$x^{4}=1$
$x^{2}=\pm1$
$x^{2}=1$ and $x^{2}=-1$
$x=\pm1$ and $x=\pm i$
Hence the four roots are
$1,-1,i,-i$.
Now
$z _{1}=1=-z _{2}$
$z _{3}=i=-z _{4}$
Hence
$z _{1}^{2}+z _{2}^{2}+z _{3}^{2}+z _{4}^{2}$
$=1+1+(i)^{2}+(-i)^{2}$
$=2-2$
$=0$ ...(i)
And also
$z _{1}+z _{2}+z _{3}+z _{4}$
$=1-1+i-i$
$=0$ ...(ii)
Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

$x^5 - 1 = 0$ has roots $1, \alpha _1, \alpha _2, \alpha _3, \alpha _4$
$\therefore (x^5 - 1) = (x- 1) (x - \alpha _1) (x - \alpha _2) (x - \alpha _3) (x- \alpha _4)$
$\Rightarrow \displaystyle \frac{x^5 -1}{x - 1} = (x - \alpha _1) (x- \alpha _2) (x - \alpha _3) (x - \alpha _4)$           ........   (1)
Putting $x = \omega$ (1) we have
$\displaystyle \frac{\omega^5 - 1}{\omega - 1} = (\omega - \alpha _1) (\omega - \alpha _2) (\omega - \alpha _3) (\omega - \alpha _4)$
$\displaystyle \frac{\omega^2 - 1}{\omega - 1} = (\omega - \alpha _1) (\omega - \alpha _2) (\omega - \alpha _3) (\omega - \alpha _4)$         ....... (2)
and putting $x = \omega^2$ in (1) we have
$\displaystyle \frac{\omega^{10} - 1}{\omega^2 - 1} = (\omega^2- \alpha _1) (\omega^2 - \alpha _2) (\omega^2 - \alpha _3) (\omega^2 - \alpha _4)$
$\Rightarrow \displaystyle \frac{\omega - 1}{\omega^2 - 1} = (\omega^2- \alpha _1) (\omega^2 - \alpha _2) (\omega^2 - \alpha _3) (\omega^2 - \alpha _4)$           ....... (3)
Dividing (2) by (3)
then $\displaystyle \frac{\omega - \alpha _1}{\omega^2 - \alpha _1} \cdot \frac{\omega - \alpha _2}{\omega^2 - \alpha _2} \cdot \frac{\omega - \alpha _3}{\omega^2 - \alpha _3} \cdot \frac{\omega - \alpha _4}{\omega^2 - \alpha _4} \cdot = \frac{(\omega^2 - 1)^2}{(\omega - 1)^2}$
                                                                                  $= \displaystyle \frac{\omega^4 + 1 - 2 \omega^2}{\omega^2 + 1 - 2 \omega}$
                                                                                   $= \displaystyle \frac{\omega + 1 - 2 \omega^2}{\omega^2 + 1 - 2 \omega}$
                                                                                   $= \displaystyle \frac{- \omega^2 - 2 \omega^2}{- \omega - 2 \omega}$
                                                                                    $= \displaystyle \frac{- 3 \omega^2}{- 3 \omega}$
                                                                                    $= \omega$

Ans: C

$S=1+2\alpha+3\alpha^{2}+...n\alpha^{n-1}$
$\alpha S=\:\:\alpha+2\alpha^{2}+3\alpha^{3}+...(n-1)\alpha^{n-1}+n\alpha^{n}$
$S(1-\alpha)=1+\alpha+\alpha^{2}+\alpha^{3}+...\alpha^{n-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{1-\alpha^{n}}{1-\alpha}-n\alpha^{n}$
Now $\alpha^{n}=1$ since it is the $n^{th}$ root of unity.
Therefore,
$S(1-\alpha)=-n$
$S=\dfrac{-n}{1-\alpha}$

