De Moivre's theorem and Roots of unity Questions in English

(A) Given $y = \cos \theta + i\sin \theta = e^{i\theta}$.
By De Moivre's theorem,$\frac{1}{y} = y^{-1} = (e^{i\theta})^{-1} = e^{-i\theta} = \cos \theta - i\sin \theta$.
Therefore,$y + \frac{1}{y} = (\cos \theta + i\sin \theta) + (\cos \theta - i\sin \theta) = 2\cos \theta$.

(D) We need to find the cube roots of $-i$.
In polar form,$-i = \cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})$.
The cube roots are given by $z_k = \cos(\frac{3\pi/2 + 2k\pi}{3}) + i\sin(\frac{3\pi/2 + 2k\pi}{3})$ for $k = 0, 1, 2$.
For $k=0$: $z_0 = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}) = i$.
For $k=1$: $z_1 = \cos(\frac{7\pi}{6}) + i\sin(\frac{7\pi}{6}) = -\frac{\sqrt{3}}{2} - \frac{1}{2}i = \frac{-\sqrt{3} - i}{2}$.
For $k=2$: $z_2 = \cos(\frac{11\pi}{6}) + i\sin(\frac{11\pi}{6}) = \frac{\sqrt{3}}{2} - \frac{1}{2}i = \frac{\sqrt{3} - i}{2}$.
Since both $\frac{-\sqrt{3} - i}{2}$ and $\frac{\sqrt{3} - i}{2}$ are roots of $(-i)^{1/3}$,the correct option is $D$.

(C) We express the complex number in polar form: $1 + i\sqrt{3} = 2\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 2e^{i\pi/3}$.
Using De Moivre's theorem: $(1 + i\sqrt{3})^9 = (2e^{i\pi/3})^9 = 2^9 \cdot e^{i(3\pi)}$.
Since $e^{i(3\pi)} = \cos(3\pi) + i\sin(3\pi) = -1 + i(0) = -1$,we have $(1 + i\sqrt{3})^9 = 2^9(-1) = -512$.
Comparing this with $a + ib$,we get $a = -512$ and $b = 0$.

(C) Given $z = \frac{1 + i\sqrt{3}}{\sqrt{3} + i}$.
Multiply numerator and denominator by the conjugate of the denominator: $z = \frac{1 + i\sqrt{3}}{\sqrt{3} + i} \times \frac{\sqrt{3} - i}{\sqrt{3} - i}$.
$z = \frac{\sqrt{3} - i + 3i - i^2\sqrt{3}}{3 + 1} = \frac{\sqrt{3} + 2i + \sqrt{3}}{4} = \frac{2\sqrt{3} + 2i}{4} = \frac{\sqrt{3} + i}{2}$.
$z = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)$.
Then $\bar{z} = \cos\left(\frac{\pi}{6}\right) - i\sin\left(\frac{\pi}{6}\right)$.
Using De Moivre's Theorem,$(\bar{z})^{100} = \left[\cos\left(\frac{\pi}{6}\right) - i\sin\left(\frac{\pi}{6}\right)\right]^{100} = \cos\left(\frac{100\pi}{6}\right) - i\sin\left(\frac{100\pi}{6}\right)$.
$(\bar{z})^{100} = \cos\left(\frac{50\pi}{3}\right) - i\sin\left(\frac{50\pi}{3}\right)$.
Since $\frac{50\pi}{3} = 16\pi + \frac{2\pi}{3}$,we have $(\bar{z})^{100} = \cos\left(\frac{2\pi}{3}\right) - i\sin\left(\frac{2\pi}{3}\right)$.
This is equal to $-\frac{1}{2} - i\frac{\sqrt{3}}{2}$.
Since both the real and imaginary parts are negative,$(\bar{z})^{100}$ lies in the $III$ quadrant.

(D) Let $z = -1 + i\sqrt{3}$. The modulus $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$.
We can write $z = 2 \left( -\frac{1}{2} + i\frac{\sqrt{3}}{2} \right) = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)$.
Using De Moivre's Theorem,$z^{20} = 2^{20} \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)^{20} = 2^{20} \left( \cos \frac{40\pi}{3} + i \sin \frac{40\pi}{3} \right)$.
Since $\frac{40\pi}{3} = 13\pi + \frac{\pi}{3} = 12\pi + \pi + \frac{\pi}{3} = 6(2\pi) + \frac{4\pi}{3}$,we have $\cos \frac{40\pi}{3} = \cos \frac{4\pi}{3} = -\frac{1}{2}$ and $\sin \frac{40\pi}{3} = \sin \frac{4\pi}{3} = -\frac{\sqrt{3}}{2}$.
Thus,$z^{20} = 2^{20} \left( -\frac{1}{2} - i\frac{\sqrt{3}}{2} \right) = 2^{19}(-1 - i\sqrt{3})$.
None of the given options match this result.

(C) We know that $i = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})$.
Applying De Moivre's theorem for the square root:
$\sqrt{i} = (i)^{1/2} = [\cos(\frac{\pi}{2} + 2n\pi) + i\sin(\frac{\pi}{2} + 2n\pi)]^{1/2}$ for $n = 0, 1$.
For $n = 0$:
$\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} = \frac{1 + i}{\sqrt{2}}$.
For $n = 1$:
$\cos(\frac{5\pi}{4}) + i\sin(\frac{5\pi}{4}) = -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} = -\frac{1 + i}{\sqrt{2}}$.
Thus,$\sqrt{i} = \pm \frac{1 + i}{\sqrt{2}}$.
Trick: Squaring the option $\pm \frac{1 + i}{\sqrt{2}}$ gives $\frac{(1 + i)^2}{2} = \frac{1 + 2i + i^2}{2} = \frac{1 + 2i - 1}{2} = i$.

