Tag: demoivre's theorem

Questions Related to demoivre's theorem

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

If $1,{\alpha _1},{\alpha _2}....{\alpha _8}$ are nine, ninth roots of unity (taken in counter-clock wises direction) then $\left| {\left( {2 - {\alpha _1}} \right)\left( {2 - {\alpha _3}} \right)\left( {2 - {\alpha _5}} \right)\left( {2 - {\alpha _7}} \right)} \right|$ is equal to

  1. $\sqrt {255} $
  2. $\sqrt {1023} $
  3. $\sqrt {511} $
  4. $\sqrt {15} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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).

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

Number of values of $z$ (real or complex) simultaneously satisfying the system of equations
$1+z+{z}^{2}+{z}^{3}+....+{z}^{17}=0$ and $1+z+{z}^{2}+{z}^{3}+.....+{z}^{13}=0$ is

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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.

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

If $1,{\alpha _1},{\alpha _2},{\alpha _3}$ are the fourth roots of unity, then the value of $\left( {1 + {\alpha _1}} \right)\left( {1 + {\alpha _2}} \right)\left( {1 + {\alpha _3}} \right)$ is equal to

  1. $-3$
  2. $-1$
  3. $0$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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


We know that the fourth roots of unity are $1,i,-1,-i$


All these roots are got by solving equation $x=(1)^{\dfrac{1}{4}}$

By using demovire's theorem.

Now,

$(1+\alpha _1)(1+\alpha _2)(1+\alpha _3)$

$\Rightarrow$  $(1+i)(1+(-1))(1+(-i))$

$\Rightarrow$  $(1+i)(0)(1-i)$

$\Rightarrow$  $0$

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

If $n^{th}$ root of unity be $1,a _{1},a _{2},...a _{n-1}$, then $\displaystyle \sum^{n-1} _{r=1}\dfrac {1}{2+a _{r}}$ is equal to

  1. $\dfrac {n.2^{n-1}}{2^{n}-1}-1$
  2. $\dfrac {n(-2)^{n-1}}{(-2)^{n}-1}-1$
  3. $\dfrac {n(-2)^{n-1}}{1+(-2)^{n+1}}-\dfrac {1}{3}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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

Let $a^{k}$ where $k=0.1.2....2013$ are the $2014^{th}$ roots of unity. If $Z _{1}$ and $Z _{2}$ be any two complex number such that $|Z _{1}|=|Z _{2}|=\dfrac{1}{\sqrt{2014}}$, then the value of $\displaystyle \sum _{ k=0 }^{ 2013 }{ { \left| { Z } _{ 1 }+{ a }^{ k }{ Z } _{ 2 } \right|  }^{ 2 } } $ is equal to

  1. $4028$
  2. $0$
  3. $2$
  4. $2014$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $A=\begin{bmatrix} a & b\ 0 & a\end{bmatrix}$ is nth root of $I _2$, then choose the correct statements.

  1. If n is odd, $a=1$, $b=0$
  2. If n is odd, $a=-1, b=0$
  3. If n is even, $a=1, b=0$
  4. If n is even, $a=-1, b=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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

The value of $\displaystyle\ \alpha^{4n-1}+\alpha^{4n-3}, n\epsilon\mathbb{N}$ and $\displaystyle\ \alpha$ is a nonreal fourth root of unity is 

  1. $0$
  2. $-1$
  3. $3$
  4. none of these

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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