Mathematics

Complex Variables and Numbers

206 Questions

Complex numbers and variables form a crucial part of advanced mathematics syllabi. This topic covers roots of unity, Argand plane geometry, and Z-transforms. These concepts are frequently tested in engineering entrance exams and UPSC mathematics optional papers.

Roots of unityArgand plane geometryZ-transform sequencesExponential complex formsComplex number quadrantsRiemann hypothesis

Complex Variables and Numbers Questions

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

The number of common roots of the 15th and  of 25th roots of unity are

  1. 1

  2. 5

  3. 6

  4. 10
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

$25th$ roots of unity are $e^{\dfrac{{i}2{m}\pi}{25}}$ where $m=1,2,3,\cdots,25$
The common roots are $e^{i\dfrac{2{n}}{5}}$ where $n=1,2,3,4,5$
Number of common roots will be $5$

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

If $\alpha $ is a non- real fifth root of unity, then the value of ${3^{\left[ {1 + a + {a^2} - {a^{ - 1}}} \right]}}$,is

  1. $9$
  2. $1$
  3. $11/3$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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

Let principle argument of complex number be re-defined between $(\pi,3\pi)$, then sum of principle arguments of roots of equation $z^{n}+z^{2}+1=0$ is

  1. $0$
  2. $3\pi$
  3. $6\pi$
  4. $12\pi$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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 $n^{th}$ roots of unity then $\dfrac{1}{1-a _1}+\dfrac{1}{1-a _2}+....+\dfrac{1}{1-a _{n-1}}$ equals?

  1. $\dfrac{2^n-1}{n}$
  2. $\dfrac{n-1}{2}$
  3. $\dfrac{n}{n-1}$
  4. $\dfrac{n}{n+1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

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

If $1,{z} _{1},{z} _{2},{z} _{n-1}$ are the ${n}^{th}$ roots of unity then the value of $\dfrac{1}{3-z _{1}}+\dfrac{1}{3-z _{2}}+.......+\dfrac{1}{3-z _{n-1}}$ is equal to 

  1. $\displaystyle\frac { { n.3 }^{ n-1 } }{ { 3 }^{ n }-1 } +\frac { 1 }{ 2 }$
  2. $\displaystyle\frac { { n.3 }^{ n-1 } }{ { 3 }^{ n }-1 } -1$
  3. $\displaystyle\frac { { n.3 }^{ n-1 } }{ { 3 }^{ n }-1 } +1$
  4. $\displaystyle\frac { { n.3 }^{ n-1 } }{ { 3 }^{ n }-1 } -\frac { 1 }{ 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$z _{1}, z _{2} , z _{3}........ z _{n-1}$ are normal $n^{th}$
$x^{n}-1= (x-1) (x-z _{1}) (x-z _{2}) (x-z _{3})....... (x-z _{n-1})$
ln $(x^{n}-1)= ln (x-1) (x-z _{2}) (x-z _{3})......... (x-z _{n-1})$
$\dfrac{nx^{n-1}}{x^{n}-1}= \dfrac{1}{x-1} + \dfrac{1}{x- z _{1}} + \dfrac{1}{x- _{2}}...... \dfrac{1}{x- z _{n-1}}$
Putting $x=3$
$\dfrac{n3^{n-1}}{3^{n-1}}= \dfrac{1}{2} + \dfrac{1}{3- z _{1}}+ \dfrac{1}{3-z _{2}}........ \dfrac{1}{3-z _{n-1}}$
$\dfrac{1}{3-z _{1}}+ \dfrac{1}{3- z _{2}}.............. \dfrac{1}{3-z _{n-1}}= \dfrac{n(3)^{n-1}}{3^{n}-1}\dfrac{1}{2}$
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 ${n^{th}}$ roots of unity then $\frac{1}{{1 - {a _1}}} + \frac{1}{{1 - {a _2}}} + .... + \frac{1}{{1 - {a _{n - 1}}}}$ equals                                                            

  1. $\frac{{{2^n} - 1}}{n}$
  2. $\frac{{n - 1}}{2}$
  3. $\frac{n}{{n - 1}}$
  4. $\frac{n}{{n + 1}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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).

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

If $1, \alpha _1, \alpha _2, \alpha _3, \alpha _4, \alpha _5, \alpha _6$, are seven, $7^{th}$ root of unity them $|(3-\alpha _1)(3-\alpha _3)(3-\alpha _5)|$ is?