(D) Given expression: $\frac{(\cos \theta + i\sin \theta)^4}{(\sin \theta + i\cos \theta)^5}$
We know that $\sin \theta + i\cos \theta = i(\cos \theta - i\sin \theta) = i(\cos \theta + i\sin \theta)^{-1}$.
Substituting this into the denominator:
$\frac{(\cos \theta + i\sin \theta)^4}{[i(\cos \theta + i\sin \theta)^{-1}]^5} = \frac{(\cos \theta + i\sin \theta)^4}{i^5(\cos \theta + i\sin \theta)^{-5}}$
Since $i^5 = i$,we have:
$\frac{(\cos \theta + i\sin \theta)^4}{i(\cos \theta + i\sin \theta)^{-5}} = \frac{1}{i} (\cos \theta + i\sin \theta)^{4 - (-5)} = \frac{1}{i} (\cos \theta + i\sin \theta)^9$
Using De Moivre's Theorem:
$\frac{1}{i} (\cos 9\theta + i\sin 9\theta) = -i(\cos 9\theta + i\sin 9\theta) = -i\cos 9\theta - i^2\sin 9\theta = \sin 9\theta - i\cos 9\theta$.

(B) Given $z = {\left( {\frac{{\sqrt 3 }}{2} + i\frac{1}{2}} \right)^5} + {\left( {\frac{{\sqrt 3 }}{2} - i\frac{1}{2}} \right)^5}$.
We know that $\frac{{\sqrt 3 }}{2} + i\frac{1}{2} = \cos \left( {\frac{\pi }{6}} \right) + i\sin \left( {\frac{\pi }{6}} \right) = e^{i\pi/6}$.
Similarly,$\frac{{\sqrt 3 }}{2} - i\frac{1}{2} = \cos \left( {\frac{\pi }{6}} \right) - i\sin \left( {\frac{\pi }{6}} \right) = e^{-i\pi/6}$.
Using De Moivre's Theorem,$z = (e^{i\pi/6})^5 + (e^{-i\pi/6})^5 = e^{i5\pi/6} + e^{-i5\pi/6}$.
This simplifies to $z = 2\cos \left( {\frac{{5\pi }}{6}} \right)$.
Since $\cos \left( {\frac{{5\pi }}{6}} \right) = -\frac{{\sqrt 3 }}{2}$,we have $z = 2 \times \left( -\frac{{\sqrt 3 }}{2} \right) = -\sqrt 3$.
Since $z$ is a real number,its imaginary part $\text{Im}(z) = 0$.

(A) Let $z = 2 - 2i$. The modulus $r = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$.
The argument $\theta = \tan^{-1}(\frac{-2}{2}) = -\frac{\pi}{4}$.
So,$z = 2\sqrt{2} \left( \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) \right) = 2\sqrt{2} \left( \cos(\frac{\pi}{4}) - i\sin(\frac{\pi}{4}) \right)$.
Using De Moivre's theorem,the roots are $z_k = (2\sqrt{2})^{1/3} \left( \cos \frac{2k\pi - \pi/4}{3} + i\sin \frac{2k\pi - \pi/4}{3} \right)$ for $k = 0, 1, 2$.
For $k=0$: $\sqrt{2} \left( \cos \frac{\pi}{12} - i\sin \frac{\pi}{12} \right)$.
For $k=1$: $\sqrt{2} \left( \cos \frac{7\pi}{12} + i\sin \frac{7\pi}{12} \right) = \sqrt{2} \left( -\sin \frac{\pi}{12} + i\cos \frac{\pi}{12} \right)$.
For $k=2$: $\sqrt{2} \left( \cos \frac{15\pi}{12} + i\sin \frac{15\pi}{12} \right) = \sqrt{2} \left( \cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4} \right) = \sqrt{2} \left( -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \right) = -1 - i$.

(C) Given expression: $E = (\cos 2\theta + i\sin 2\theta )^{ - 5} (\cos 3\theta - i\sin 3\theta )^6 (\sin \theta - i\cos \theta )^3$.
First,simplify the terms using De Moivre's Theorem:
$(\cos 2\theta + i\sin 2\theta )^{ - 5} = \cos(-10\theta) + i\sin(-10\theta)$.
$(\cos 3\theta - i\sin 3\theta )^6 = (\cos(-3\theta) + i\sin(-3\theta))^6 = \cos(-18\theta) + i\sin(-18\theta)$.
$(\sin \theta - i\cos \theta )^3 = [-i(\cos \theta + i\sin \theta )]^3 = (-i)^3 (\cos \theta + i\sin \theta )^3 = i(\cos 3\theta + i\sin 3\theta)$.
Multiplying these:
$E = [\cos(-10\theta) + i\sin(-10\theta)] [\cos(-18\theta) + i\sin(-18\theta)] [i(\cos 3\theta + i\sin 3\theta)]$.
$E = i [\cos(-10\theta - 18\theta + 3\theta) + i\sin(-10\theta - 18\theta + 3\theta)]$.
$E = i [\cos(-25\theta) + i\sin(-25\theta)] = i(\cos 25\theta - i\sin 25\theta)$.

(A) Given $a = \sqrt{2i}$.
We can write $2i$ in polar form as $2(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})$.
Then $a = \sqrt{2} (\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})^{1/2}$.
Using De Moivre's theorem,$a = \sqrt{2} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$.
$a = \sqrt{2} (\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}) = 1 + i$.
Alternatively,squaring the options: $(1+i)^2 = 1^2 + i^2 + 2i = 1 - 1 + 2i = 2i$. Thus,$a = 1+i$ is correct.

