To find the magnitude of a complex number we use the formula:

\(\displaystyle |z|=\sqrt{a^2+b^2}\),

where our complex number is in the form \(\displaystyle (a+bi)\).

Therefore,

\(\displaystyle |z|=\sqrt{1^2+3^2}=\sqrt{10}\)

Note for any complex number z, we have:

\(\displaystyle {}|z^n|=|z|^n\).

Let \(\displaystyle {}z=1+i3\).  Hence \(\displaystyle {}|z|=\sqrt{10}\)

Therefore:

\(\displaystyle {}|z^{7}|=|z|^{7}=(\sqrt{10})^{7}=(\sqrt{10})^{6}(\sqrt{10})\)

This gives the result.

To find the magnitude of a complex number we use the following formula:

\(\displaystyle |z|=\sqrt{a^2+b^2}\), where \(\displaystyle (a+bi)\).

Therefore we get,

\(\displaystyle |z|=\sqrt{2^2+3^2}=\sqrt{13}\).

Now to find

\(\displaystyle |z|^2=(\sqrt{13})^2\)

\(\displaystyle z=13\).

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 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 this complex number to polar form:

\(\displaystyle r = \sqrt{2^2 + (2\sqrt3)^2 } = \sqrt{4 +12} = \sqrt{16} = 4\)

\(\displaystyle \sin \theta = \frac{ -2\sqrt3 }{4} = -\frac{\sqrt3}{2}\)

\(\displaystyle \theta = \sin ^{-1} (-\frac{\sqrt3}{2}) = \frac{4 \pi }{3} \enspace or \enspace \frac{ 5 \pi }{3}\)

Since the real part is positive and the imaginary part is negative, this is in quadrant IV, so the angle is \(\displaystyle \frac{5 \pi }{3}\)

So we are evaluating \(\displaystyle (4(\cos \frac{5 \pi }{3} + i \sin \frac{5 \pi }{3} ) )^6\)

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.

\(\displaystyle 4^6 (\cos \frac{ 5 \pi }{3} \cdot 6 + i \sin \frac{ 5 \pi }{3} \cdot 6 ) = 4,096 ( \cos 10 \pi + i \sin 10 \pi)\)

\(\displaystyle 10 \pi\) is coterminal with \(\displaystyle 0\) since it is an even multiple of \(\displaystyle \pi\)

\(\displaystyle 4,096( \cos 0 + i \sin 0 ) = 4,096 ( 1 + 0i) = 4,096\)

First convert the complex number into polar form:

\(\displaystyle r = \sqrt{3^2 + (-3)^2 } = \sqrt{9+9} = \sqrt{2 \cdot 9 } = 3 \sqrt2\)

\(\displaystyle \sin \theta = \frac{3}{3 \sqrt2} = \frac{ 1}{\sqrt2}\)

\(\displaystyle \theta = \sin ^{-1 } (\frac{1}{\sqrt2}) = \frac{\pi}{4} \enspace or \enspace \frac{ 3 \pi }{4}\)

Since the real part is negative but the imaginary part is positive, the angle should be in quadrant II, so it is \(\displaystyle \frac{ 3 \pi }{4}\)

We are evaluating \(\displaystyle (3\sqrt2 (\cos \frac{ 3 \pi }{4} + i \sin \frac{ 3 \pi }{4} ) ) ^ {12}\)

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.

\(\displaystyle (3\sqrt2)^{12 } (\cos \frac{ 3 \pi }{4} \cdot 12 + i \sin \frac{3 \pi }{4} \cdot 12 )\) simplify and take the exponent

\(\displaystyle 34,012,224 (\cos 9\pi + i \sin 9 \pi )\)

\(\displaystyle 9\pi\) is coterminal with \(\displaystyle \pi\) since it is an odd multiple of pi

\(\displaystyle 34, 012, 224 ( \cos \pi + i \sin \pi ) = 34, 012, 224 ( -1 + 0i) = -34,012, 224\)

First convert this complex number to polar form:

\(\displaystyle r = \sqrt{ (\sqrt3)^2 + 1^2 } = \sqrt{ 3 + 1 } = \sqrt4 = 2\)

\(\displaystyle \sin \theta = \frac{ 1}{ 2}\) so \(\displaystyle \theta = \frac{\pi }{6} \enspace or \enspace \frac{5 \pi }{6}\)

Since this number has positive real and imaginary parts, it is in quadrant I, so the angle is \(\displaystyle \frac{ \pi }{6}\)

So we are evaluating \(\displaystyle (2(\cos \frac{ \pi }{6} + i \sin \frac{ \pi }{6} ))^3\)

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.

\(\displaystyle 2^3 (\cos \frac{3\pi}{6} + i \sin \frac{ 3\pi }{6} ) = 8 ( \cos \frac{\pi}{2} + i \sin \frac{ \pi }{2} ) = 8(0 + i ) = 8i\)

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 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: 

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