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

(D) Given expression: $E = \frac{(\cos \theta + i \sin \theta)^4}{(\sin \theta + i \cos \theta)^5}$
Using De Moivre's Theorem,the numerator is $(\cos \theta + i \sin \theta)^4 = \cos 4 \theta + i \sin 4 \theta$.
For the denominator,rewrite $\sin \theta + i \cos \theta$ as $i(\cos \theta - i \sin \theta) = i(\cos(-\theta) + i \sin(-\theta))$.
So,$(\sin \theta + i \cos \theta)^5 = i^5 (\cos(-\theta) + i \sin(-\theta))^5 = i (\cos(-5 \theta) + i \sin(-5 \theta)) = i(\cos 5 \theta - i \sin 5 \theta)$.
Thus,$E = \frac{\cos 4 \theta + i \sin 4 \theta}{i(\cos 5 \theta - i \sin 5 \theta)} = \frac{1}{i} \cdot \frac{\cos 4 \theta + i \sin 4 \theta}{\cos 5 \theta - i \sin 5 \theta}$.
Using the property $\frac{\cos \alpha + i \sin \alpha}{\cos \beta + i \sin \beta} = \cos(\alpha - \beta) + i \sin(\alpha - \beta)$,we have:
$E = -i \cdot \frac{\cos 4 \theta + i \sin 4 \theta}{\cos(-5 \theta) + i \sin(-5 \theta)} = -i (\cos(4 \theta - (-5 \theta)) + i \sin(4 \theta - (-5 \theta)))$
$E = -i (\cos 9 \theta + i \sin 9 \theta) = -i \cos 9 \theta - i^2 \sin 9 \theta = \sin 9 \theta - i \cos 9 \theta$.

(C) Given that $z$ is a complex cube root of unity,we have $z^3 = 1$ and $1+z+z^2 = 0$.
From $1+z+z^2 = 0$,we get $z^2+1 = -z$.
Now,consider the expression $\left(z^3+\frac{1}{z^3}\right)^2+\left(z^4+\frac{1}{z^4}\right)^2$.
Substituting $z^3 = 1$,we get $\left(1+\frac{1}{1}\right)^2+\left(z^3 \cdot z+\frac{1}{z^3 \cdot z}\right)^2$.
$= (1+1)^2 + \left(z+\frac{1}{z}\right)^2$.
$= 4 + \left(\frac{z^2+1}{z}\right)^2$.
Substituting $z^2+1 = -z$,we get $4 + \left(\frac{-z}{z}\right)^2$.
$= 4 + (-1)^2 = 4 + 1 = 5$.

(C) Given $w = \frac{-1 + i \sqrt{3}}{2}$,which is the complex cube root of unity,denoted as $\omega$.
We know that $1 + \omega + \omega^2 = 0$,which implies $\omega^2 = -1 - \omega$.
We need to evaluate $(3 + \omega + 3 \omega^2)^4$.
Substitute $\omega^2 = -1 - \omega$ into the expression:
$(3 + \omega + 3(-1 - \omega))^4 = (3 + \omega - 3 - 3 \omega)^4$.
$= (-2 \omega)^4$.
$= 16 \omega^4$.
Since $\omega^3 = 1$,we have $\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$.
Therefore,$16 \omega^4 = 16 \omega$.

(B) Let $\alpha = \omega$ and $\beta = \omega^2$,where $\omega$ is an imaginary cube root of unity.
We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
Now,substitute these into the expression $\alpha^4 + \beta^{28} + \frac{1}{\alpha \beta}$:
$= \omega^4 + (\omega^2)^{28} + \frac{1}{\omega \cdot \omega^2}$
$= \omega^4 + \omega^{56} + \frac{1}{\omega^3}$
Since $\omega^3 = 1$,we have $\omega^4 = \omega^3 \cdot \omega = \omega$ and $\omega^{56} = (\omega^3)^{18} \cdot \omega^2 = 1^{18} \cdot \omega^2 = \omega^2$.
Thus,the expression becomes $\omega + \omega^2 + \frac{1}{1} = \omega + \omega^2 + 1$.
Since $1 + \omega + \omega^2 = 0$,the value is $0$.

(C) The complex cube roots of unity are $1, \omega, \omega^2$. Since $\alpha$ and $\beta$ are the complex cube roots,we have $\alpha = \omega$ and $\beta = \omega^2$ (or vice versa).
Given expression: $\alpha^3 + \beta^3 + \alpha^{-2} \times \beta^{-2} = \alpha^3 + \beta^3 + \frac{1}{(\alpha \beta)^2}$.
Since $\alpha = \omega$ and $\beta = \omega^2$,we have $\alpha^3 = \omega^3 = 1$ and $\beta^3 = (\omega^2)^3 = \omega^6 = 1$.
Also,$\alpha \beta = \omega \times \omega^2 = \omega^3 = 1$.
Substituting these values: $1 + 1 + \frac{1}{(1)^2} = 1 + 1 + 1 = 3$.

(C) We know that $1+\omega+\omega^2=0$,so $1+\omega=-\omega^2$.
Substituting this into the given expression:
$(1+\omega)^7 = (-\omega^2)^7$
$= (-1)^7 \times (\omega^2)^7$
$= -1 \times \omega^{14}$
Since $\omega^3=1$,we have $\omega^{14} = \omega^{12} \times \omega^2 = (\omega^3)^4 \times \omega^2 = 1^4 \times \omega^2 = \omega^2$.
Thus,$(1+\omega)^7 = -\omega^2$.
Using $1+\omega+\omega^2=0$,we have $-\omega^2 = 1+\omega$.
Comparing $1+\omega$ with $A+B\omega$,we get $A=1$ and $B=1$.