(B) Given expression: $L.H.S. = {\left[ \frac{1 + \cos \phi + i\sin \phi }{1 + \cos \phi - i\sin \phi } \right]^n}$
Using half-angle identities $1 + \cos \phi = 2\cos^2(\phi/2)$ and $\sin \phi = 2\sin(\phi/2)\cos(\phi/2)$:
$L.H.S. = {\left[ \frac{2\cos^2(\phi/2) + 2i\sin(\phi/2)\cos(\phi/2)}{2\cos^2(\phi/2) - 2i\sin(\phi/2)\cos(\phi/2)} \right]^n}$
$= {\left[ \frac{2\cos(\phi/2)(\cos(\phi/2) + i\sin(\phi/2))}{2\cos(\phi/2)(\cos(\phi/2) - i\sin(\phi/2))} \right]^n}$
$= {\left[ \frac{\cos(\phi/2) + i\sin(\phi/2)}{\cos(\phi/2) - i\sin(\phi/2)} \right]^n}$
Using Euler's formula $e^{i\theta} = \cos \theta + i\sin \theta$:
$= {\left[ \frac{e^{i\phi/2}}{e^{-i\phi/2}} \right]^n} = {(e^{i\phi/2 + i\phi/2})^n} = {(e^{i\phi})^n}$
$= e^{in\phi} = \cos n\phi + i\sin n\phi$.

(D) Let $z = \frac{1 + \cos \theta + i\sin \theta}{i + \sin \theta + i\cos \theta}$.
Simplifying the numerator: $1 + \cos \theta + i\sin \theta = 2\cos^2(\theta/2) + 2i\sin(\theta/2)\cos(\theta/2) = 2\cos(\theta/2) [\cos(\theta/2) + i\sin(\theta/2)]$.
Simplifying the denominator: $i + \sin \theta + i\cos \theta = i(1 + \cos \theta) + \sin \theta = i[2\cos^2(\theta/2)] + 2\sin(\theta/2)\cos(\theta/2) = 2\cos(\theta/2) [\sin(\theta/2) + i\cos(\theta/2)]$.
Since $\sin(\theta/2) + i\cos(\theta/2) = i(\cos(\theta/2) - i\sin(\theta/2)) = i(\cos(\theta/2) + i\sin(\theta/2))^{-1}$,we have:
$z = \frac{2\cos(\theta/2) [\cos(\theta/2) + i\sin(\theta/2)]}{2\cos(\theta/2) i [\cos(\theta/2) - i\sin(\theta/2)]} = \frac{1}{i} \frac{\cos(\theta/2) + i\sin(\theta/2)}{\cos(\theta/2) - i\sin(\theta/2)} = -i [\cos(\theta/2) + i\sin(\theta/2)]^2 = -i(\cos \theta + i\sin \theta)$.
Thus,$z^4 = (-i)^4 (\cos \theta + i\sin \theta)^4 = 1 \cdot (\cos 4\theta + i\sin 4\theta) = \cos 4\theta + i\sin 4\theta$.
Comparing this with $\cos n\theta + i\sin n\theta$,we get $n = 4$.

(D) We have the expression $Z = {\left( \frac{\cos \theta + i\sin \theta}{\sin \theta + i\cos \theta} \right)^4}$.
Note that $\sin \theta + i\cos \theta = i(\cos \theta - i\sin \theta)$.
Thus,$Z = \frac{(\cos \theta + i\sin \theta)^4}{i^4(\cos \theta - i\sin \theta)^4}$.
Since $i^4 = 1$,we use De Moivre's Theorem: $(\cos \theta + i\sin \theta)^n = \cos n\theta + i\sin n\theta$.
$Z = \frac{\cos 4\theta + i\sin 4\theta}{\cos 4\theta - i\sin 4\theta}$.
Multiply the numerator and denominator by the conjugate $(\cos 4\theta + i\sin 4\theta)$:
$Z = \frac{(\cos 4\theta + i\sin 4\theta)^2}{\cos^2 4\theta + \sin^2 4\theta}$.
Since $\cos^2 4\theta + \sin^2 4\theta = 1$,we get $Z = (\cos 4\theta + i\sin 4\theta)^2 = \cos 8\theta + i\sin 8\theta$.

(C) Let $z = -\sqrt{3} + i$.
First,express $z$ in polar form: $z = 2\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = 2(\cos 150^\circ + i \sin 150^\circ)$.
Using De Moivre's Theorem,$z^{53} = 2^{53}(\cos 150^\circ + i \sin 150^\circ)^{53}$.
$z^{53} = 2^{53}(\cos(53 \times 150^\circ) + i \sin(53 \times 150^\circ))$.
$53 \times 150^\circ = 7950^\circ$.
Dividing $7950^\circ$ by $360^\circ$: $7950 = 22 \times 360 + 30$.
So,$z^{53} = 2^{53}(\cos 30^\circ + i \sin 30^\circ)$.
$z^{53} = 2^{53}\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right)$.