We have $1 + \alpha +\alpha^2 +\alpha^3 +\alpha^4 = 0$

$x^{n}-1=(x-1)(x-\alpha _{1})(x-\alpha _{2})(x-\alpha _{3})...(x-\alpha _{n-1})$
$\dfrac{x^{n}-1}{x-1}=(x-\alpha _{1})(x-\alpha _{2})(x-\alpha _{3})...(x-\alpha _{n-1})$
$(x-\alpha _{1})(x-\alpha _{2})(x-\alpha _{3})...(x-\alpha _{n-1})=1+x+x^{2}+...x^{n-1}$
Substituting  $x=-3$.
$(3+\alpha _{1})(3+\alpha _{2})(3+\alpha _{3})...(3+\alpha _{n-1})=1-3+3^{2}+...(-1)^{n-1}3^{n-1}$
Therefore $(3+\alpha _{1})(3+\alpha _{2})(3+\alpha _{3})...(3+\alpha _{n-1})$
$=\dfrac{1-(-3)^{n}}{1-(-3)}$
Now, $n$ is odd, therefore
$=\dfrac{3^{n}+1}{4}$

$S=1+2\alpha+3\alpha^{2}+....n\alpha^{n-1}$
$S\alpha=\alpha+2\alpha^{2}+3\alpha^{3}...(n-1)\alpha^{n-1}+n\alpha^{n}$
$S(1-\alpha)=1+\alpha+\alpha^{2}+...\alpha^{n-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{\alpha^{n}-1}{\alpha-1}-n\alpha^{n}$
Since $\alpha$ is the nth root of unity, hence $\alpha^{n}=1$
Thus
$S(1-\alpha)=\dfrac{\alpha^{n}-1}{\alpha-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{1-1}{\alpha-1}-n$
$S(1-\alpha)=-n$
$S=-\dfrac{n}{1-\alpha}$

Given, $z _\gamma = \cos{\frac{2 \pi \gamma}{5}} + i \sin{\frac{2 \pi \gamma}{5}}$ where $\gamma = 1, 2, 3, 4, 5.$
Thus, $z _\gamma = e^(i \frac{2 \pi \gamma}{5})$
Thus, $z _1z _2z _3z _4z _5 = e^(\frac{2 \pi}{5} + \frac{4 \pi}{5} + \frac{6 \pi}{5} + \frac{8 \pi}{5} + \frac{10 \pi}{5}) $
$= e^{6 \pi}$
$= \cos(6 \pi) + i \sin(6 \pi)$
$= 1$

If $z$ be the fourth roots of unity then, $z^4=1$
$\Rightarrow (z^4-1)=0\Rightarrow (z^2-1)(z^2+1)=0$
$\Rightarrow z=\pm 1, \pm i,$ where $i^2=-1$
$\therefore z _1^2+z _2^2+z _3^2+z _4^2=1+1+i^2+i^2=2-2=0$

$a = cos (2\pi/7)+i  sin(2\pi/7)$
$\Longrightarrow a^7 = [cos(2\pi/7)+i  sin(2\pi/7)]^7$
$= cos 2\pi + i sin 2\pi = 1$          (1)
$S = \alpha + \beta = (a + a^2 + a^4) + (a^3 + a^5 + a^6)$
$= a +a^2 +a^3 +a^4 + a^5 +a^6 = \frac{a(1-a^6)}{1-a}$
$= \frac{a-a^7}{1-a} = \frac{a-1}{1-a} = -1$                           (2)
$P= \alpha \beta = (a+a^2+a^4)(a^3+a^5+a^6)$
$= a^4+a^6 +a^7+a^5+a^7+a^8+a^7+a^9+a^{10}$
$= a^4+a^6+1+a^5+1+a+1+a^2+a^3$         [From Eq. (1)] 
$= 3+(a+ a^2+ a^3+ a^4+ a^5 + a^6)$ 
$= 3+S = 3-1=2$              [From Eq. (2)]
Therefore, the required equation is 
$x^2 -Sx + P = 0$
$\Longrightarrow x^2 + x + 2=0$


Ans: D

Real (z) $\displaystyle =\frac{z+\bar{z}}{2}=\frac{1}{2}\left ( z+\frac{1}{z} \right )$