(D) We know that $1+\omega+\omega^2 = 0$,which implies $1+\omega^2 = -\omega$ and $\omega^3 = 1$.
Given expression: $E = (3+5\omega+3\omega^2)^2 + (3+3\omega+5\omega^2)^2$
Rearranging terms using $3(1+\omega+\omega^2) = 0$:
$E = (3(1+\omega+\omega^2) + 2\omega)^2 + (3(1+\omega+\omega^2) + 2\omega^2)^2$
$E = (0 + 2\omega)^2 + (0 + 2\omega^2)^2$
$E = 4\omega^2 + 4\omega^4$
Since $\omega^3 = 1$,we have $\omega^4 = \omega$.
$E = 4\omega^2 + 4\omega = 4(\omega^2 + \omega)$
Since $1+\omega+\omega^2 = 0$,we have $\omega^2 + \omega = -1$.
$E = 4(-1) = -4$.

(B) Given equation is $x^{2}+x+1=0$.
The roots of this equation are the complex cube roots of unity,$\omega$ and $\omega^{2}$.
Let $\alpha = \omega$ and $\beta = \omega^{2}$.
We need to find $\alpha^{16} + \beta^{16} = \omega^{16} + (\omega^{2})^{16}$.
Since $\omega^{3} = 1$,we have $\omega^{16} = (\omega^{3})^{5} \cdot \omega = 1^{5} \cdot \omega = \omega$.
Similarly,$\omega^{32} = (\omega^{3})^{10} \cdot \omega^{2} = 1^{10} \cdot \omega^{2} = \omega^{2}$.
Thus,$\alpha^{16} + \beta^{16} = \omega + \omega^{2}$.
Since $1 + \omega + \omega^{2} = 0$,it follows that $\omega + \omega^{2} = -1$.

(D) Let $z = \frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}$.
Using $\sin \theta = \cos(\frac{\pi}{2}-\theta)$ and $\cos \theta = \sin(\frac{\pi}{2}-\theta)$,let $\alpha = \frac{\pi}{2}-\frac{\pi}{8} = \frac{3\pi}{8}$.
Then $z = \frac{1+\cos \alpha + i \sin \alpha}{1+\cos \alpha - i \sin \alpha} = \frac{2 \cos^2 \frac{\alpha}{2} + 2i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \cos^2 \frac{\alpha}{2} - 2i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = \frac{\cos \frac{\alpha}{2} + i \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - i \sin \frac{\alpha}{2}} = \frac{e^{i \alpha/2}}{e^{-i \alpha/2}} = e^{i \alpha}$.
So,$z^n = e^{i n \alpha} = e^{i n (3\pi/8)} = \cos(\frac{3n\pi}{8}) + i \sin(\frac{3n\pi}{8})$.
For $z^n = 1$,we require $\sin(\frac{3n\pi}{8}) = 0$ and $\cos(\frac{3n\pi}{8}) = 1$.
This implies $\frac{3n\pi}{8} = 2k\pi$ for some integer $k$.
$3n = 16k \implies n = \frac{16k}{3}$.
For the smallest positive integer $n$,set $k=3$,which gives $n = 16$.

(D) Given,$2x = -1 + i\sqrt{3}$.
$x = \frac{-1 + i\sqrt{3}}{2} = \omega$,where $\omega$ is a complex cube root of unity.
We know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$.
Substituting $x = \omega$ into the expression:
$(1 - \omega^2 + \omega)^6 - (1 - \omega + \omega^2)^6$
$= (-\omega^2 - \omega^2)^6 - (-\omega - \omega)^6$
$= (-2\omega^2)^6 - (-2\omega)^6$
$= 2^6 \cdot \omega^{12} - 2^6 \cdot \omega^6$
Since $\omega^3 = 1$,we have $\omega^6 = 1$ and $\omega^{12} = 1$.
$= 64(1) - 64(1) = 0$.

(A) Given that $\omega^{3} = 1$ and $1+\omega+\omega^{2} = 0$.
We know that $1+\omega^{2} = -\omega$ and $1+\omega = -\omega^{2}$.
Also,$\omega^{3} = 1, \omega^{4} = \omega, \omega^{8} = \omega^{2}, \omega^{16} = \omega$.
The expression is $P = (1-\omega+\omega^{2})(1-\omega^{2}+\omega)(1-\omega+\omega^{2})(1-\omega^{2}+\omega) \ldots$ ($2n$ factors).
Substituting $1+\omega^{2} = -\omega$ and $1+\omega = -\omega^{2}$:
$P = (-\omega-\omega)(-\omega^{2}-\omega^{2})(-\omega-\omega)(-\omega^{2}-\omega^{2}) \ldots$ ($2n$ factors).
$P = (-2\omega)(-2\omega^{2})(-2\omega)(-2\omega^{2}) \ldots$ ($2n$ factors).
There are $n$ pairs of $(-2\omega)(-2\omega^{2}) = 4\omega^{3} = 4(1) = 4$.
Thus,$P = (4)^{n} = 2^{2n}$.

(B) Given $x+iy=(-1+i\sqrt{3})^{2010}$.
We know that $\omega = \frac{-1+i\sqrt{3}}{2}$ is the cube root of unity,so $-1+i\sqrt{3} = 2\omega$.
Substituting this into the expression:
$x+iy = (2\omega)^{2010} = 2^{2010} \cdot \omega^{2010}$.
Since $\omega^3 = 1$,we have $\omega^{2010} = (\omega^3)^{670} = 1^{670} = 1$.
Therefore,$x+iy = 2^{2010} \cdot 1 = 2^{2010} + i(0)$.
Comparing the real parts,we get $x = 2^{2010}$.

(A) The given equation is $\alpha^{2}+\alpha+1=0$.
This is the characteristic equation for the cube roots of unity,where $\alpha$ is either $\omega$ or $\omega^{2}$.
We know that $\omega^{3}=1$.
For $\alpha=\omega$,we have $\alpha^{31} = \omega^{31} = (\omega^{3})^{10} \cdot \omega = 1^{10} \cdot \omega = \omega = \alpha$.
For $\alpha=\omega^{2}$,we have $\alpha^{31} = (\omega^{2})^{31} = \omega^{62} = (\omega^{3})^{20} \cdot \omega^{2} = 1^{20} \cdot \omega^{2} = \omega^{2} = \alpha$.
In both cases,$\alpha^{31} = \alpha$.