(B) Let $\theta = \frac{\pi}{10}$. The expression is $\left[ \frac{1 - \cos \theta + i\sin \theta}{1 - \cos \theta - i\sin \theta} \right]^{10}$.
Using the identities $1 - \cos \theta = 2\sin^2 \frac{\theta}{2}$ and $\sin \theta = 2\sin \frac{\theta}{2} \cos \frac{\theta}{2}$:
Numerator $= 2\sin^2 \frac{\theta}{2} + i(2\sin \frac{\theta}{2} \cos \frac{\theta}{2}) = 2\sin \frac{\theta}{2} (\sin \frac{\theta}{2} + i\cos \frac{\theta}{2}) = 2\sin \frac{\theta}{2} [i(\cos \frac{\theta}{2} - i\sin \frac{\theta}{2})] = 2i\sin \frac{\theta}{2} e^{-i\theta/2}$.
Denominator $= 2\sin^2 \frac{\theta}{2} - i(2\sin \frac{\theta}{2} \cos \frac{\theta}{2}) = 2\sin \frac{\theta}{2} (\sin \frac{\theta}{2} - i\cos \frac{\theta}{2}) = 2\sin \frac{\theta}{2} [-i(\cos \frac{\theta}{2} + i\sin \frac{\theta}{2})] = -2i\sin \frac{\theta}{2} e^{i\theta/2}$.
Dividing the two: $\frac{2i\sin \frac{\theta}{2} e^{-i\theta/2}}{-2i\sin \frac{\theta}{2} e^{i\theta/2}} = -e^{-i\theta} = -(\cos \theta - i\sin \theta)$.
Raising to the power of $10$: $(-1)^{10} (\cos \theta - i\sin \theta)^{10} = 1 \cdot (\cos 10\theta - i\sin 10\theta) = \cos \pi - i\sin \pi = -1 - 0 = -1$.

(A) Using De Moivre's theorem,we know that $(\cos \theta + i\sin \theta)^n = \cos n\theta + i\sin n\theta$.
Let $z = \cos \theta + i\sin \theta$. Then $\cos n\theta - i\sin n\theta = z^{-n}$ and $\cos n\theta + i\sin n\theta = z^n$.
The expression becomes:
$\frac{(z^{-2})^4 (z^4)^{-5}}{(z^3)^{-2} (z^{-3})^{-9}}$
$= \frac{z^{-8} \cdot z^{-20}}{z^{-6} \cdot z^{27}}$
$= \frac{z^{-28}}{z^{21}}$
$= z^{-28-21} = z^{-49}$
$= \cos(-49\theta) + i\sin(-49\theta)$
$= \cos 49\theta - i\sin 49\theta$.

(C) We know that $\sin \theta = \cos \left( \frac{\pi }{2} - \theta \right)$ and $\cos \theta = \sin \left( \frac{\pi }{2} - \theta \right)$.
Substituting these into the expression,we get $(\sin \theta + i\cos \theta )^n = \left[ \cos \left( \frac{\pi }{2} - \theta \right) + i\sin \left( \frac{\pi }{2} - \theta \right) \right]^n$.
By De Moivre's theorem,$(\cos \phi + i\sin \phi )^n = \cos(n\phi) + i\sin(n\phi)$.
Applying this with $\phi = \frac{\pi }{2} - \theta$,we obtain $\cos n\left( \frac{\pi }{2} - \theta \right) + i\sin n\left( \frac{\pi }{2} - \theta \right)$.

(A) Let $z = \frac{1 + \cos(\pi/8) + i\sin(\pi/8)}{1 + \cos(\pi/8) - i\sin(\pi/8)}$.
Using the identities $1 + \cos \theta = 2\cos^2(\theta/2)$ and $\sin \theta = 2\sin(\theta/2)\cos(\theta/2)$:
$z = \frac{2\cos^2(\pi/16) + 2i\sin(\pi/16)\cos(\pi/16)}{2\cos^2(\pi/16) - 2i\sin(\pi/16)\cos(\pi/16)}$
$z = \frac{2\cos(\pi/16) [\cos(\pi/16) + i\sin(\pi/16)]}{2\cos(\pi/16) [\cos(\pi/16) - i\sin(\pi/16)]}$
$z = \frac{\cos(\pi/16) + i\sin(\pi/16)}{\cos(\pi/16) - i\sin(\pi/16)}$
Using Euler's formula $e^{i\theta} = \cos \theta + i\sin \theta$,we have $z = \frac{e^{i\pi/16}}{e^{-i\pi/16}} = e^{i\pi/8}$.
Then $z^8 = (e^{i\pi/8})^8 = e^{i\pi} = \cos \pi + i\sin \pi = -1$.

(A) Given ${x_n} = \cos \left( \frac{\pi }{4^n} \right) + i \sin \left( \frac{\pi }{4^n} \right)$.
Using the property of complex numbers,${x_1} \cdot {x_2} \cdot {x_3} \dots \infty = \prod_{n=1}^{\infty} \left( \cos \left( \frac{\pi }{4^n} \right) + i \sin \left( \frac{\pi }{4^n} \right) \right)$.
This is equal to $\cos \left( \sum_{n=1}^{\infty} \frac{\pi }{4^n} \right) + i \sin \left( \sum_{n=1}^{\infty} \frac{\pi }{4^n} \right)$.
The sum of the infinite geometric series is $\sum_{n=1}^{\infty} \frac{\pi }{4^n} = \frac{\pi/4}{1 - 1/4} = \frac{\pi/4}{3/4} = \frac{\pi}{3}$.
Therefore,the expression becomes $\cos \left( \frac{\pi}{3} \right) + i \sin \left( \frac{\pi}{3} \right)$.
$= \frac{1}{2} + i \frac{\sqrt{3}}{2} = \frac{1 + i\sqrt{3}}{2}$.

