystyle (r,\theta )\), the  \(\displaystyle y\)-coordinate is  \(\displaystyle y= r \sin \theta\).  We can find this coordinate by substituting \(\displaystyle r = 2.1, \theta = \frac{7\pi }{3}\):

\(\displaystyle y = r \sin \theta = 2.1 \cdot \sin \frac{7\pi }{3} \approx 2.1 \cdot 0.8660 \approx 1.82\)

Therefore the rectangular coordinates of the point \(\displaystyle \left(2.1, \frac{7\pi }{3}\right )\) are \(\displaystyle (1.05, 1.82)\).

To convert this number to rectangular form, first think about what \(\displaystyle \cos240^\circ\) and \(\displaystyle \sin240^\circ\) are equal to. Because \(\displaystyle 240^\circ=180^\circ+60^\circ\), we can use a 30-60-90^o reference triangle in the 3rd quadrant to determine these values. 

\(\displaystyle \cos240^\circ=-\frac{1}{2}\)

\(\displaystyle \sin240^\circ=-\frac{\sqrt{3}}{2}\)

Now plug these in and continue solving:

\(\displaystyle z=4(\cos240^\circ+i\sin240^\circ)\)

\(\displaystyle z=4(-\frac{1}{2}-i\frac{\sqrt{3}}{2})\)

\(\displaystyle z=-2-2\sqrt{3}i\)

This problem has given us formulas, so we just need to plug in \(\displaystyle x=7\) and \(\displaystyle y=13\) and solve. 

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle r=\sqrt{7^2+13^2}\)

\(\displaystyle r=\sqrt{218}\)

\(\displaystyle \theta=\tan^-^1\left (\frac{y}{x} \right )\)

\(\displaystyle \theta=\tan^-^1\left (\frac{13}{7} \right )\)

\(\displaystyle \theta=61.7^\circ\)

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

\(\displaystyle z=\sqrt{218}(\cos61.7^\circ+i\sin61.7^\circ)\)