(A) We know that $1+\omega+\omega^{2}=0$ and $\omega^{3}=1$.
Given expression: $(1+\omega)(1+\omega^{2})(1+\omega^{4})(1+\omega^{8})$
$= (1+\omega)(1+\omega^{2})(1+\omega)(1+\omega^{2})$
$= [(1+\omega)(1+\omega^{2})]^{2}$
$= [1+\omega^{2}+\omega+\omega^{3}]^{2}$
$= [1+(\omega^{2}+\omega)+1]^{2}$
$= [1+(-1)+1]^{2}$
$= [1]^{2} = 1$.

(C) Given,$\alpha^{2}-\alpha+1=0$.
Multiplying by $(\alpha+1)$,we get $(\alpha+1)(\alpha^{2}-\alpha+1)=0$,which implies $\alpha^{3}+1=0$,so $\alpha^{3}=-1$.
Therefore,$\alpha^{6}=1$.
We need to find $\alpha^{2011}$.
Since $2011 = 6 \times 335 + 1$,we have $\alpha^{2011} = (\alpha^{6})^{335} \times \alpha^{1} = (1)^{335} \times \alpha = \alpha$.
Thus,$\alpha^{2011} = \alpha$.

(A) We are given the sum $S = \sum_{k=1}^{6}\left(\sin \frac{2 k \pi}{7}-i \cos \frac{2 k \pi}{7}\right)$.
Factor out $-i$ from the expression:
$S = -i \sum_{k=1}^{6}\left(\cos \frac{2 k \pi}{7} + i \sin \frac{2 k \pi}{7}\right)$.
Let $\omega = e^{i \frac{2 \pi}{7}} = \cos \frac{2 \pi}{7} + i \sin \frac{2 \pi}{7}$. Then the sum becomes:
$S = -i \sum_{k=1}^{6} \omega^k$.
This is a geometric series with $6$ terms,where the first term is $\omega$ and the common ratio is $\omega$:
$S = -i \left( \frac{\omega(1 - \omega^6)}{1 - \omega} \right) = -i \left( \frac{\omega - \omega^7}{1 - \omega} \right)$.
Since $\omega^7 = e^{i 2 \pi} = 1$,we have:
$S = -i \left( \frac{\omega - 1}{1 - \omega} \right) = -i (-1) = i$.

(D) We have,$(1-\omega+\omega^{2})(1+\omega-\omega^{2}) \quad \dots(1)$
We know that,$1+\omega+\omega^{2}=0$.
From this,we can write:
$1+\omega^{2} = -\omega$,so $(1-\omega+\omega^{2}) = -\omega - \omega = -2\omega$.
$1+\omega = -\omega^{2}$,so $(1+\omega-\omega^{2}) = -\omega^{2} - \omega^{2} = -2\omega^{2}$.
Substituting these values into Eq. $(1)$:
$(-2\omega)(-2\omega^{2}) = 4\omega^{3}$.
Since $\omega^{3}=1$,the expression becomes $4(1) = 4$.

(B) Given the quadratic equation $x^2-2x+4=0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, we get:
$x = \frac{2 \pm \sqrt{4-16}}{2} = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3}$.
Converting to polar form:
$x = 2 \left( \frac{1}{2} \pm i\frac{\sqrt{3}}{2} \right) = 2 \left( \cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3} \right)$.
Let $\alpha = 2 e^{i\pi/3}$ and $\beta = 2 e^{-i\pi/3}$.
Then $\alpha^n + \beta^n = (2 e^{i\pi/3})^n + (2 e^{-i\pi/3})^n = 2^n (e^{in\pi/3} + e^{-in\pi/3})$.
Using Euler's formula $e^{i\theta} + e^{-i\theta} = 2 \cos \theta$:
$\alpha^n + \beta^n = 2^n (2 \cos \frac{n\pi}{3}) = 2^{n+1} \cos \frac{n\pi}{3}$.
Thus, the correct option is $B$.

(A) The given quadratic equation is $x^2-2x+4=0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, we get $x = \frac{2 \pm \sqrt{4-16}}{2} = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3}$.
Let $\alpha = 1+i\sqrt{3}$ and $\beta = 1-i\sqrt{3}$.
Converting to polar form, $\alpha = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2e^{i\pi/3}$ and $\beta = 2(\cos \frac{\pi}{3} - i \sin \frac{\pi}{3}) = 2e^{-i\pi/3}$.
Then $\alpha^n + \beta^n = (2e^{i\pi/3})^n + (2e^{-i\pi/3})^n = 2^n(e^{in\pi/3} + e^{-in\pi/3})$.
Using Euler's formula $e^{i\theta} + e^{-i\theta} = 2 \cos \theta$, we get $\alpha^n + \beta^n = 2^n(2 \cos \frac{n\pi}{3}) = 2^{n+1} \cos \frac{n\pi}{3}$.
Comparing this with the given expression $k \cos \frac{n\pi}{3}$, we find $k = 2^{n+1}$.

(C) The given quadratic equation is $x^2-2x+4=0$.
Since $\alpha$ and $\beta$ are the roots,we have $\alpha+\beta=2$ and $\alpha\beta=4$.
The roots of the equation are given by $x = \frac{2 \pm \sqrt{4-16}}{2} = 1 \pm i\sqrt{3}$.
In polar form,$1+i\sqrt{3} = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 2e^{i\pi/3}$ and $1-i\sqrt{3} = 2e^{-i\pi/3}$.
Thus,$\alpha = 2e^{i\pi/3}$ and $\beta = 2e^{-i\pi/3}$.
Then $\alpha^9 = (2e^{i\pi/3})^9 = 2^9 e^{i3\pi} = 2^9(\cos 3\pi + i\sin 3\pi) = 2^9(-1) = -2^9$.
Similarly,$\beta^9 = (2e^{-i\pi/3})^9 = 2^9 e^{-i3\pi} = 2^9(\cos(-3\pi) + i\sin(-3\pi)) = 2^9(-1) = -2^9$.
Therefore,$\alpha^9+\beta^9 = -2^9 - 2^9 = -2 \times 2^9 = -2^{10}$.