(C) Given expression: $\frac{(\cos \alpha + i\sin \alpha )^4}{(\sin \beta + i\cos \beta )^5}$
Using De Moivre's theorem,the numerator is $\cos 4\alpha + i\sin 4\alpha$.
For the denominator,we write $\sin \beta + i\cos \beta = i(\cos \beta - i\sin \beta)$.
So,the denominator becomes $i^5(\cos \beta - i\sin \beta)^5 = i(\cos 5\beta - i\sin 5\beta)$.
Thus,the expression is $\frac{\cos 4\alpha + i\sin 4\alpha}{i(\cos 5\beta - i\sin 5\beta)} = \frac{1}{i}(\cos 4\alpha + i\sin 4\alpha)(\cos 5\beta + i\sin 5\beta)$.
Since $\frac{1}{i} = -i$,we have $-i[\cos (4\alpha + 5\beta) + i\sin (4\alpha + 5\beta)]$.
$= -i\cos (4\alpha + 5\beta) - i^2\sin (4\alpha + 5\beta) = \sin (4\alpha + 5\beta) - i\cos (4\alpha + 5\beta)$.

(A) We know that $i = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})$.
Using De Moivre's theorem,$i^{1/3} = (\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}))^{1/3} = \cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})$.
Evaluating the trigonometric values,we get $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$.
Thus,$i^{1/3} = \frac{\sqrt{3}}{2} + i\frac{1}{2} = \frac{\sqrt{3} + i}{2}$.

(C) Let $w = 1 + i\sqrt{3}$.
Converting to polar form,$r = |w| = \sqrt{1^2 + (\sqrt{3})^2} = 2$.
$\theta = \arg(w) = \tan^{-1}(\sqrt{3}/1) = \frac{\pi}{3}$.
So,$w = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})$.
Then $z = w^{100} = 2^{100}(\cos \frac{100\pi}{3} + i \sin \frac{100\pi}{3})$.
Since $\frac{100\pi}{3} = 33\pi + \frac{\pi}{3}$,we have $\cos(\frac{100\pi}{3}) = \cos(33\pi + \frac{\pi}{3}) = -\cos(\frac{\pi}{3}) = -\frac{1}{2}$ and $\sin(\frac{100\pi}{3}) = \sin(33\pi + \frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}$.
Thus,$z = 2^{100}(-\frac{1}{2} - i\frac{\sqrt{3}}{2})$.
$\text{Re}(z) = -2^{99}$ and $\text{Im}(z) = -2^{99}\sqrt{3}$.
Therefore,$\frac{\text{Re}(z)}{\text{Im}(z)} = \frac{-2^{99}}{-2^{99}\sqrt{3}} = \frac{1}{\sqrt{3}}$.

(A) Let $z = \frac{1 + \sin \theta + i\cos \theta}{1 + \sin \theta - i\cos \theta}$.
Using the identities $\sin \theta = \cos(\frac{\pi}{2} - \theta)$ and $\cos \theta = \sin(\frac{\pi}{2} - \theta)$, let $\alpha = \frac{\pi}{2} - \theta$.
Then $z = \frac{1 + \cos \alpha + i\sin \alpha}{1 + \cos \alpha - i\sin \alpha}$.
Using half-angle formulas $1 + \cos \alpha = 2\cos^2(\frac{\alpha}{2})$ and $\sin \alpha = 2\sin(\frac{\alpha}{2})\cos(\frac{\alpha}{2})$:
$z = \frac{2\cos^2(\frac{\alpha}{2}) + i(2\sin(\frac{\alpha}{2})\cos(\frac{\alpha}{2}))}{2\cos^2(\frac{\alpha}{2}) - i(2\sin(\frac{\alpha}{2})\cos(\frac{\alpha}{2}))} = \frac{\cos(\frac{\alpha}{2}) + i\sin(\frac{\alpha}{2})}{\cos(\frac{\alpha}{2}) - i\sin(\frac{\alpha}{2})}$.
Using Euler's formula $e^{ix} = \cos x + i\sin x$, this is $\frac{e^{i\alpha/2}}{e^{-i\alpha/2}} = e^{i\alpha}$.
Thus, $z^n = (e^{i\alpha})^n = e^{in\alpha} = \cos(n\alpha) + i\sin(n\alpha)$.
Substituting $\alpha = \frac{\pi}{2} - \theta$, we get $\cos(n(\frac{\pi}{2} - \theta)) + i\sin(n(\frac{\pi}{2} - \theta)) = \cos(\frac{n\pi}{2} - n\theta) + i\sin(\frac{n\pi}{2} - n\theta)$.

(C) We express $1+i$ and $1-i$ in polar form:
$1+i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)$
$1-i = \sqrt{2} \left( \cos \frac{\pi}{4} - i \sin \frac{\pi}{4} \right)$
Using De Moivre's theorem:
$(1+i)^n + (1-i)^n = (\sqrt{2})^n \left( \cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4} \right) + (\sqrt{2})^n \left( \cos \frac{n\pi}{4} - i \sin \frac{n\pi}{4} \right)$
$= 2 \cdot (\sqrt{2})^n \cos \frac{n\pi}{4}$
$= 2^1 \cdot 2^{n/2} \cos \frac{n\pi}{4}$
$= 2^{(n/2) + 1} \cos \frac{n\pi}{4}$
$= (\sqrt{2})^{n+2} \cos \frac{n\pi}{4}$

(A) Given $\frac{1}{x} + x = 2\cos \theta$.
Multiplying by $x$,we get $x^2 - 2x\cos \theta + 1 = 0$.
Using the quadratic formula,$x = \frac{2\cos \theta \pm \sqrt{4\cos^2 \theta - 4}}{2} = \cos \theta \pm i\sin \theta$.
By De Moivre's Theorem,$x^n = (\cos \theta \pm i\sin \theta)^n = \cos n\theta \pm i\sin n\theta$.
Similarly,$\frac{1}{x^n} = x^{-n} = (\cos \theta \pm i\sin \theta)^{-n} = \cos n\theta \mp i\sin n\theta$.
Adding these,we get $x^n + \frac{1}{x^n} = (\cos n\theta \pm i\sin n\theta) + (\cos n\theta \mp i\sin n\theta) = 2\cos n\theta$.

