N th root of unity - class-XII

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

For a non-real fifth root of unity alpha, 1 + alpha + alpha^2 + alpha^3 + alpha^4 = 0. The expression 1 + alpha + alpha^2 - alpha^-1 = 1 + alpha + alpha^2 - alpha^4 = 1 + alpha + alpha^2 - (-1 - alpha - alpha^2 - alpha^3) = ... This simplifies to 3^2 = 9.

The roots of z^n + z^2 + 1 = 0 are symmetric about the origin in the complex plane. When summing principal arguments, the positive and negative arguments cancel out, resulting in a sum of 0.

$p, q, r, s, t$ 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 $5$

The polynomial whose roots are the n-th roots of unity except 1 is x^(n-1) + x^(n-2) + ... + 1 = 0. Substituting x = 1 into the derivative or evaluating the sum using standard properties of roots of unity yields (n-1)/2.

The sum of 1/(1 - a_k) where a_k are the n-th roots of unity excluding 1 is given by the formula (n-1)/2. This is derived from the logarithmic derivative of (z^n - 1)/(z - 1).

The product |(2 - a_1)(2 - a_3)(2 - a_5)(2 - a_7)| for 9th roots of unity can be evaluated using the property of the polynomial z^9 - 1 = (z - 1)(z - a_1)...(z - a_8).

The roots of the first equation are the 18th roots of unity excluding 1. The roots of the second are the 14th roots of unity excluding 1. The common roots are the roots of unity that are both 18th and 14th roots, which are the gcd(18, 14) = 2nd roots of unity (excluding 1). This logic needs careful checking of the number of solutions.

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

This is a known summation identity for roots of unity. The sum of 1/(x + a_r) is related to the derivative of the polynomial P(z) = z^n - 1.

Expanding the sum |Z_1 + a^k Z_2|^2 yields sum(|Z_1|^2 + |Z_2|^2 + Z_1 bar(Z_2) a^k + bar(Z_1) Z_2 a^(-k)). Since sum(a^k) for k=0 to n-1 (where n=2014) is 0 for roots of unity, the cross terms vanish. We are left with 2014 * (|Z_1|^2 + |Z_2|^2) = 2014 * (1/2014 + 1/2014) = 2.

For two n-th roots of unity to subtend a right angle at the origin, their arguments must differ by pi/2. This means (2pi/n) * m = pi/2 for some integer m, which implies n = 4m. Thus, n must be of the form 4k.

For a matrix A = [[a, b], [0, a]] to be an nth root of I, A^n = I. This implies a^n = 1. If n is odd, a=1, b=0 is a solution. If n is even, a=1 or a=-1 are possible.

$x^{4}=1$
$x^{2}=\pm1$
$x=\pm i$ 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. 

This is identical to the standard roots of unity summation identity, where the sum evaluates to (n-1)/2 by considering the factorization of x^n - 1.

The equation |z - w|^2 + |z - w^2|^2 = lambda represents a circle if the constant term after simplification is positive. For w being the cube root of unity, this simplifies to a circle for certain values of lambda.

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.

The fourth roots of unity are 1, -1, i, and -i. Evaluating option C gives (i / sqrt(2)), which does not equal any of these standard roots when raised to the fourth power, as (i / sqrt(2))^4 = (-1/4) != 1. Therefore, option C is not a root of unity.

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
$=-\alpha^{2}+\alpha^{2}-1$
$=-1$

The nth roots of unity are given by e^(i 2 pi k / n) for k = 0, 1, ..., n-1. These roots are equally spaced points on the unit circle, subtending an angle of 2 pi / n at the origin. If two roots subtend a right angle (pi / 2 radians) at the origin, then 2 pi / n must divide pi / 2 or be related such that n is a multiple of 4, making n of the form 4k.

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) $

The product of (x - alpha_i) for all roots of unity is related to the polynomial x^n - 1. For n=7, the roots are 1, alpha_1, ..., alpha_6. The product (3 - alpha_1)(3 - alpha_2)...(3 - alpha_6) equals (3^7 - 1) / (3 - 1) = 2186 / 2 = 1093. The question asks for the product of three specific terms, which is the square root of the full product.

$x^{2n}=1$
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.

Expanding the square |z1 + w^k z2|^2 gives |z1|^2 + |z2|^2 + z1*conj(w^k)conj(z2) + conj(z1)w^k z2. Summing over k=0 to n-1, the cross terms involve the sum of powers of w, which is 0 for n-th roots of unity (w not equal to 1). Thus, the sum becomes n(|z1|^2 + |z2|^2).

${ z }^{ 7 }=1\ \Rightarrow z={ 1 }^{ \frac { 1 }{ 7 }  }={ \left( \cos { 0 } +i\sin { 0 }  \right)  }^{ \frac { 1 }{ 7 }  }$
$\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 $1$. 

The expression |z1 w^k z2|^2 simplifies to |z1|^2 * |w^k|^2 * |z2|^2. Since |w^k| = 1 for any root of unity, this is |z1|^2 * |z2|^2. Summing this n times yields n * |z1 z2|^2.

$Z _1 = e^{i\dfrac{2k _1\pi}{n}}$
$Z _2 = e^i{\frac{2k _2\pi}{n}}$
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 $4k$

All three identities are standard properties of n-th roots of unity. (i) Product of (1 - a_i) is n. (ii) Sum of roots is 0. (iii) The partial fraction sum is a known identity derived from the derivative of the polynomial x^n - 1.

$x^n-1=(x-1)(x-z _1)(x-z _2)....(x-z _{n-1})$
$\Rightarrow \dfrac {x^n-1}{x-1}=(x-z _1)(x-z _2)....(x-z _{n-1})$
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
$(a+b)+(2a+b)w+(3a+b)w^{2}+...(na+b)w^{n-1}$
$=a(1+2w+3w^{2}+...nw^{n-1})+b(1+w+w^{2}+...w^{n-1})$
$=a(1+2w+3w^{2}+...nw^{n-1})+b(\cfrac{1-w^{n}}{1-w})$
$=a(1+2w+3w^{2}+...nw^{n-1})+0$

The 5th roots of unity are the roots of the equation x^5 - 1 = 0, which can be factored as (x - 1)(x^4 + x^3 + x^2 + x + 1) = 0. For the non-unity roots a_1, a_2, a_3, a_4, the polynomial can also be expressed as (x - 1)(x - a_1)(x - a_2)(x - a_3)(x - a_4) = x^5 - 1. Substituting x = -1 into both sides gives (-1 - a_1)(-1 - a_2)(-1 - a_3)(-1 - a_4) = (-1)^5 - 1 = -2, which factors out to give the product (1 + a_1)...(1 + a_4) = 1.

A complex number is both an nth root and an mth root of unity if and only if it is an LCM(n, m)th root of unity, or more precisely, the common roots are the kth roots of unity where k is the greatest common divisor of n and m. Here, the common roots of the 15th and 25th roots of unity are the gcd(15, 25) = 5th roots of unity. Thus, there are 5 common roots.

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 )}$
$=\frac{\left ( w^{2} -1\right )/\left ( w-1 \right )}{\left ( w-1 \right )/\left ( w^{2} -1\right )}$
$\left [ \because w^{3}=1,w^{5} =w^{3}w^{2}=w^{2}\right ]$
$=\frac{\left ( w^{2}-1 \right )^{2}}{\left ( w-1 \right )^{2}}$
$=\frac{\left ( w-1 \right )^{2}\left ( w+1 \right )^{2}}{\left ( w-1 \right )^{2}}$
$=1+2w+w^{2}$
$=w\left [ \because 1+w^{2} =-w\right ]$