(A) The given equation is $x^9-x^5+x^4-1=0$.
Factoring the expression: $x^5(x^4-1) + 1(x^4-1) = 0$,which gives $(x^5+1)(x^4-1) = 0$.
For $x^4-1=0$,the roots are $x=1, -1, i, -i$.
For $x^5+1=0$,the roots are $x=-1$ and $e^{\pm i\pi/5}, e^{\pm 3i\pi/5}$.
The distinct roots are $x=1, -1, i, -i, e^{i\pi/5}, e^{-i\pi/5}, e^{i3\pi/5}, e^{-i3\pi/5}$.
Real roots are $1, -1$,so $n=2$.
Complex roots with argument on the imaginary axis (argument $\pi/2$ or $3\pi/2$) are $i, -i$,so $m=2$.
Complex roots with argument in the $2^{nd}$ quadrant (argument between $\pi/2$ and $\pi$) is $e^{i3\pi/5}$,so $k=1$.
Therefore,$m \cdot n \cdot k = 2 \times 2 \times 1 = 4$.
Wait,re-evaluating the roots: $x^5 = -1 = e^{i(\pi + 2k\pi)}$. Roots are $e^{i\pi/5}, e^{i3\pi/5}, e^{i\pi}, e^{i7\pi/5}, e^{i9\pi/5}$.
Real roots: $1, -1$ (from $x^4=1$) and $-1$ (from $x^5=-1$). Distinct real roots are $1, -1$,so $n=2$.
Complex roots: $i, -i$ (arg $\pi/2, 3\pi/2$),$e^{i\pi/5}$ (arg $36^{\circ}$),$e^{i3\pi/5}$ (arg $108^{\circ}$),$e^{i7\pi/5}$ (arg $252^{\circ}$),$e^{i9\pi/5}$ (arg $324^{\circ}$).
$m=2$ (roots $i, -i$). $k=1$ (root $e^{i3\pi/5}$).
$m \cdot n \cdot k = 2 \times 2 \times 1 = 4$.
Given the options,there might be a misinterpretation of distinct roots. If we count total roots including multiplicity: $n=3$ (real roots $1, -1, -1$),$m=2$,$k=1$. Then $3 \times 2 \times 1 = 6$.

(C) Given equation is $x^3-x^2-x-2=0$.
By testing $x=2$,we get $8-4-2-2=0$,so $(x-2)$ is a factor.
Dividing by $(x-2)$,we get $(x-2)(x^2+x+1)=0$.
The non-real roots $\alpha$ and $\beta$ are the roots of $x^2+x+1=0$.
These are the complex cube roots of unity,$\omega$ and $\omega^2$,where $\omega^3=1$ and $1+\omega+\omega^2=0$.
We need to evaluate $\alpha^{2020}+\beta^{2020}+\alpha^{2020} \cdot \beta^{2020}$.
Since $\alpha=\omega$ and $\beta=\omega^2$,we have $\alpha^{2020}=\omega^{2020}=\omega^{3 \times 673 + 1}=\omega$ and $\beta^{2020}=(\omega^2)^{2020}=\omega^{4040}=\omega^{3 \times 1346 + 2}=\omega^2$.
Thus,$\alpha^{2020}+\beta^{2020}+\alpha^{2020} \cdot \beta^{2020} = \omega + \omega^2 + \omega \cdot \omega^2 = \omega + \omega^2 + \omega^3 = \omega + \omega^2 + 1 = 0$.
However,looking at the options,$1+\alpha+\beta = 1 + \omega + \omega^2 = 0$.
Therefore,the expression equals $0$,which is equivalent to $1+\alpha+\beta$.

(C) Given that $\alpha$ is a non-real root of $x^7=1$,we have $\alpha^7=1$ and $\alpha \neq 1$.
Expanding the expression:
$\alpha(1+\alpha)(1+\alpha^2+\alpha^4) = \alpha(1+\alpha^2+\alpha^4+\alpha+\alpha^3+\alpha^5)$
$= \alpha + \alpha^3 + \alpha^5 + \alpha^2 + \alpha^4 + \alpha^6$
$= \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6$
This is a geometric series sum:
$= \frac{\alpha(1-\alpha^6)}{1-\alpha} = \frac{\alpha-\alpha^7}{1-\alpha}$
Since $\alpha^7=1$,we have:
$= \frac{\alpha-1}{1-\alpha} = -1$.

(B) Let $z = 8i = 8(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) = 8e^{i(\frac{\pi}{2} + 2k\pi)}$ for $k = 0, 1, 2$.
Taking the cube root,$z^{\frac{1}{3}} = 2e^{i(\frac{\pi}{6} + \frac{2k\pi}{3})}$.
For $k=0$: $2e^{i\frac{\pi}{6}} = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) = 2(\frac{\sqrt{3}}{2} + i \frac{1}{2}) = \sqrt{3} + i$.
For $k=1$: $2e^{i(\frac{\pi}{6} + \frac{2\pi}{3})} = 2e^{i\frac{5\pi}{6}} = 2(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}) = 2(-\frac{\sqrt{3}}{2} + i \frac{1}{2}) = -\sqrt{3} + i$.
For $k=2$: $2e^{i(\frac{\pi}{6} + \frac{4\pi}{3})} = 2e^{i\frac{9\pi}{6}} = 2e^{i\frac{3\pi}{2}} = 2(0 - i) = -2i$.
Thus,the values are $\pm \sqrt{3} + i, -2i$.

(D) Let $z = (1+i \sqrt{3})^{3/4}$.
We know that for a complex number $w = z^n$,where $n = p/q$,the product of the $q$ values is given by $(-1)^{q-1} (z^p)$.
Alternatively,let $z = r^{3/4} e^{i(3\theta/4 + 3k\pi/2)}$ for $k = 0, 1, 2, 3$.
The product of these four values is $P = \prod_{k=0}^{3} r^{3/4} e^{i(3\theta/4 + 3k\pi/2)} = (r^{3/4})^4 e^{i \sum_{k=0}^{3} (3\theta/4 + 3k\pi/2)}$.
Here,$r = |1+i\sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = 2$.
So,$r^3 = 2^3 = 8$.
The product is $8 \times e^{i(3\theta + 3\pi/2(0+1+2+3))} = 8 \times e^{i(3\theta + 9\pi)}$.
Since $e^{i9\pi} = -1$,the product is $8 \times e^{i3\theta} \times (-1) = -8(e^{i\theta})^3$.
Given $1+i\sqrt{3} = 2e^{i\pi/3}$,we have $e^{i\theta} = e^{i\pi/3}$.
Thus,$(e^{i\theta})^3 = e^{i\pi} = -1$.
Therefore,the product is $-8 \times (-1) = 8$.

