According to question,

$15th$ roots of unity are $e^{\dfrac{{i}2{k}\pi}{15}}$ where $k=1,2,\cdots,15$

$p, q, r, s, t$ are all fifth root of unity
$\Rightarrow p^5=q^5= r^5= s^5= t^5=1$   
$ \Rightarrow p^{10}=q^{10}= r^{10}= s^{10}= t^{10}=1$
Hence, the required sum is $5$

Let $1,\alpha _1,\alpha _2,\alpha _3$ are the fourth rotts of unity.

$x^{4}=1$
$x^{2}=\pm1$
$x=\pm i$ and $x=\pm 1$
Hence
$\alpha^{4n-1}+\alpha^{4n-3}$
$=\alpha^{4n}[\alpha^{-1}+\alpha^{-3}]$
$=[\alpha^{-1}+\alpha^{-3}]$
$=\alpha^{-1}[1+\alpha^{-2}]$
$=\alpha^{-3}[\alpha^{2}+1]$
$=\alpha^{-3}[(\pm i)^{2}+1]$
$=0$

Since $1,{ \alpha  } _{ 1 },{ \alpha  } _{ 2 },{ \alpha  } _{ 3 },{ \alpha  } _{ 4 }$ are roots of the equation ${ x }^{ 5 }-1=0$

$z _{1}$ and $z _{2}$ subtend right angle at the origin. 

The $n^{th}$ roots of unity lie on a circle, where the angle between any 2 consecutive roots is $\dfrac { 2\pi  }{ n } $.
Hence, $ \dfrac{2r\pi}{n} = \dfrac{\pi}{2}$
$\Rightarrow 4r=n$
Hence, option D is correct.

From above concept

We have, $\displaystyle { \left| { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } \right|  }^{ 2 }=\left( { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } \right) \left( \overline { { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } }  \right) $

$x^{5}=-1$
Hence
$\alpha^{5}=-1$
Therefore
$\alpha^{10n+2}+\alpha^{5n+2}+\alpha^{5n}$
$=(\alpha^{5n})^{2}\alpha^{2}+(\alpha^{5n}).\alpha^{2}+\alpha^{5n}$
$=(-1)^{2}\alpha^{2}+(-1)\alpha^{2}+\alpha^{5n}$ .... Since n is odd
$=-\alpha^{2}+\alpha^{2}-1$
$=-1$

We have, $\displaystyle { x }^{ n }-1=\left( x-1 \right) \left( x-\omega  \right) \left( x-{ \omega  }^{ 2 } \right) ...\left( x-{ \omega  }^{ n-1 } \right) $

$x^{2n}=1$
Now if  $n$ is odd we have
$x^{n}=\pm1 $
$x^{n}=1$ and $x^{n}=-1$
$x^{n}=-1$
$x=-1$
Now if $n$ is odd
$x^{n}-1=0$
$(x-1)(1+x+x^{2}+..x^{n-1})=0$
Hence $x=1$ and  the equation $1+x+x^{2}+..x^{n-1}=0$ gives $nth$  roots of unity.
Hence at most $2$ real roots.
Similarly if $n$  is even.
Then
$x^{n}=\pm1 $
$x^{n}=1$ and $x^{n}=-1$
Now 
$x^{n}=-1$ will given imaginary roots.
$x^{n}=1$ can be further simplified in to
$x^{\frac{n}{2}}=\pm1 $ and so on.
Hence we will get remaining pairs of imaginary roots ans two real roots $1$ and  $-1$  at the end.
Hence at-most $2$ real roots.

$\alpha$ is non real root of $x^6 = 1$

$\displaystyle { z }^{ 5 }+{ z }^{ 4 }+{ z }^{ 3 }+{ z }^{ 2 }+z+1=0\ \Rightarrow \frac { { z }^{ 6 }-1 }{ z-1 } =0\ \Rightarrow z\neq 1\quad & \quad { z }^{ 6 }=1=\cos { 0 } +i\sin { 0 } \ \Rightarrow z={ \left( \cos { 0 } +i\sin { 0 }  \right)  }^{ \frac { 1 }{ 6 }  }=\cos { \frac { 2k\pi  }{ 6 }  } +i\sin { \frac { 2k\pi  }{ 6 }  } \ \Rightarrow z=\cos { \frac { k\pi  }{ 3 }  } +i\sin { \frac { k\pi  }{ 3 }  } $
where $k=0,1,2,3,4,5$.

