Tag: n th root of unity

Questions Related to n th root of unity

lf $a=\displaystyle \cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}, \alpha=a+a^{2}+a^{4}$ and $\beta=a^{3}+a^{5}+a^{6}$, then $\alpha, \beta$ are the roots of the equation

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $x^{2}+x+1=0$

  2. $x^{2}+x+2=0$

  3. $x^{2}+2x+2=0$

  4. $x^{2}+2x+3=0$

Show answer
Correct Option: B
Explanation:

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

$=a+a^ 2+a^ 3+a^ 4+a^ 5+a^ 6\ =\dfrac{a(a^ 6-1)}{(a-1)}\ =\dfrac{(a^ 7-a)}{(a-1)}\ =\dfrac{(1-a)}{(a-1)}$
$=-1$
$\alpha \beta =(a+a^ 2+a^ 4)(a^ 3+a^ 5+a^ 6)\ =(a^ 4+a^ 6+a^ 7+a^ 5+a^ 7+a^ 8+a^ 7+a^ 9+a^ {10})\ =(a^ 4+a^ 6+1+a^ 5+1+a+1+a^ 2+a^ 3)\ =(3+a+a^ 2+a^ 3+a^ 4+a^ 5+a^ 6)\ =(3-1)$
$=2 $
The equation can be written as
 $x^ 2-(\alpha +\beta )x +\alpha \beta  =x^ 2+x+2$
Hence, option B is correct.

If the expression $z^5 =32$ can be factorised into linear and quadratic factors over real coefficients as $(z^5 - 32)=(z - 2) (z^2-pz+4)(z^2-qz+4)$, where p > q, then the value of $p^2-  2q$
finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $8$

  2. $4$

  3. $-4$

  4. $-8$

Show answer
Correct Option: A
Explanation:

Given,

$ { z }^{ 5 }=32 $ can be factorized as
$(z-2)({ z }^{ 2 }-pz+4)({ z }^{ 2 }-qz+4)$ where $ p>q.$
 To find the value of $ { p }^{ 2 }-2q$
 Solution,
${ z }^{ 5 }=32\ { z }^{ 5 }-32=0$
$ { z }^{ 5 }-{ 2 }^{ 5 }=0$
$ z=2$ 
Will be one of the factor of given equation.
$ \quad \quad \quad \quad \quad \quad \quad { z }^{ 4 }+{ 2z }^{ 3 }+{ 4z }^{ 2 }+8z+16\ \therefore \quad z-2\sqrt { { z }^{ 5 }-32\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  } \ \quad \quad \quad \cfrac { \pm { z }^{ 5 }\mp { 2z }^{ 4 } }{ { \quad \quad \quad \quad \quad \quad 2z }^{ 4 }-32 } \ \quad \quad \quad \quad \quad \quad \cfrac { \pm { 2z }^{ 4 }\mp 4{ z }^{ 3 } }{ \quad \quad \quad \quad \quad \quad \quad 4{ z }^{ 3 }-32 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cfrac { \pm 4{ z }^{ 3 }\mp 8{ z }^{ 2 } }{ \quad \quad \quad \quad \quad \quad \quad 8{ z }^{ 2 }-32 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cfrac { \pm 8{ z }^{ 2 }\mp 16z }{ \quad \quad \quad \quad \quad \quad \quad 16z-32 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cfrac { \pm 16z\mp 32 }{ 0 } \ (z-2)({ z }^{ 4 }+2{ z }^{ 3 }+4{ z }^{ 2 }+8z+16)\longrightarrow (1)\ { z }^{ 5 }-32=(z-2)({ z }^{ 2 }-pz+4)({ z }^{ 2 }-qz+4)\quad \ { z }^{ 5 }-32=(z-2)({ z }^{ 4 }+(p+q)z^{ 3 }+(8+pq){ z }^{ 2 }+16-(4p+4q)z)\longrightarrow (2)$
On comparing $ (1)& (2)$ we get
$ p=-2-q\ 8+(-2-q)q=4\ -2q-{ q }^{ 2 }=-4\ { q }^{ 2 }+2q-4=0\ d=4+16=20\ q=\cfrac { -2\pm 2\sqrt { 5 }  }{ 5 } \ q=-1\pm \sqrt { 5 } \ { \parallel  }^{ rly }p=-1\pm \sqrt { 5 } $
But$\quad p>q\therefore p=-1+\sqrt { 5 } & q=-1-\sqrt { 5 } $
value of ${ p }^{ 2 }-2q={ \left( -1+\sqrt { 5 }  \right)  }^{ 2 }+2{ \left( -1-\sqrt { 5 }  \right)  }\ =1+5-2\sqrt { 5 } +2+2\sqrt { 5 } \ { p }^{ 2 }-2q=8\ $

Suppose A is a complex number and $ n \in N, $ such that $A^{n} = (A + 1)^{n} =1, $ then the least value of $n$ is

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $3$

  2. $6$

  3. $9$

  4. $12$

Show answer
Correct Option: B
Explanation:

$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\ $

If $1,$$\alpha _{1},\alpha _{2,} \alpha _{3},\alpha _{4}$ be the  roots of $z^{5}-1=0$ and $\omega $ be an imaginary cube root of unity, 


then  $ \displaystyle \left ( \frac{\omega -\alpha _{1}}{\omega ^{2}-\alpha _{1}} \right )\left ( \frac{\omega -\alpha _{2}}{\omega ^{2}-\alpha _{2}} \right )\left ( \frac{\omega -\alpha _{3}}{\omega ^{2}-\alpha _{3}} \right )\left ( \frac{\omega -\alpha _{4}}{\omega ^{2}-\alpha _{4}} \right )$ is ?

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $\omega $

  2. $\omega^{2}$

  3. $1$

  4. $2$

Show answer
Correct Option: A
Explanation:

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