$1+w+w^{2}+...w^{3}-\dfrac{1}{w}$
$=\dfrac{w+w^{2}+w^{3}+...w^{4}-1}{w}$

$^nC _1 + ^nC _2.\alpha + ^nC _3.\alpha^2 + ......... + ^nC _n.\alpha^{n1}$

$1+\alpha+...+\alpha^{5}=0$ [sum of n roots of unity]
$\Rightarrow 1+\alpha +{ \alpha  }^{ 3 }+{ \alpha  }^{ 5 }=-\left( { \alpha  }^{ 2 }+{ \alpha  }^{ 4 } \right) $
$\displaystyle \Rightarrow 1+\alpha +{ \alpha  }^{ 3 }+{ \alpha  }^{ 5 }=-{ \alpha  }^{ 2 }\left( { 1+\alpha  }^{ 2 } \right) $
$\displaystyle \Rightarrow \frac { 1+\alpha +{ \alpha  }^{ 3 }+{ \alpha  }^{ 5 } }{ { 1+\alpha  }^{ 2 } } =-{ \alpha  }^{ 2 }$

$s=1+2\alpha+3\alpha^{2}- -n\alpha^{n-1}$
$s\alpha=\alpha+2\alpha^{2}+3\alpha^{3}- - n\alpha
\So, s-s\alpha=1+\alpha+\alpha^{2}- -\alpha^{n-1}-n\alpha^{n}$
(Sum of roots of unity)
$\Rightarrow s (1-\alpha)=-n\alpha^{n}$
$\Rightarrow s=\dfrac{-n\alpha^{n}}{1-\alpha}=\dfrac{-n}{1-\alpha}$

as $\alpha^{n}=1$
$\alpha$ being of unity root

Since $2 + i \sqrt 3$ is one root, then other root will be $2 - i \sqrt 3$.
$\therefore x^2 + px + q = 0$ is given equatiion
$\therefore$ Sum of roots $= 2 + i \sqrt 3 + 2 - i \sqrt 3 = p$
$\therefore p = - 4$
Product of roots $q = 4 + 3 = 7$

Since $\omega $ is the $n^{th}$ root of unity, hence $\omega^{n}=1$.
Now inverse of $\omega ^{k}$ where $k<n$ is
$=\dfrac{1}{\omega ^{k}}$
$=\dfrac{\omega^{n}}{\omega^{k}}$
$=\omega^{n-k}$.
Hence inverse of $\omega^{k}$
$={\omega^{-k}}$
$=\omega^{n-k}$. 

The fourth roots of unity are $1,i,-1, -i $. Hence, they form a square in the Argand plane. 
We can see that by plotting the points on the graph with coordinate system according to that of complex numbers.

Since $1, \omega, \omega^2, \omega^3, ......... \omega^{n - 1}$ are nth roots of unity, therefore, we have the identity
$(x - 1)(x - \omega)(x - \omega^2) ...... (x - \omega^{n - 1})$
$= x^n - 1$
or $(x - \omega)(x - \omega^2) ...... (x - \omega^{n - 1}) = \displaystyle \frac{x^n - 1}{x - 1}$
$= x^{n - 1} + x^{n - 2} + ...... + x + 1$
Putting x = 1 on both sides, we get $(1 - \omega) (1 - \omega^2) (1 - \omega^{n - 1}) = n$

$2 - i$ and $\sqrt {5} + 2i$ are other roots.
So, Product is $(2 + i)(2 - i)(\sqrt {5} + 2i)(\sqrt {5} - 2i)$
$= 5\times 9 = 45$

If $\alpha _1, \alpha _2, ... , \alpha _m$ are the roots of polynomial equation
$\quad f(x) = a _0x^m + a _1x^{m-1} + ... + a _{m-1}x + a _m = 0$
Then,
$\quad f(x) = a _0(x-\alpha _1) ... (x-\alpha _m).$

${ x }^{ n }-1=0$ has n roots (of unity).
Thus, ${ x }^{ n }-1=(x-1)(x-\omega )(x-{ \omega  }^{ 2 })...(x-{ \omega  }^{ n-1 })$.
Substitute $x=5$: 
${ 5 }^{ n }-1=(5-1)(5-\omega )(5-{ \omega  }^{ 2 })...(5-{ \omega  }^{ n-1 })$
=> $(5-\omega )(5-{ \omega  }^{ 2 })...(5-{ \omega  }^{ n-1 })=\dfrac { { 5 }^{ n }-1 }{ 4 } $
Hence, option D is correct.