(B) Given $i{z^4} + 1 = 0$.
$i{z^4} = -1$.
${z^4} = \frac{-1}{i} = i$.
We know that $i = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$.
Using De Moivre's theorem,$z = {\left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)}^{1/4}$.
$z = \cos \frac{\pi}{8} + i \sin \frac{\pi}{8}$.
Thus,the correct option is $B$.

(D) Let the two numbers be $x$ and $y$.
According to the problem,$y = x^2$ and $x = y^2$.
Substituting the first into the second,we get $x = (x^2)^2 = x^4$.
This implies $x^4 - x = 0$,or $x(x^3 - 1) = 0$.
The roots are $x = 0$ or $x^3 = 1$.
If $x = 0$,then $y = 0^2 = 0$.
If $x^3 = 1$,the roots are $1, \omega, \omega^2$.
For the pair $(\omega, \omega^2)$,we have $(\omega)^2 = \omega^2$ and $(\omega^2)^2 = \omega^4 = \omega \cdot \omega^3 = \omega \cdot 1 = \omega$.
Thus,the numbers are $\omega$ and $\omega^2$.

(D) We know that for a complex cube root of unity $\omega$,the following properties hold: $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
From $1 + \omega + \omega^2 = 0$,we have $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$.
Substituting these into the given expression:
$(1 + \omega - \omega^2)(1 - \omega + \omega^2) = (-\omega^2 - \omega^2)(-\omega - \omega)$
$= (-2\omega^2)(-2\omega)$
$= 4\omega^3$
Since $\omega^3 = 1$,the expression simplifies to $4(1) = 4$.

(C) Let $x = (27)^{1/3}$.
Then $x^3 = 27$,which implies $x^3 - 27 = 0$.
This can be factored as $(x - 3)(x^2 + 3x + 9) = 0$.
The roots are $x = 3$ and the roots of $x^2 + 3x + 9 = 0$.
Using the quadratic formula,$x = \frac{-3 \pm \sqrt{9 - 36}}{2} = \frac{-3 \pm i\sqrt{27}}{2} = 3 \left( \frac{-1 \pm i\sqrt{3}}{2} \right)$.
Since $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$,the roots are $3, 3\omega, 3\omega^2$.

(C) We know that $\omega$ is a cube root of unity,so $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
Since $n$ is not a multiple of $3$,$n$ can be of the form $3k + 1$ or $3k + 2$ for some integer $k \ge 0$.
Case $1$: If $n = 3k + 1$,then $\omega^n = \omega^{3k+1} = (\omega^3)^k \cdot \omega = 1^k \cdot \omega = \omega$ and $\omega^{2n} = \omega^{6k+2} = (\omega^3)^{2k} \cdot \omega^2 = 1^{2k} \cdot \omega^2 = \omega^2$.
Thus,$1 + \omega^n + \omega^{2n} = 1 + \omega + \omega^2 = 0$.
Case $2$: If $n = 3k + 2$,then $\omega^n = \omega^{3k+2} = (\omega^3)^k \cdot \omega^2 = \omega^2$ and $\omega^{2n} = \omega^{6k+4} = (\omega^3)^{2k+1} \cdot \omega = \omega$.
Thus,$1 + \omega^n + \omega^{2n} = 1 + \omega^2 + \omega = 0$.
In both cases,the sum is $0$.

(B) The cube roots of unity are $1, \omega, \omega^2$,where $\omega = e^{i\frac{2\pi}{3}}$ and $\omega^2 = e^{i\frac{4\pi}{3}}$.
Since $\omega^2$ is the square of $\omega$,and $(\omega^2)^2 = \omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$,it follows that the square of either imaginary cube root is the other imaginary cube root.

(A) We know that for a cube root of unity,$1 + \omega + \omega^2 = 0$.
Therefore,$1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$.
Substituting these values into the expression:
$(1 + \omega)^3 - (1 + \omega^2)^3 = (-\omega^2)^3 - (-\omega)^3$
$= -\omega^6 - (-\omega^3)$
$= -(\omega^3)^2 + \omega^3$
Since $\omega^3 = 1$,we have:
$= -(1)^2 + 1 = -1 + 1 = 0$.

(B) The imaginary cube roots of unity are $\omega$ and $\omega^2$.
Let $\alpha = \omega$ and $\beta = \omega^2$.
We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
Substituting these values into the expression $\alpha^4 + \beta^4 + \frac{1}{\alpha\beta}$:
$= \omega^4 + (\omega^2)^4 + \frac{1}{\omega \cdot \omega^2}$
$= \omega^4 + \omega^8 + \frac{1}{\omega^3}$
$= \omega + \omega^2 + \frac{1}{1}$
$= \omega + \omega^2 + 1$
$= 0$.

(C) We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
Given expression: $(1 - \omega)(1 - \omega^2)(1 - \omega^4)(1 - \omega^8)$.
Since $\omega^3 = 1$,we have $\omega^4 = \omega$ and $\omega^8 = \omega^2$.
Substituting these values,the expression becomes: $(1 - \omega)(1 - \omega^2)(1 - \omega)(1 - \omega^2) = [(1 - \omega)(1 - \omega^2)]^2$.
Expanding the inner product: $(1 - \omega - \omega^2 + \omega^3) = (1 - (\omega + \omega^2) + 1) = (1 - (-1) + 1) = 3$.
Thus,the expression is $3^2 = 9$.