(D) For the quadratic equation $x^2+x+1=0$,the roots are the complex cube roots of unity,$\omega$ and $\omega^2$.
Thus,$\alpha = \omega$ and $\beta = \omega^2$.
We need to find the equation with roots $\alpha^{2021}$ and $\beta^{2021}$.
Since $\omega^3 = 1$,we have $\alpha^{2021} = \omega^{2021} = (\omega^3)^{673} \cdot \omega^2 = \omega^2$.
Similarly,$\beta^{2021} = (\omega^2)^{2021} = \omega^{4042} = (\omega^3)^{1347} \cdot \omega = \omega$.
The new roots are $\omega^2$ and $\omega$.
The quadratic equation with roots $\omega$ and $\omega^2$ is given by $x^2 - (\omega + \omega^2)x + \omega \cdot \omega^2 = 0$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$ and $\omega^3 = 1$.
Substituting these values,we get $x^2 - (-1)x + 1 = 0$,which simplifies to $x^2 + x + 1 = 0$.

(B) Let $z_1 = -i+\sqrt{3}$ and $z_2 = -i-\sqrt{3}$.
We can write $z_1 = -i(1+i\sqrt{3})$ and $z_2 = i(1-i\sqrt{3})$.
Alternatively,note that $z_1 = -i(1+i\sqrt{3}) = -2i(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = -2i e^{i\pi/3}$.
$z_1^{300} = (-2i)^{300} (e^{i\pi/3})^{300} = 2^{300} (i)^{300} e^{i100\pi} = 2^{300} (1) (1) = 2^{300}$.
Similarly,$z_2 = -i(1-i\sqrt{3}) = -2i(\frac{1}{2} - i\frac{\sqrt{3}}{2}) = -2i e^{-i\pi/3}$.
$z_2^{300} = (-2i)^{300} (e^{-i\pi/3})^{300} = 2^{300} (i)^{300} e^{-i100\pi} = 2^{300} (1) (1) = 2^{300}$.
Thus,$z_1^{300} + z_2^{300} = 2^{300} + 2^{300} = 2 \times 2^{300} = 2^{301}$.
Hence,option $(B)$ is correct.

(C) Given that $x^6 = 1$,we have $x^6 - 1 = 0$.
This can be factored as $(x^2-1)(x^4+x^2+1) = 0$ or $(x-1)(x^5+x^4+x^3+x^2+x+1) = 0$.
Since $\alpha$ is a non-real root of $x^6=1$,it satisfies the equation $x^5+x^4+x^3+x^2+x+1=0$.
Therefore,$\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$.
We can rearrange the terms as:
$\alpha^5+\alpha^3+\alpha+1 = -(\alpha^4+\alpha^2)$
$\alpha^5+\alpha^3+\alpha+1 = -\alpha^2(\alpha^2+1)$
Dividing both sides by $(\alpha^2+1)$,we get:
$\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1} = -\alpha^2$.

(C) Given $f(x, y) = (x+y)(x \omega + y \omega^2)(x \omega^2 + y \omega)$.
We know that $(x \omega + y \omega^2)(x \omega^2 + y \omega) = x^2 \omega^3 + xy \omega^2 + xy \omega^4 + y^2 \omega^3 = x^2 + xy(\omega^2 + \omega) + y^2 = x^2 - xy + y^2$ (since $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$).
Thus,$f(x, y) = (x+y)(x^2 - xy + y^2) = x^3 + y^3$.
Substituting $x = 2$ and $y = 3$:
$f(2, 3) = 2^3 + 3^3 = 8 + 27 = 35$.

(D) Given,$z^2+z+1=0$.
Since $z^2+z+1=0$,we have $z^2+1=-z$.
Dividing by $z$,we get $z+\frac{1}{z}=-1$.
Thus,$\left(z+\frac{1}{z}\right)^3 = (-1)^3 = -1$.
Also,$z^3=1$ (since $z$ is a cube root of unity).
Then $z^4 = z^3 \cdot z = z$.
So,$z^4+\frac{1}{z^4} = z+\frac{1}{z} = -1$.
Therefore,$\left(z^4+\frac{1}{z^4}\right)^3 = (-1)^3 = -1$.
Adding these results,$\left(z+\frac{1}{z}\right)^3+\left(z^4+\frac{1}{z^4}\right)^3 = -1 + (-1) = -2$.

(C) Given the quadratic equation $x^2-x+1=0$.
The roots are $x = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2} = -\omega$ and $-\omega^2$,where $\omega$ is the complex cube root of unity.
Let $\alpha = -\omega$ and $\beta = -\omega^2$.
We need to calculate $\alpha^{2009} + \beta^{2009} = (-\omega)^{2009} + (-\omega^2)^{2009}$.
This simplifies to $-(\omega^{2009} + \omega^{4018})$.
Since $\omega^3 = 1$,we have $\omega^{2009} = \omega^{3 \times 669 + 2} = \omega^2$ and $\omega^{4018} = \omega^{3 \times 1339 + 1} = \omega$.
Thus,$\alpha^{2009} + \beta^{2009} = -(\omega^2 + \omega)$.
Using the identity $1 + \omega + \omega^2 = 0$,we get $\omega^2 + \omega = -1$.
Therefore,$\alpha^{2009} + \beta^{2009} = -(-1) = 1$.