The roots of the equation $x^{2n + 1}- 1 = 0$ are
$1, \displaystyle cos \frac{2 \pi}{2n + 1} + i  sin  \frac{2 \pi}{2n + 1}, cos \frac{4 \pi}{2n + 1} + i  sin \frac{4 \pi}{2n + 1}, ........, cos \frac{4 n \pi}{2n + 1} + i  sin  \frac{4n  \pi}{2n + 1}$
Therefore $\displaystyle x^{2n+1} - 1 = (x - 1) \left ( x - cos \frac{2 \pi}{2n + 1} - i  sin \frac{2 \pi}{2n + 1} \right ) \left ( x - cos \frac{4 \pi}{2n + 1} - i  sin \frac{4 \pi}{2n + 1} \right ) ....... \left ( x - cos \frac{4 n\pi}{2n + 1} - i  sin \frac{4n \pi}{2n + 1} \right )$
Further since
$\displaystyle cos \left ( \frac{(2n + 1) - r}{2n + 1} \right ) 2\pi = cos \frac{2 r \pi}{2n + 1}$
and $\displaystyle sin \left ( \frac{(2n + 1) - r}{2n + 1} \right ) 2\pi = -sin \frac{2 r \pi}{2n + 1}$
it follows that
$\displaystyle \left ( x - cos \frac{2 \pi}{2n + 1} - i  sin \frac{2 \pi}{2n + 1}\right ) \left ( x - cos \frac{4 \pi}{2n + 1} - i  sin \frac{4 \pi}{2n + 1}\right )$
$= x^2 - 2x  cos \displaystyle \frac{2 \pi}{2n + 1} + 1$
$\left ( x - cos \frac{4 \pi}{2n + 1} - i  sin \frac{4 \pi}{2n + 1}\right )\left ( x - cos \frac{(4n - 2) \pi}{2n + 1} - i  sin \frac{(4n - 2) \pi}{2n + 1}\right )$
$=x^2 - 2x  cos \displaystyle \frac{4 \pi}{2n + 1} + 1$
$\left ( x - cos \frac{2 n\pi}{2n + 1} - i  sin \frac{2 n\pi}{2n + 1}\right )\left ( x - cos \frac{(2n + 2) \pi}{2n + 1} - i  sin \frac{(2n + 2)}{(2n + 1)} \pi\right )$
$= x^2 - 2x   cos \frac{2 n \pi}{2n + 1} + 1$
Thus the polynomial $x^{2n + 1} - 1$ can be rewritten thus
$x^{2n + 1} - 1 = (x - 1) \displaystyle \left ( x^2 - 2x  cos  \frac{2 \pi}{2n + 1} + 1\right ) \left ( x^2 - 2x  cos  \frac{4 \pi}{2n + 1} + 1\right )........ \left ( x^2 - 2x  cos  \frac{2 n\pi}{2n + 1} + 1\right ) $
or $\displaystyle \frac{x^{2n + 1} - 1}{x - 1} = \left ( x^2 - 2x  cos  \frac{2 \pi}{2n + 1} + 1\right ) \left ( x^2 - 2x  cos  \frac{4 \pi}{2n + 1} + 1\right ) ......... \left ( x^2 - 2x  cos  \frac{2 n\pi}{2n + 1} + 1\right ) $
Taking $\displaystyle \lim _{x \rightarrow 1}$ on both sides
$(2n + 1) = 2^{2n} sin^2 \displaystyle \frac{\pi}{2n + 1} sin^2 \frac{2 \pi}{2n + 1} ..... sin^2 \frac{n \pi}{2n + 1}$
Hence, $\displaystyle sin \frac{\pi}{2n + 1} sin \frac{2 \pi}{2n + 1} ...... sin \frac{n \pi}{2n + 1} = \frac{\sqrt{(2n + 1)}}{2^n}$