For $k=0$,
$z=1$ but $z\neq 1$

For $k=1$,
$\displaystyle z=\cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  } =\frac { 1+i\sqrt { 3 }  }{ 2 } $

For  $k=2$,
$\displaystyle z=\cos { \frac { 2\pi  }{ 3 }  } +i\sin { \frac { 2\pi  }{ 3 }  } =\frac { -1+i\sqrt { 3 }  }{ 2 } $

For  $k=3$,
$\displaystyle z=\cos { \frac { 3\pi  }{ 3 }  } +i\sin { \frac { 3\pi  }{ 3 }  } =-1$

For  $k=4$,
$\displaystyle z=\cos { \frac { 4\pi  }{ 3 }  } +i\sin { \frac { 4\pi  }{ 3 }  } =\frac { -1-i\sqrt { 3 }  }{ 2 } $

For $k=4$,
$\displaystyle z=\cos { \frac { 5\pi  }{ 3 }  } +i\sin { \frac { 5\pi  }{ 3 }  } =\frac { 1-i\sqrt { 3 }  }{ 2 } $

Hence, all the options A,B,C and D are correct.

$(13-w)(13-w^{n-1})$
$=(13-w)(13-\dfrac{w^{n}}{w})$
$=(13-w)(13-\overline{w})$
$=169-13(w+\overline{w})+w\overline{w}$
$=169-13(Re(w))+1$
$=170-13(Re(w))$
Hence answer is none of these.

${ z }^{ 7 }=1\ \Rightarrow z={ 1 }^{ \frac { 1 }{ 7 }  }={ \left( \cos { 0 } +i\sin { 0 }  \right)  }^{ \frac { 1 }{ 7 }  }$
$\Rightarrow z=\cos { \frac { 2k\pi  }{ 7 }  } +i\sin { \frac { 2k\pi  }{ 7 }  } $       ...{ De Moivre's Theorem}
Where $k=0,1,2,3,4,5,6$

For $k=0$
$z=1$
And $z=\cos { \frac { 2k\pi  }{ 7 }  } +i\sin { \frac { 2k\pi  }{ 7 }  } $ for $k=1,2,3,4,5,6$

Ans:B

$z^{n-1}=\overline{z}$
Or 
$z^{n}=1$
This describes $nth$ roots of unity.
Hence let $z=e^{i\theta}$
$e^{in\theta}=e^{i2k\pi}$
Hence
$\theta=\dfrac{2k\pi}{n}$ Where $k\epsilon N$ and $0\leq K\leq n$
Hence
$z=cos\theta+isin\theta$
$=cos(\dfrac{2k\pi}{n})+isin(\dfrac{2k\pi}{n})$

$z _{1}= e^{i\dfrac{2k _{1}\pi}{n}}$         $z _2= e^{i\dfrac{2k _{2}\pi }{n}}$

Given, $z _{1}= z _{2}e^{i\ ^{\pi }/ _{2}}$

$\Rightarrow e^{i\left ( \dfrac{2k _{1}-2k _{2}}{n} \right ){\pi }} = e^{i\ ^{\pi }/ _{2}}$
$\Rightarrow \dfrac{2(k _{1}-k _{2})\pi }{n} = \dfrac{\pi }{2}$
$or\ 2= 4(k _{1}-k _{2})=4k$

Clearly,
$\alpha = e^{i0} = 1$
$\beta = e \frac{i2\pi}{4} = i$
$\gamma = e \frac{i4\pi}{4} = -1$
$\delta = e \frac{i6\pi}{4} = -i$
So, $\dfrac {a\alpha+b\beta+c\gamma+d\delta}{ a\gamma+b\delta+c\alpha+d\beta}=\dfrac {a+bi-c-di}{- a-bi+c+di}$
$=-1$
Similarly second expression is nothing but reciprocal of first
$=\frac{1}{-1}=-1$
Ans $ = -1-1 = -2$

For the fourth roots of unity, consider the following equation
$(x^{4}-1)=0$
or
$x^{4}=1$
or
$x^{2}=\pm1$ Therefore,
$x^{2}=1$ or $x^{2}=-1$. Hence
$x=\pm1$ and $x=\pm i$
Hence order of $(-i)$ is $1$. 