(D) Given $2 \alpha = -1 - i \sqrt{3}$ and $2 \beta = -1 + i \sqrt{3}$.
Note that $\alpha = \omega$ and $\beta = \omega^2$ (or vice versa),where $\omega$ is the complex cube root of unity.
Thus,$\alpha + \beta = -1$ and $\alpha \beta = 1$.
We need to evaluate $5 \alpha^4 + 5 \beta^4 + \frac{7}{\alpha \beta}$.
Since $\alpha^3 = 1$ and $\beta^3 = 1$,we have $\alpha^4 = \alpha$ and $\beta^4 = \beta$.
So,$5 \alpha^4 + 5 \beta^4 + \frac{7}{\alpha \beta} = 5(\alpha + \beta) + \frac{7}{1}$.
Substituting $\alpha + \beta = -1$,we get $5(-1) + 7 = -5 + 7 = 2$.

(C) Given the equation $\alpha_1 + \alpha_2 z + \alpha_3 z^2 + \ldots + \alpha_n z^{n-1} + z^n = 0$.
Since $z = \cos \theta + i \sin \theta = e^{i \theta}$,we have $z^n = \cos n \theta + i \sin n \theta$.
Substituting $z$ into the equation: $\alpha_1 + \alpha_2 e^{i \theta} + \alpha_3 e^{i 2 \theta} + \ldots + \alpha_n e^{i (n-1) \theta} + e^{i n \theta} = 0$.
Multiply the entire equation by $e^{-i n \theta}$:
$\alpha_1 e^{-i n \theta} + \alpha_2 e^{-i (n-1) \theta} + \ldots + \alpha_n e^{-i \theta} + 1 = 0$.
Taking the real part of this equation:
$\alpha_1 \cos (-n \theta) + \alpha_2 \cos (-(n-1) \theta) + \ldots + \alpha_n \cos (-\theta) + 1 = 0$.
Since $\cos (-x) = \cos x$,this simplifies to:
$\alpha_1 \cos n \theta + \alpha_2 \cos (n-1) \theta + \ldots + \alpha_n \cos \theta + 1 = 0$.
Therefore,$\alpha_1 \cos n \theta + \alpha_2 \cos (n-1) \theta + \ldots + \alpha_n \cos \theta = -1$.

(D) Let $\omega = \frac{\sqrt{3}+i}{2} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.
Then the given expression is $z = (e^{i\pi/6})^5 + (e^{-i\pi/6})^5$.
$z = e^{i5\pi/6} + e^{-i5\pi/6}$.
Using the identity $e^{i\theta} + e^{-i\theta} = 2\cos(\theta)$, we get $z = 2\cos\left(\frac{5\pi}{6}\right)$.
Since $\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$, we have $z = 2\left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3}$.
Since $z = -\sqrt{3} + 0i$, the imaginary part $\operatorname{Im}(z) = 0$.

(B) Given the complex equation:
$i z^3+4 z^2-z+4 i=0$
Grouping the terms:
$z^2(i z+4) - 1(z - 4i) = 0$
Wait,let us factorize correctly:
$i z^3 + 4 z^2 - z + 4 i = 0$
$z^2(i z + 4) + i(i z + 4) = 0$
$(i z + 4)(z^2 + i) = 0$
This gives $z = \frac{-4}{i} = 4i$ or $z^2 = -i$.
For $z^2 = -i$,we write $-i$ in polar form: $z^2 = \cos(\frac{3\pi}{2}) + i \sin(\frac{3\pi}{2})$.
Using De Moivre's theorem,$z = \cos(\frac{3\pi}{4} + k\pi) + i \sin(\frac{3\pi}{4} + k\pi)$ for $k=0, 1$.
For $k=0$,$z = \cos(\frac{3\pi}{4}) + i \sin(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}$.
For $k=1$,$z = \cos(\frac{7\pi}{4}) + i \sin(\frac{7\pi}{4}) = \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}$.
The magnitudes are $|4i| = 4$ and $|z| = \sqrt{(\pm \frac{1}{\sqrt{2}})^2 + (\mp \frac{1}{\sqrt{2}})^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$.
Thus,the roots with minimum magnitude are $\pm(\frac{1-i}{\sqrt{2}})$.
Comparing with options,$\frac{1-i}{\sqrt{2}}$ is the correct choice.

(D) Given $z=x+iy$ and $x^2+y^2=1$, we can write $z=e^{i\phi}$ where $\phi = \tan^{-1}(\frac{y}{x})$.
$z_1 = ze^{i\theta} = e^{i\phi}e^{i\theta} = e^{i(\phi+\theta)}$.
Then $z_1^{2n} = e^{i2n(\phi+\theta)}$.
Consider the expression $\frac{z_1^{2n}-1}{z_1^{2n}+1} = \frac{e^{i2n(\phi+\theta)}-1}{e^{i2n(\phi+\theta)}+1}$.
Multiply numerator and denominator by $e^{-in(\phi+\theta)}$:
$= \frac{e^{in(\phi+\theta)} - e^{-in(\phi+\theta)}}{e^{in(\phi+\theta)} + e^{-in(\phi+\theta)}} = \frac{2i \sin(n(\phi+\theta))}{2 \cos(n(\phi+\theta))}$.
$= i \tan(n(\phi+\theta)) = i \tan(n(\theta+\tan^{-1}(\frac{y}{x})))$.

(D) Given $Z = \cos \theta + i \sin \theta$.
By De Moivre's Theorem,$Z^{2n} = \cos(2n\theta) + i \sin(2n\theta)$.
Substituting this into the expression:
$\frac{1 + Z^{2n}}{1 - Z^{2n}} = \frac{1 + \cos(2n\theta) + i \sin(2n\theta)}{1 - (\cos(2n\theta) + i \sin(2n\theta))} = \frac{1 + \cos(2n\theta) + i \sin(2n\theta)}{1 - \cos(2n\theta) - i \sin(2n\theta)}$.
Using half-angle identities $1 + \cos(2A) = 2 \cos^2 A$ and $1 - \cos(2A) = 2 \sin^2 A$,and $\sin(2A) = 2 \sin A \cos A$:
$= \frac{2 \cos^2(n\theta) + 2i \sin(n\theta) \cos(n\theta)}{2 \sin^2(n\theta) - 2i \sin(n\theta) \cos(n\theta)} = \frac{2 \cos(n\theta) [\cos(n\theta) + i \sin(n\theta)]}{2i \sin(n\theta) [-i \sin(n\theta) + \cos(n\theta)]}$.
$= \frac{\cos(n\theta)}{i \sin(n\theta)} = -i \cot(n\theta)$.

