Tag: finding nth roots of a complex number

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

Solutions of the equation $z^{7}-1=0$ are given by

  1. $\displaystyle z=-1,z=\cos \frac{2k\pi }{7}+i\sin \frac{2k\pi }{7},k=0,1,2, 3, 4, 5$
  2. $\displaystyle z=1\; and \; z=\cos \frac{2k\pi }{7}+i\sin \frac{2k\pi }{7},k=1,2,3, 4, 5, 6$
  3. $\displaystyle z=-1,z=\cos \frac{k\pi }{7}+i\sin \frac{k\pi }{7},k=0,1,2, 3, 4, 5$
  4. $\displaystyle z=1 \; and \; z=\cos \frac{k\pi }{7}+i\sin \frac{k\pi }{7},k=0,1,2,3, 4, 5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

Solve the equation $\displaystyle z^{n-1}=\bar{z},n\epsilon N.$

  1. $\displaystyle z =\sin \frac{2m\pi }{n}+i\cos \frac{2m\pi }{n}$
  2. $\displaystyle z =\sin \frac{2m\pi }{n}-i\cos \frac{2m\pi }{n}$
  3. $\displaystyle z =\cos \frac{2m\pi }{n}-i\sin \frac{2m\pi }{n}$
  4. $\displaystyle z =\cos \frac{2m\pi }{n}+i\sin \frac{2m\pi }{n}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$z^{n-1}=\overline{z}$
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
$z=cos\theta+isin\theta$
$=cos(\dfrac{2k\pi}{n})+isin(\dfrac{2k\pi}{n})$

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

lf $z _{1},z _{2}$ are $n^{th}$ roots of unity which are ends of a line segment that subtends $\displaystyle \frac{\pi}{2}$ at the origin. 

then $\mathrm{n}$ is of the form.

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

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

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

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

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

If $\alpha,\ \beta,\ \gamma$ and $\Delta $ are the roots of the equation $x^{4}-1=0$, then the value of $\displaystyle \frac{a\alpha+b\beta+c\gamma+d\Delta}{a\gamma+b\Delta +c\alpha+d\beta}+\frac{a\gamma+b\Delta +c\alpha+d\beta}{a\alpha+b\beta+c\gamma+d\Delta }$ is

  1. $ 3\beta$
  2. $0$
  3. $ 2\gamma$
  4. $-2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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$

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

The number of roots of the equation $z^{15}=1$ satisfying $|\arg(z)|<\pi/2$ is

  1. 6

  2. 7

  3. 8

  4. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
For the nth root of unity of complex number $z$ i.e. $z^n = 1$, there are 'n' total roots.
In the present case, $n=15$, thus, we have 15 roots.

$|arg(z)|<\cfrac {\pi}{2}$ 
$\Rightarrow -\cfrac {\pi}{2} < arg(z) < \cfrac {\pi}{2}$

Each root is at equal angular distance i.e. $\dfrac{2\pi}{15}$      ...(because $\dfrac{2\pi}{n}$).

$\therefore$ between $-\cfrac {\pi}{2}$ and $\cfrac {\pi}{2}$, their will be 7 roots (imcluding 1).
Hence, the correct option is B. 
Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

lf 1, $a _{1},\ a _{2},...,\ a _{n-1}$ are $n^{th}$ roots of unity then $\displaystyle \frac{1}{1-a _{1}}+\frac{1}{1-a _{2}}+\ldots+\frac{1}{1-a _{n-1}}$ equals?

  1. $\displaystyle \frac{2^{n}-1}{n}$
  2. $\displaystyle \frac{n-1}{2}$
  3. $\displaystyle \frac{n}{n-1}$
  4. $\displaystyle \frac{n}{n+1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We know that $w,w^{2},1$ are the cube roots of unity
Hence,
$\cfrac{1}{1-w}+\cfrac{1}{1-w^{2}}$
$=\cfrac{1}{1-w}+\cfrac{1}{(1-w)(1+w)}$
Now,
$=\cfrac{1}{1-w}(1+\cfrac{1}{1+w})$
$=\cfrac{1}{1-w}(\cfrac{2+w}{1+w})$
$=\cfrac{2+w}{1-w^{2}}$
We know 
$1+w+w^{2}=0$
$2+w+w^{2}=1$ ...(by adding 1 on both sides).
$2+w=1-w^{2}$ substituting we get
$=\cfrac{2+w}{2+w}$
$=1$
$=\cfrac{3-1}{2}$
Hence, the correct alternative is $\cfrac{n-1}{2}$
Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $w \neq 1$ is $n^{th}$ root of unity, then value of $ \displaystyle \sum _{k=0}^{n-1} \left| z _{1} w^{k} z _{2} \right| ^{2}$ is

  1. $n( \left| z _{1} z _{2}\right| ^{2})$
  2. $ \left| z _{1}\right| ^{2}+\left| z _{2}\right| ^{2}$
  3. $( \left| z _{1}\right|+\left| z _{2}\right|) ^{2}$
  4. $n ( \left| z _{1}\right|+\left| z _{2}\right|) ^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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

Let $z _1$ and $z _2$ be ${ n }^{ th }$ roots of unity which subtend a right angle at the origin. Then n must be of the form

  1. 4k + 1

  2. 4k + 2

  3. 4k + 3

  4. 4k
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

If 1, ${ a } _{ 1 },{ a } _{ 2 },....{ a } _{ n-1 }$ are the nth roots of unity then 
i) $\left( 1-{ a } _{ 1 } \right) \left( 1-{ a } _{ 2 } \right) \left( 1-{ a } _{ 3 } \right) ......\left( 1-{ a } _{ n-1 } \right) =n$
ii) $1+{ a } _{ 1 }+{ a } _{ 2 }+....+{ a } _{ n-1 }=0$
iii) $\dfrac { 1 }{ 2-{ a } _{ 1 } } +\dfrac { 1 }{ 2-{ a } _{ 2 } } +....+\dfrac { 1 }{ 2-{ a } _{ n-1 } } =\dfrac { \left( n-2 \right) { 2 }^{ n-1 }+1 }{ { 2 }^{ n }-1 } $

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.