(B) Given that $\omega$ is a cube root of unity,we know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
Using $1 + \omega^2 = -\omega$,the first term becomes $(1 - \omega + \omega^2)^5 = (-\omega - \omega)^5 = (-2\omega)^5 = -32\omega^5$.
Since $\omega^5 = \omega^3 \cdot \omega^2 = \omega^2$,this simplifies to $-32\omega^2$.
Using $1 + \omega = -\omega^2$,the second term becomes $(1 + \omega - \omega^2)^5 = (-\omega^2 - \omega^2)^5 = (-2\omega^2)^5 = -32\omega^{10}$.
Since $\omega^{10} = (\omega^3)^3 \cdot \omega = \omega$,this simplifies to $-32\omega$.
Adding the two terms: $-32\omega^2 - 32\omega = -32(\omega^2 + \omega)$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 + \omega = -1$.
Therefore,$-32(-1) = 32$.

(C) Given that $x = a, y = b\omega, z = c\omega^2$.
Substituting these values into the expression:
$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = \frac{a}{a} + \frac{b\omega}{b} + \frac{c\omega^2}{c}$
$= 1 + \omega + \omega^2$
Since $\omega$ is a complex cube root of unity,we know that $1 + \omega + \omega^2 = 0$.
Therefore,the value is $0$.

(C) Given the expression: $(x - y)(x\omega - y)(x\omega^2 - y)$
First,multiply the first two terms:
$(x - y)(x\omega - y) = x^2\omega - xy - xy\omega + y^2$
Now,multiply the result by the third term $(x\omega^2 - y)$:
$= (x^2\omega - xy - xy\omega + y^2)(x\omega^2 - y)$
$= x^3\omega^3 - x^2y\omega - x^2y\omega^2 + xy^2 - x^2y\omega^3 + xy^2\omega + xy^2\omega^2 - y^3$
$= x^3(1) - x^2y(\omega + \omega^2 + 1) + xy^2(1 + \omega + \omega^2) - y^3$
Since $1 + \omega + \omega^2 = 0$:
$= x^3 - x^2y(0) + xy^2(0) - y^3$
$= x^3 - y^3$

(B) Given the expression: $P = (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots$ to $2n$ factors.
Since $\omega^3 = 1$,we have $\omega^4 = \omega$ and $\omega^8 = \omega^2$.
Thus,the expression becomes: $P = (1 + \omega)(1 + \omega^2)(1 + \omega)(1 + \omega^2) \dots$ to $2n$ factors.
This can be grouped into $n$ pairs of $(1 + \omega)(1 + \omega^2)$.
We know that $1 + \omega + \omega^2 = 0$,so $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$.
Therefore,$(1 + \omega)(1 + \omega^2) = (-\omega^2)(-\omega) = \omega^3 = 1$.
Since there are $n$ such pairs,the product is $1^n = 1$.

(B) Let $z = \left( \cos \frac{\pi }{3} + i\sin \frac{\pi }{3} \right)^{3/4}$.
Using De Moivre's Theorem,$z = \left( e^{i\pi/3} \right)^{3/4} = e^{i\pi/4}$.
The roots are given by $z_k = \cos \left( \frac{\pi + 2k\pi}{4} \right) + i\sin \left( \frac{\pi + 2k\pi}{4} \right)$ for $k = 0, 1, 2, 3$.
These are the $4$th roots of $e^{i\pi} = -1$.
The product of the $n$th roots of a complex number $w$ is given by $(-1)^{n-1} w$.
Here,$n = 4$ and $w = -1$.
Therefore,the product of the roots is $(-1)^{4-1} \times (-1) = (-1)^3 \times (-1) = (-1) \times (-1) = 1$.

(B) Given that $\alpha$ and $\beta$ are complex cube roots of unity,we have $\alpha = \omega$ and $\beta = \omega^2$ (or vice versa),where $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
We need to calculate $xyz = (a + b)(a\omega + b\omega^2)(a\omega^2 + b\omega)$.
First,multiply $y$ and $z$:
$yz = (a\omega + b\omega^2)(a\omega^2 + b\omega) = a^2\omega^3 + ab\omega^2 + ab\omega^4 + b^2\omega^3$.
Since $\omega^3 = 1$ and $\omega^4 = \omega$,this simplifies to:
$yz = a^2(1) + ab(\omega^2 + \omega) + b^2(1) = a^2 + ab(-1) + b^2 = a^2 - ab + b^2$.
Now,multiply by $x$:
$xyz = (a + b)(a^2 - ab + b^2) = a^3 + b^3$.

(B) Given $x = a + b$,$y = a\omega + b\omega^2$,and $z = a\omega^2 + b\omega$.
We know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
Expanding the cubes:
$x^3 = (a + b)^3 = a^3 + b^3 + 3ab(a + b)$
$y^3 = (a\omega + b\omega^2)^3 = a^3\omega^3 + b^3\omega^6 + 3a^2b\omega^4 + 3ab^2\omega^5 = a^3 + b^3 + 3a^2b\omega + 3ab^2\omega^2$
$z^3 = (a\omega^2 + b\omega)^3 = a^3\omega^6 + b^3\omega^3 + 3a^2b\omega^5 + 3ab^2\omega^4 = a^3 + b^3 + 3a^2b\omega^2 + 3ab^2\omega$
Adding these:
$x^3 + y^3 + z^3 = 3(a^3 + b^3) + 3a^2b(1 + \omega + \omega^2) + 3ab^2(1 + \omega^2 + \omega)$
Since $1 + \omega + \omega^2 = 0$,the terms with $a^2b$ and $ab^2$ vanish.
Thus,$x^3 + y^3 + z^3 = 3(a^3 + b^3)$.