$1,\alpha ,{ \alpha  }^{ 2 }...{ \alpha  }^{ n-1 }$ are the nth roots of unity. Thus, they are solutions of the equation: ${ x }^{ n }-1=0$. 
Thus, product of roots $= { (-1) }^{ n }(\dfrac { -1 }{ 1 } )$
(i.e. ${ (-1) }^{ n }$constant term / coefficient of ${ x }^{ n }$)
Thus, the product = ${ (-1) }^{ n }(\dfrac { -1 }{ 1 } )={ (-1) }^{ n }(\dfrac { -1 }{ 1 } )={ (-1) }^{ n+1 }={ (-1) }^{ 2 }{ (-1) }^{ n-1 }=1{ (-1) }^{ n-1 }={ (-1) }^{ n-1 }$
Hence, (A) is correct.

Given : $\displaystyle \alpha = \cos\dfrac{8\pi}{11}+i\sin\dfrac{8\pi }{11}$
$\displaystyle \therefore \alpha ^{11}= \cos 8\pi +i\sin 8\pi = 1$
$\displaystyle \sum _{n= 0}^{10}\alpha ^{n}= \dfrac{1-\alpha ^{11}}{1-\alpha }= 0$ (sum of 11th roots of unity)
Now $\displaystyle Re(z)= \dfrac{z+\bar{z}}{2}=\dfrac{sum \ of\  11 \ roots\  of\  unity -1}{2}= -1/2 $

$S=1+2\alpha+3\alpha^{2}+...n\alpha^{n-1}$
Here $S$ forms an A.G.P
Therefore
$S=1+2\alpha+3\alpha^{2}+...n\alpha^{n-1}$
$S(\alpha)=\alpha+2\alpha^{2}+3\alpha^{3}+...(n-1)\alpha^{n-1}+n\alpha^{n}$
Hence
$S(1-\alpha)=1+\alpha+\alpha^{2}+....\alpha^{n-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{1-\alpha^{n}}{1-\alpha}-n\alpha^{n}$
$S(1-\alpha)=0-n$
$S=\dfrac{-n}{1-\alpha}$
Hence, option 'A' is correct.

$\displaystyle \frac{2^{2n + 1} - 1}{x - 1} = \left ( x^2 - 2x   cos  \frac{2 \pi}{2n + 1} + 1\right ) \left ( x^2 - 2x  cos \frac{4 \pi}{2n + 1} + 1\right ) ....... \left ( x^2 - 2x   cos \frac{2n  \pi}{2n + 1}
 + 1\right )$
Taking $\displaystyle \lim _{x \rightarrow - 1}$ on both sides, we have
$1 = 4^n cos^2 \displaystyle \frac{\pi}{2n + 1} cos^2 \frac{2 \pi}{2n + 1} ..... cos^2 \frac{n \pi}{2n + 1}$
$\therefore \displaystyle cos \frac{\pi}{2n + 1} cos \frac{2\pi}{2n + 1} ..... cos \frac{n \pi}{2n + 1} = \frac{1}{2^n}$

$\left| 1+r+{ r }^{ 2 }+{ r }^{ -2 }-{ r }^{ -1 } \right| =\left| 1+r+{ r }^{ 2 }+{ r }^{ 3 }-{ r }^{ 4 } \right| $

$\displaystyle 2\sum ab= \left ( \sum a \right )^{2}-\sum a^{2}$
$\displaystyle \therefore 2\sum \sum \left ( \alpha _{i}\alpha _{j} \right )^{5}= \left ( \alpha _{1}^{5}+\alpha _{2}^{5}+\cdots  \right )^{2}-\left ( \alpha _{1}^{10}+\alpha _{2}^{10}+\cdots  \right )$
$\displaystyle = 0-0$ ($\displaystyle \because \sum \alpha _{i}^{r}= 100$ if $\displaystyle r= 100k$
and $ \sum \alpha _{i}^{r} = 0$ if $\displaystyle r\neq 100k$)
Here both 5 and 10 are not multiples of 100.