$Z _1 = e^{i\dfrac{2k _1\pi}{n}}$
$Z _2 = e^i{\frac{2k _2\pi}{n}}$
Now, $Z _1, Z _2$ subtend a right angle at origin
$\Rightarrow \frac {Z _1}{|Z _1|}=\frac {Z _2}{|Z _2|}e^{i(\frac{\pi}{2})}$

$\Rightarrow e^{i(k _1-k _2) \frac{2\pi}{n}} = e^{i(\frac{\pi}{2})} $

Hence, 

$ (k _1-k _2)\frac{2\pi}{n} = \frac{\pi}{2} $

$ \Rightarrow k _1 -k _2 = \frac{n}{4} $

As $k _1, k _2$ are integers, $n$ must be of the form $4k$

$x^n-1=(x-1)(x-z _1)(x-z _2)....(x-z _{n-1})$
$\Rightarrow \dfrac {x^n-1}{x-1}=(x-z _1)(x-z _2)....(x-z _{n-1})$
Putting $x=\omega$, we have
$\Pi _{r=1}^{n-1}(\omega-z _r)=\dfrac {\omega^n-1}{\omega-1}=\left{\begin{matrix}0 & if \ n=3k, k\epsilon Z \ 1, & if\  n=3k+1, k\epsilon Z \ 1+\omega, & if\   n=3k+2, k\epsilon Z\end{matrix}\right.$

Upon expanding, we get
$(a+b)+(2a+b)w+(3a+b)w^{2}+...(na+b)w^{n-1}$
$=a(1+2w+3w^{2}+...nw^{n-1})+b(1+w+w^{2}+...w^{n-1})$
$=a(1+2w+3w^{2}+...nw^{n-1})+b(\cfrac{1-w^{n}}{1-w})$
$=a(1+2w+3w^{2}+...nw^{n-1})+0$

It is given that, $p$ and $q$ are prime numbers.
Hence the only common $pth$ and $qth$ root of unity will be the number 1.
Thus there will be no common imaginary $pth$ and $qth$ root of unity.
Hence answer is zero.

Let $x={ \left( 16 \right)  }^{ { 1 }/{ 4 } }$

${ x }^{ 4 }=16$

${ x }^{ 4 }-16=0$

${ x }^{ 4 }-{ 2 }^{ 4 }=0$

$\left( { x }^{ 2 }-{ 2 }^{ 2 } \right) \left( { x }^{ 2 }+{ 2 }^{ 2 } \right) =0$

$\left( x-2 \right) \left( x+2 \right) \left( { x }^{ 2 }+4 \right) =0$

$x-2=0,x+2=0,{ x }^{ 2 }+4=0$

$x=2,-2$ or ${ x }^{ 2 }=-4\Longrightarrow x=\pm \sqrt { -4 } =\pm 2i$

$\therefore x=\pm 2,\pm 2i$

$(z^{6}-28)^{2}-784-512=0$
$(z^{6}-28)^{2}=1296$
$z^{6}-28=\pm36$
$z^{6}=64$ and $z^{6}=-8$
$z^{3}=\pm8$
$z=2$ and $z=-2$ ...(i)
$z^{6}=2^{3}.e^{i(2k-1)\pi}$
$z=2^{\frac{1}{2}}(e^{i\frac{(2k-1)\pi}{6}})$ where $k=1,2,3..6$.

Given $\displaystyle z^{n}= 1,z= 1,a _{1}, a _{2},..., a _{n-1}$
Let $\displaystyle a= \frac{1}{1-z}\Rightarrow z= 1-\frac{1}{a}$
 $\displaystyle \therefore \left ( 1-\frac{1}{a} \right )^{n}= 1$
$\displaystyle \Rightarrow \left ( a-1 \right )^{n}-a^{n}= 0$
$\displaystyle \Rightarrow -C _{1}a^{n-1}+C _{2}a^{n-2}+...+\left ( -1 \right )^{n}= 0$ where  $\displaystyle a= \frac{1}{1-a _{1}}, \frac{1}{1-a _{2}}.....\frac{1}{1-a _{n-1}}$

$\displaystyle \Rightarrow \frac{1}{1-a _{1}}+\frac{1}{1-a _{2}}+.....+\frac{1}{1-a _{n-1}}= \frac{^{n}c _{2}}{n}= \frac{n-1}{2}$

Given that $ 1,{ \alpha  } _{ 1 },{ \alpha  } _{ 2 },....,{ \alpha  } _{ n }$ are $n$th roots of unity