(B) Let the given expression be $E = \frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} + \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2}$.
Multiply the numerator and denominator of the first term by $\omega$:
$\frac{\omega(a + b\omega + c\omega^2)}{\omega(b + c\omega + a\omega^2)} = \frac{\omega(a + b\omega + c\omega^2)}{b\omega + c\omega^2 + a\omega^3} = \frac{\omega(a + b\omega + c\omega^2)}{a + b\omega + c\omega^2} = \omega$.
Multiply the numerator and denominator of the second term by $\omega^2$:
$\frac{\omega^2(a + b\omega + c\omega^2)}{\omega^2(c + a\omega + b\omega^2)} = \frac{\omega^2(a + b\omega + c\omega^2)}{c\omega^2 + a\omega^3 + b\omega^4} = \frac{\omega^2(a + b\omega + c\omega^2)}{c\omega^2 + a + b\omega} = \omega^2$.
Thus,$E = \omega + \omega^2$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.

(A) The cube roots of unity are $1, \omega, \omega^2$.
These roots can be represented as points on the Argand plane: $1 = (1, 0)$,$\omega = (\cos \frac{2\pi}{3}, \sin \frac{2\pi}{3})$,and $\omega^2 = (\cos \frac{4\pi}{3}, \sin \frac{4\pi}{3})$.
These points lie on the unit circle $|z| = 1$ and are separated by an angle of $\frac{2\pi}{3}$ radians $(120^\circ)$ from each other.
Since the points are equally spaced on the circle,they form the vertices of an equilateral triangle.
Alternatively,using the property that $z_1, z_2, z_3$ form an equilateral triangle if $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$,we substitute $z_1=1, z_2=\omega, z_3=\omega^2$:
$1^2 + \omega^2 + (\omega^2)^2 = 1 + \omega^2 + \omega^4 = 1 + \omega^2 + \omega = 0$.
And $1(\omega) + \omega(\omega^2) + \omega^2(1) = \omega + \omega^3 + \omega^2 = \omega + 1 + \omega^2 = 0$.
Since both sides are equal to $0$,the triangle is equilateral.

(C) Let $z = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$.
This is the complex cube root of unity,denoted by $\omega$.
We know that $\omega^3 = 1$.
We need to calculate $\omega^{1000}$.
Since $1000 = 3 \times 333 + 1$,we have $\omega^{1000} = (\omega^3)^{333} \times \omega^1$.
Substituting $\omega^3 = 1$,we get $\omega^{1000} = (1)^{333} \times \omega = \omega$.
Therefore,$\omega^{1000} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$.

(D) Let $p = -q$ where $q > 0$. The cube roots of $p$ are $\alpha = -q^{1/3}$,$\beta = -q^{1/3}\omega$,and $\gamma = -q^{1/3}\omega^2$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$ is the complex cube root of unity.
Substituting these into the expression:
$\frac{x(-q^{1/3}) + y(-q^{1/3}\omega) + z(-q^{1/3}\omega^2)}{x(-q^{1/3}\omega) + y(-q^{1/3}\omega^2) + z(-q^{1/3}\omega^3)} = \frac{-(x + y\omega + z\omega^2)}{-\omega(x + y\omega + z\omega^2)} = \frac{1}{\omega} = \omega^2$.
Since $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$,none of the given options match.

(A) Given $z = \frac{\sqrt{3}}{2} + i\frac{1}{2} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)$.
Using De Moivre's theorem,$z^{69} = \left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right)^{69}$.
$z^{69} = \cos\left(\frac{69\pi}{6}\right) + i\sin\left(\frac{69\pi}{6}\right)$.
$z^{69} = \cos\left(\frac{23\pi}{2}\right) + i\sin\left(\frac{23\pi}{2}\right)$.
Since $\frac{23\pi}{2} = 11\pi + \frac{\pi}{2}$,we have $\cos\left(11\pi + \frac{\pi}{2}\right) = 0$ and $\sin\left(11\pi + \frac{\pi}{2}\right) = \sin\left(\pi + \frac{\pi}{2}\right) = -1$.
Therefore,$z^{69} = 0 + i(-1) = -i$.

(B) Given equation is $x^4 - 1 = 0$.
This can be factored as $(x^2 - 1)(x^2 + 1) = 0$.
This implies $x^2 - 1 = 0$ or $x^2 + 1 = 0$.
For $x^2 - 1 = 0$,we have $x^2 = 1$,so $x = \pm 1$.
For $x^2 + 1 = 0$,we have $x^2 = -1$,so $x = \pm i$.
Thus,the roots are $1, -1, i, -i$.

(D) The product is given by $\omega \cdot \omega^2 \cdot \omega^3 \cdots \omega^n = \omega^{1 + 2 + 3 + \cdots + n} = \omega^{n(n + 1)/2}$.
For $n = 1$,the product is $\omega^1 = \omega$.
For $n = 2$,the product is $\omega^{2(3)/2} = \omega^3 = 1$.
For $n = 3$,the product is $\omega^{3(4)/2} = \omega^6 = 1$.
For $n = 4$,the product is $\omega^{4(5)/2} = \omega^{10} = \omega^1 = \omega$.
Since $\omega = \frac{-1 + i\sqrt{3}}{2}$,the value $-\frac{1 - i\sqrt{3}}{2} = \frac{-1 + i\sqrt{3}}{2} = \omega$. Thus,the product results in values $1$ and $\omega$.