(C) The $n^{\text{th}}$ roots of a complex number $z = r(\cos \theta + i \sin \theta)$ are given by $z_k = r^{1/n} \operatorname{cis} \left( \frac{\theta + 2k\pi}{n} \right)$ for $k = 0, 1, \dots, n-1$.
For $z = -1$,we have $r = 1$ and $\theta = \pi$.
Thus,the $15^{\text{th}}$ roots are $\operatorname{cis} \left( \frac{\pi + 2k\pi}{15} \right)$ for $k = 0, 1, \dots, 14$.
For $k = 6$,the root is $\operatorname{cis} \left( \frac{\pi + 12\pi}{15} \right) = \operatorname{cis} \left( \frac{13\pi}{15} \right)$.

(C) Let $z = 1 \pm i \sqrt{3}$. We can write $z$ in polar form as $2(\cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3})$.
Then,$(1+i \sqrt{3})^n + (1-i \sqrt{3})^n = [2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})]^n + [2(\cos \frac{\pi}{3} - i \sin \frac{\pi}{3})]^n$.
Using De Moivre's Theorem,this becomes $2^n(\cos \frac{n \pi}{3} + i \sin \frac{n \pi}{3}) + 2^n(\cos \frac{n \pi}{3} - i \sin \frac{n \pi}{3})$.
Simplifying the expression,we get $2^n(2 \cos \frac{n \pi}{3}) = 2^{n+1} \cos \frac{n \pi}{3}$.

(C) We have the expression $S = \sum_{k=1}^6 \left(\sin \frac{2 \pi k}{7} - i \cos \frac{2 \pi k}{7}\right)$.
Factoring out $-i$,we get $S = -i \sum_{k=1}^6 \left(\cos \frac{2 \pi k}{7} + i \sin \frac{2 \pi k}{7}\right)$.
Using Euler's formula $e^{i\theta} = \cos \theta + i \sin \theta$,we have $S = -i \sum_{k=1}^6 e^{i \frac{2 \pi k}{7}}$.
Let $\omega = e^{i \frac{2 \pi}{7}}$. Then the sum is $S = -i \sum_{k=1}^6 \omega^k$.
Since $\omega$ is a $7^{th}$ root of unity,$1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = 0$.
Therefore,$\sum_{k=1}^6 \omega^k = -1$.
Substituting this into the expression for $S$,we get $S = -i(-1) = i$.

(C) Given $z = \frac{\sqrt{3}+i}{2} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.
$z^{101} = e^{i(101\pi/6)} = e^{i(16\pi + 5\pi/6)} = e^{i5\pi/6} = \cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} + \frac{i}{2}$.
Also,$i^{103} = i^{100} \cdot i^3 = 1 \cdot (-i) = -i$.
Thus,$z^{101} + i^{103} = -\frac{\sqrt{3}}{2} + \frac{i}{2} - i = -\frac{\sqrt{3}}{2} - \frac{i}{2} = -\left(\frac{\sqrt{3}+i}{2}\right) = -z$.
Therefore,$\left(z^{101} + i^{103}\right)^{105} = (-z)^{105} = -z^{105}$.
Since $z = e^{i\pi/6}$,$z^{105} = e^{i(105\pi/6)} = e^{i(17\pi + 3\pi/6)} = e^{i(17\pi + \pi/2)} = e^{i17\pi} \cdot e^{i\pi/2} = (-1) \cdot i = -i$.
Wait,let us re-evaluate: $-z^{105} = -\left(e^{i\pi/6}\right)^{105} = -e^{i(17\pi + \pi/2)} = -(-1 \cdot i) = i$.
Actually,$z^3 = e^{i(3\pi/6)} = e^{i\pi/2} = i$. Thus,$-z^{105} = -(-z^3) = z^3$ is incorrect. Let us re-calculate: $(-z)^{105} = -z^{105} = -(e^{i\pi/6})^{105} = -e^{i(17\pi + \pi/2)} = -(-1 \cdot i) = i$. Since $z^3 = i$,the answer is $z^3$.

(C) Let $z = 1+\sin \theta+i \cos \theta$. We can write this as $z = 1+\cos(\frac{\pi}{2}-\theta) + i\sin(\frac{\pi}{2}-\theta)$.
Using the identities $1+\cos(2A) = 2\cos^2(A)$ and $\sin(2A) = 2\sin(A)\cos(A)$,where $A = \frac{\pi}{4}-\frac{\theta}{2}$,we get:
$z = 2\cos^2(\frac{\pi}{4}-\frac{\theta}{2}) + 2i\sin(\frac{\pi}{4}-\frac{\theta}{2})\cos(\frac{\pi}{4}-\frac{\theta}{2})$
$z = 2\cos(\frac{\pi}{4}-\frac{\theta}{2}) [\cos(\frac{\pi}{4}-\frac{\theta}{2}) + i\sin(\frac{\pi}{4}-\frac{\theta}{2})]$
Then $z^n = 2^n \cos^n(\frac{\pi}{4}-\frac{\theta}{2}) [\cos(\frac{n\pi}{4}-\frac{n\theta}{2}) + i\sin(\frac{n\pi}{4}-\frac{n\theta}{2})]$.
Similarly,the conjugate term is $\bar{z}^n = 2^n \cos^n(\frac{\pi}{4}-\frac{\theta}{2}) [\cos(\frac{n\pi}{4}-\frac{n\theta}{2}) - i\sin(\frac{n\pi}{4}-\frac{n\theta}{2})]$.
Adding these two expressions:
$z^n + \bar{z}^n = 2^n \cos^n(\frac{\pi}{4}-\frac{\theta}{2}) [2\cos(\frac{n\pi}{4}-\frac{n\theta}{2})]$
$= 2^{n+1} \cos^n(\frac{\pi}{4}-\frac{\theta}{2}) \cos(\frac{n\pi}{4}-\frac{n\theta}{2})$.