$\sum _{1 \leq i}^{ }  \sum _{j \leq 100}^{ } \alpha _{i}^{5} \alpha _{j}^{5}=(\alpha _{1}^{5}+\alpha _{2}^{5}.......+\alpha _{100}^{5})^{2}-(\alpha _{1}^{10}+\alpha _{2}^{10}.......+\alpha _{100}^{10})$
=0-0=0

$a={ e }^{ i2\pi /7 }\ a^ 7=1\ \alpha +\beta $

Given,

$Since\quad { z }^{ n }=1\ \Rightarrow \quad { \left| z \right|  }^{ n }=1\ \quad \quad \quad \left| z \right| =1\ similarly,\ { \left( z+1 \right)  }^{ n }=1\ \Rightarrow \quad { \left| z+1 \right|  }^{ n }=1\ \left| z+1 \right| =1\ Let\quad z=a+ib\ \left| z \right| =\left| z+1 \right| \quad \Rightarrow \quad { a }^{ 2 }+{ b }^{ 2 }={ \left( a+1 \right)  }^{ 2 }+{ b }^{ 2 }\ { a }^{ 2 }+{ b }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2a+1\ \Rightarrow \quad 2a+1=0\ \therefore \quad a=\frac { -1 }{ 2 } \ putting\quad the\quad value\quad of\quad a\quad in\quad eq.\ \Rightarrow \quad { \left( \frac { -1 }{ 2 }  \right)  }^{ 2 }+{ b }^{ 2 }=1\ \Rightarrow \quad { b }^{ 2 }=\frac { 3 }{ 4 } \ \Rightarrow \quad b=\pm \frac { \sqrt { 3 }  }{ 2 } \ Now,\quad z+1=\quad \frac { 1 }{ 2 } \pm \frac { \sqrt { 3 }  }{ 2 } \ \Rightarrow \quad z+1={ e }^{ \pm \frac { zni }{ 3 }  }\ { \left( z+1 \right)  }^{ n }=\quad { e }^{ \pm \frac { zni }{ 3 }  }\ For\quad { \left( z+1 \right)  }^{ n }\quad to\quad be\quad 1\quad cos\quad \pm \frac { zn }{ 3 } =1\quad and\quad sin\quad \pm \frac { zn }{ 3 } =0\ This\quad can\quad only\quad happen\quad if\quad \pm \frac { zn }{ 3 } =2ak\quad for\quad integer\quad k.\ Solving\quad for\quad n,\quad we\quad get:\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \pm \frac { zn }{ 3 } =2ak\quad \Rightarrow \quad n=6k\ \quad \quad \quad \quad \quad \quad \quad k=\frac { 6 }{ n } \ least\quad value=\quad 6\ $

We have,
$z^{5}-1=\left ( z-1 \right )\left ( z-\alpha _{1} \right )\left ( z-\alpha _{2} \right )\left ( z-\alpha 3 \right )\left ( z-\alpha _{4} \right )$
putting z=w,
$\Rightarrow w^{5}-1=\left ( w-1 \right )\left ( w-\alpha _{1} \right )\left ( w-\alpha _{2} \right )\left ( w-\alpha _{3} \right )\left ( w-\alpha _{4} \right )..............(1)$
Similarly putting $z=w^{2}$
$\left ( w^{2} \right )^{5}-1=w^{10}-1=\left ( w^{2}-1 \right )\left ( w^{2}-\alpha _{1} \right )\left ( w^{2}-\alpha _{2}\right )\left ( w^{2} -\alpha _{3}\right )\left ( w^{2}-\alpha _{4} \right ).........(2)$
So the required expression reduces to 
$ \displaystyle =\frac{\left ( w^{5} -1\right )/\left ( w-1 \right )}{\left ( w^{10}-1 \right )/\left ( w^{2} -1\right )}$
$=\frac{\left ( w^{2} -1\right )/\left ( w-1 \right )}{\left ( w-1 \right )/\left ( w^{2} -1\right )}$
$\left [ \because w^{3}=1,w^{5} =w^{3}w^{2}=w^{2}\right ]$
$=\frac{\left ( w^{2}-1 \right )^{2}}{\left ( w-1 \right )^{2}}$
$=\frac{\left ( w-1 \right )^{2}\left ( w+1 \right )^{2}}{\left ( w-1 \right )^{2}}$
$=1+2w+w^{2}$
$=w\left [ \because 1+w^{2} =-w\right ]$