  1. $\sqrt{2186}$
  2. $\sqrt{1093}$
  3. $\sqrt{1023}$
  4. $\sqrt{511}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$x^7 -1 = 0$
$(x-1)(x-\alpha _1).....(x-\alpha _6) = x^7-1$
Putting $x=3$, we get
$2.(x-\alpha _1).(x-\alpha _2)...(x-\alpha _6) = 2186$
$|(x-\alpha _1)|^2 |(x- \alpha _3)|^2 |(x-\alpha)|^2 = 1093$
$|(x-\alpha _1)(x- \alpha _3)(x-\alpha _5)| = \sqrt{1093}$

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

If $z _{1}$ and $z _{2}$ be the $n^{th}$ 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

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.

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

If $1,{ \alpha  } _{ 1 },{ \alpha  } _{ 2 },{ \alpha  } _{ 3 }$ and $\alpha _4$ be the roots of $x^5-1=0$, then $\displaystyle \frac { \omega -{ \alpha  } _{ 1 } }{ { \omega  }^{ 2 }-{ \alpha  } _{ 1 } } .\frac { \omega -{ \alpha  } _{ 2 } }{ { \omega  }^{ 2 }-{ \alpha  } _{ 2 } } .\frac { \omega -{ \alpha  } _{ 3 } }{ { \omega  }^{ 2 }-{ \alpha  } _{ 3 } } .\frac { \omega -{ \alpha  } _{ 4 } }{ { \omega  }^{ 2 }-{ \alpha  } _{ 4 } } =$ 

  1. $1$
  2. $\omega$
  3. ${ \omega }^{ 2 }$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since $1,{ \alpha  } _{ 1 },{ \alpha  } _{ 2 },{ \alpha  } _{ 3 },{ \alpha  } _{ 4 }$ are roots of the equation ${ x }^{ 5 }-1=0$

Thus ${ x }^{ 5 }-1=\left( x-1 \right) \left( x-{ \alpha  } _{ 1 } \right) \left( x-{ \alpha  } _{ 2 } \right) \left( x-{ \alpha  } _{ 3 } \right) \left( x-{ \alpha  } _{ 4 } \right)$
$ \Rightarrow \displaystyle\frac { { x }^{ 5 }-1 }{ \left( x-1 \right)  } =\left( x-{ \alpha  } _{ 1 } \right) \left( x-{ \alpha  } _{ 2 } \right) \left( x-{ \alpha  } _{ 3 } \right) \left( x-{ \alpha  } _{ 4 } \right) $........1
Putting $x=w$ in 1 we get,
$\Rightarrow \displaystyle\frac { { 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) $
$\Rightarrow \displaystyle\frac { { w }^{ 2 }-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) $............2
Putting $x={ w }^{ 2 }$ in 1 we get
$\Rightarrow \displaystyle\frac { { 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) $
$\Rightarrow \displaystyle\frac { { w }-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) $..........3
Dividing 2 by 3 we get, 
$\Rightarrow \displaystyle\frac { \left( { w }-{ \alpha  } _{ 1 } \right) \left( { w }-{ \alpha  } _{ 2 } \right) \left( { w }-{ \alpha  } _{ 3 } \right) \left( { w }-{ \alpha  } _{ 4 } \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)  } =\frac { { \left( { w }^{ 2 }-1 \right)  }^{ 2 } }{ { \left( { w }-1 \right)  }^{ 2 } } $
$=\displaystyle\frac { { w }^{ 4 }+1-2{ w }^{ 2 } }{ { w }^{ 2 }+1-2w } =\frac { { w }+1-2{ w }^{ 2 } }{ { w }^{ 2 }+1-2w } =\frac { -{ w }^{ 2 }-2{ w }^{ 2 } }{ -w-2w } =w$

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 from 

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

$z _{1}$ and $z _{2}$ subtend right angle at the origin. 

Then 

$arg \dfrac{z _{1}}{z _{2}} = \dfrac{\pi}{2}$

$\therefore \dfrac{z _{1}}{z _{2}} = \cos \dfrac{\pi}{2} +i\sin\dfrac{\pi}{2} 0+i =i$

$\left (\dfrac{z _{1}}{z _{2}}\right )^n = i^n = 1$    ( nth root of unity )

$\therefore n = 4k$    (as $i^4 =1$)

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  $n^{th} $ roots of unity then $\frac{1}{1-a _{1}} +\frac{1}{1-a _{2}}+...+\frac{1}{1-a _{n-1}}$ equals

  1. $\frac{2^{n}-1}{n}$
  2. $\frac{n-1}{2}$
  3. $\frac{n}{n-1}$
  4. $\frac{n}{n+1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.