$\displaystyle \sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7}$
$\displaystyle = -i^{2}\sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7}$
$\displaystyle = -i\left ( \cos \frac{2\pi k}{7}+i\sin \frac{2\pi k}{7} \right )= -i e^{i2\pi k/7}= -iz^{k}$
where, $\displaystyle z= \cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7}$ or $\displaystyle z^{7}= 1$
or $\displaystyle z= \left ( 1 \right )^{1/7} \therefore 1+z+2^{2}+\cdots +z^{6}= 0$ ...(1)
Above being the sum of seven, seventh roots of unity
$\displaystyle \sum = -i \sum _{i= 1}^{k} z^{k}= -1\left ( z+z^{2}+\cdots +z^{6} \right )= -i\left ( -1 \right )= i$ by (I)
Alternative Method. $\displaystyle \sin \theta -i\cos \theta $
$\displaystyle = -i^{2}\sin \theta -i\cos \theta = -i\left ( \cos \theta +i\sin \theta  \right )$
$\displaystyle = -ie^{i\theta }$ where $\displaystyle \theta = \frac{2\pi }{7}$ or $\displaystyle 7\theta = 2\pi $
Hence the given sigma is
$\displaystyle \sum _{k= 1}^{6}-ie^{ik\theta }= -i\left [ e^{i\theta +}e^{2i\theta }+\cdots +e^{6i\theta } \right ]$
$\displaystyle = -ie^{i\theta }\left [ \frac{1-\left ( e^{i\theta } \right )^{6}}{1-e^{i\theta }} \right ]= -i\left [ \frac{e^{i\theta }-e^{7i\theta }}{1-e^{i\theta }} \right ]$ (G.P.)
$\displaystyle = -i\left [ \frac{e^{i\theta }-1}{1-e^{i\theta }} \right ]= -i\left ( -1 \right )= i$
$\displaystyle \because 7i\theta = 2\pi i \therefore e^{7i\theta }= \left ( \cos 2\pi +i\sin 2\pi  \right )= 1.$

${ (cosx+i sinx) }^{ n }=cos(nx)+i sin(nx)$
Comparing imaginary parts: $sin(nx)= _{ 1 }^{ n }{ C }sinx{ .cos }^{ n-1 }x- _{ 3 }^{ n }{ C }sin^{ 3 }x{ .cos }^{ n-3 }x+.....$
Divide both sides by $sin^{ n }x$
$\dfrac { sin(nx) }{ sin^{ n }x } = _{ 1 }^{ n }{ C }cot^{ n-1 }x- _{ 3 }^{ n }{ C }cot^{ n-3 }x+...$
Replace $n$ with $2n+1:$ $\dfrac { sin((2n+1)x) }{ sin^{ 2n+1 }x } = _{ 1 }^{ 2n+1 }{ C }cot^{ 2n }x- _{ 3 }^{ 2n+1 }{ C }cot^{ 2n-2 }x+...$
$ = _{ 1 }^{ 2n+1 }{ C }(cot^{ 2 }x)^{ n }- _{ 3 }^{ 2n+1 }{ C }(cot^{ 2 }x)^{ n-1 }+.... ----1)$
So $cot^{ 2 }x _{ i }$ are the roots of the polynomial for LHS=0 where $cot^{ 2 }x _{ i }=cot^{ 2 }(\dfrac { \pi i }{ 2n+1 } )$ where i=1,2,3,...n.
Sum of the roots is the ratio of the first two coefficients on the RHS.
Thus $\sum _{ i=1 }^{ n }{ cot^{ 2 }(\dfrac { \pi i }{ 2n+1 } )= } -\dfrac { - _{ 3 }^{ 2n+1 }{ C } }{ _{ 1 }^{ 2n+1 }{ C } } =\dfrac { (2n+1)2n(2n-1) }{ 6 } .\dfrac { 1 }{ 2n+1 } =\dfrac { n(2n-1) }{ 3 } $
Hence, (c) is correct.