(C) Given,$z = \frac{\sqrt{3} + i}{2} = \cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)$.
Using De Moivre's Theorem,$z^{101} = \cos \left(\frac{101\pi}{6}\right) + i \sin \left(\frac{101\pi}{6}\right)$.
Since $\frac{101\pi}{6} = 17\pi - \frac{\pi}{6}$,we have $z^{101} = \cos \left(17\pi - \frac{\pi}{6}\right) + i \sin \left(17\pi - \frac{\pi}{6}\right) = -\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right) = \frac{-\sqrt{3} + i}{2}$.
Also,$i^{103} = i^{100} \times i^3 = 1 \times (-i) = -i$.
Thus,$z^{101} + i^{103} = \frac{-\sqrt{3} + i}{2} - i = \frac{-\sqrt{3} - i}{2} = -\left(\frac{\sqrt{3} + i}{2}\right) = -z$.
Therefore,$\left(z^{101} + i^{103}\right)^{105} = (-z)^{105} = -z^{105}$.
$z^{105} = \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right)^{105} = \cos \frac{105\pi}{6} + i \sin \frac{105\pi}{6} = \cos \frac{35\pi}{2} + i \sin \frac{35\pi}{2}$.
Since $\frac{35\pi}{2} = 18\pi - \frac{\pi}{2}$,we have $z^{105} = \cos \left(-\frac{\pi}{2}\right) + i \sin \left(-\frac{\pi}{2}\right) = 0 - i = -i$.
Finally,$-z^{105} = -(-i) = i$.

(C) We can rewrite the expression as: $\sum_{n=1}^{20} \left[ \sin \left( \frac{2n\pi}{21} \right) - i \cos \left( \frac{2n\pi}{21} \right) \right] = -i \sum_{n=1}^{20} \left[ \cos \left( \frac{2n\pi}{21} \right) + i \sin \left( \frac{2n\pi}{21} \right) \right]$.
Using Euler's formula,$e^{i\theta} = \cos \theta + i \sin \theta$,the expression becomes $-i \sum_{n=1}^{20} e^{i(2n\pi/21)}$.
Let $\omega = e^{i(2\pi/21)}$. Then the sum is $-i \sum_{n=1}^{20} \omega^n$.
This is a geometric series with $20$ terms,where the first term is $\omega$ and the common ratio is $\omega$.
The sum is $\omega \frac{1-\omega^{20}}{1-\omega}$.
Since $\omega^{21} = e^{i(2\pi)} = 1$,we have $\omega^{20} = \omega^{-1} = \frac{1}{\omega}$.
Thus,the sum is $-i \left( \frac{\omega - \omega^{21}}{1-\omega} \right) = -i \left( \frac{\omega - 1}{1-\omega} \right) = -i (-1) = i$.

(B) Let $z = \frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}$.
Using the half-angle formulas $1+\cos \theta = 2 \cos^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$:
$z = \frac{2 \cos^2 \frac{\pi}{16} - 2i \sin \frac{\pi}{16} \cos \frac{\pi}{16}}{2 \cos^2 \frac{\pi}{16} + 2i \sin \frac{\pi}{16} \cos \frac{\pi}{16}}$
$z = \frac{2 \cos \frac{\pi}{16} (\cos \frac{\pi}{16} - i \sin \frac{\pi}{16})}{2 \cos \frac{\pi}{16} (\cos \frac{\pi}{16} + i \sin \frac{\pi}{16})}$
$z = \frac{\cos \frac{\pi}{16} - i \sin \frac{\pi}{16}}{\cos \frac{\pi}{16} + i \sin \frac{\pi}{16}} = \frac{e^{-i \frac{\pi}{16}}}{e^{i \frac{\pi}{16}}} = e^{-i \frac{2\pi}{16}} = e^{-i \frac{\pi}{8}}$.
Now,$z^8 = (e^{-i \frac{\pi}{8}})^8 = e^{-i \pi}$.
Using Euler's formula $e^{i \theta} = \cos \theta + i \sin \theta$:
$e^{-i \pi} = \cos(-\pi) + i \sin(-\pi) = -1 + 0 = -1$.

(D) Given $Z_r = \cos \left(\frac{\pi}{2^r}\right) + i \sin \left(\frac{\pi}{2^r}\right) = e^{i \frac{\pi}{2^r}}$.
The product is $P = Z_1 Z_2 Z_3 \ldots = e^{i \frac{\pi}{2^1}} \cdot e^{i \frac{\pi}{2^2}} \cdot e^{i \frac{\pi}{2^3}} \ldots$
Using the property of exponents,$P = e^{i \left( \frac{\pi}{2} + \frac{\pi}{4} + \frac{\pi}{8} + \ldots \right)}$.
The exponent is a geometric series with first term $a = \frac{\pi}{2}$ and common ratio $r = \frac{1}{2}$.
The sum of the infinite geometric series is $S = \frac{a}{1 - r} = \frac{\pi/2}{1 - 1/2} = \frac{\pi/2}{1/2} = \pi$.
Therefore,$P = e^{i \pi} = \cos \pi + i \sin \pi = -1 + i(0) = -1$.

(C) Let $z = \sqrt{\sqrt{2}+1} + i\sqrt{\sqrt{2}-1}$.
First,calculate $z^2 = (\sqrt{2}+1) - (\sqrt{2}-1) + 2i\sqrt{(\sqrt{2}+1)(\sqrt{2}-1)}$.
$z^2 = 2 + 2i\sqrt{2-1} = 2 + 2i$.
Now,$z^8 = (z^2)^4 = (2 + 2i)^4$.
$z^8 = [2(1+i)]^4 = 16(1+i)^4$.
Since $(1+i)^2 = 1 + 2i + i^2 = 2i$,we have $(1+i)^4 = (2i)^2 = 4i^2 = -4$.
Therefore,$z^8 = 16 \times (-4) = -64$.