Mathematics

Complex Variables and Numbers

157 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 maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

Let $z _1$ and $z _2$ are two complex numbers such that $(1-i)z _1=2z _2$ and $arg(z _1z _2)=\dfrac{\pi}{2}$ then $arg(z _2)$ is equals to:

  1. $\dfrac{3 \pi}{8}$
  2. $\dfrac{\pi}{8}$
  3. $\dfrac{5 \pi}{8}$
  4. $\dfrac{-7 \pi}{8}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$(1-i)z _1=2z _2$

$\cfrac{z _2}{z _1}=\cfrac{1}{2}-\cfrac{i}{2}$
Let $z _1=r _1e^{i\theta _1}\ and\ z _=r _1r^{\theta _2}$
$arg(\cfrac{z _2}{z _1})=\tan^{-1}\cfrac{-1/2}{1/2}=-\pi/4$
$\theta _2-\theta _1=-\pi/4$  and $arg(z _1z _2)=\pi/2$ (given)
$\implies \theta _2-\theta _1=-\pi/4$  and $\theta _2
+\theta _1=\pi/2$
$\implies 2\theta _2=\pi/4,\theta _2=\pi/8$
$\implies arg(z _2)=\pi/8$

Multiple choice maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

The complex number $\dfrac{1 + 2i}{1 - i}$ lies in which quadrant of the complex plane.

  1. First

  2. Second

  3. Third

  4. Fourth
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\dfrac{1+2i}{1-i}$
$\Rightarrow \dfrac{1+2i}{1-i}\times \dfrac{1+i}{1+i}$
$=\dfrac{1+i+2i-2i^2}{1-i^2}=\dfrac{1+3i-2}{2}$
$=\dfrac{-1+3i}{2}$
$\therefore$ It lies in $2^{nd}$ Quadrant.
Multiple choice maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

If $arg(z) < 0$, then $arg(-z)-arg(z)=$

  1. $\pi$
  2. $-\pi$
  3. $\dfrac{\pi}{2}$
  4. $-\dfrac{\pi}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $Z=re^{i\theta _1}$

$-Z=-re^{i\theta _1}$
$\implies -a\cos\theta _1-ib\sin\theta _1$
$\implies -a\cos(\pi+\theta _1)-ib\sin(\pi+\theta _1)$
$\implies re^{i(\pi+\theta _1)}$
$arg(-Z)-arg(Z)$
$\implies \pi+\theta _1-\theta _1\ \implies \pi$

Multiple choice maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

Which of the given alternatives represent a point in Argand plane, equidistant from roots of the equation $(z+1)^4= 16z^4$?

  1. $(0,0)$
  2. $\left(-\dfrac{1}{3},0\right)$
  3. $\left(\dfrac{1}{3},0\right)$
  4. $\left(0,\dfrac{2}{\sqrt5}\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider the given equation $(z+1)^4=16z^4$
$ \Rightarrow (z+1)^4=16z^4$

$ \Rightarrow z+1=(2^4z^4)^{\frac{1}{4}}$

$ \Rightarrow |z+1|=2|z|$

We know that $z=x+iy$

Therefore $|x+iy+1|=2|x+iy|$

$ \Rightarrow \sqrt{(x+1)^2+y^2}=2\sqrt{x^2+y^2}$

$ \Rightarrow (x+1)^2+y^2=4(x^2+y^2)$

$ \Rightarrow x^2+2x+1+y^2=4x^2+4y^2$

$ \Rightarrow 3x^2+3y^2-2x-1=0$

Divide throughout by 3 we get,
$ \Rightarrow x^2+y^2-\dfrac{2}{3}x-\dfrac{1}{3}=0$, which represents a circle.

We know that for the circle equation of the form $x^2+y^2+2gx+2hy+c=0$ the center of the circle is given by $(-g,-h)$

We have $3x^2+3y^2-2x-1=0$ where $g=-\dfrac{1}{3}, h=0$.

Hence the center is $(\dfrac{1}{3},0)$ which is equidistant from the root of the equation.

Multiple choice maths introduction to three dimensional geometry midpoint of line segment mid point formula mid-point of a line segment

If $z = \cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}$, then

  1. $\left| z \right| = 1,\arg z = \dfrac{\pi }{4}$
  2. $\left| z \right| = 1,\arg z = \dfrac{\pi }{6}$
  3. $\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = \dfrac{{5\pi }}{{24}}$
  4. $\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = {\tan ^{ - 1}}\dfrac{1}{{\sqrt 2 }}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

if $z=\cos \left(\dfrac{\pi}{6}\right)+i\sin \left(\dfrac{\pi}{6}\right)$ then.

As $z=|z|e^{i arq (z)}$
$\therefore z=\cos \left(\dfrac{\pi}{6}\right)+i\sin \left(\dfrac{\pi}{6}\right)=e^{i\left(\pi/6\right)}\quad [\because e^{i\theta}=\cos\theta+i\sin \theta]$
$\Rightarrow |z|=1,  arq=\dfrac{\pi}{6}$

Multiple choice maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

The complex no. $\dfrac{1+2i}{1-i}$ lies in which quadrant of the complex plane

  1. first

  2. second

  3. third

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

To find the quadrant, multiply the numerator and denominator by the conjugate (1+i). (1+2i)(1+i) / (1-i)(1+i) = (1+i+2i-2) / 2 = (-1+3i) / 2 = -0.5 + 1.5i. This point lies in the second quadrant.

Multiple choice the nth roots of unity complex numbers maths

Find the number of values of complex numbers $\omega$ satisfying the system of equations ${ z }^{ 3 }=-{ \left( \overline { \omega  }  \right)  }^{ 7 }$ and ${ z }^{ 5 }.{ \omega  }^{ 11 }=1$

  1. $2$
  2. $4$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving the system of equations z^3 = -(conjugate(w)^7) and z^5 * w^11 = 1 involves substituting magnitudes and arguments. The system yields two distinct solutions for the complex number w.

Multiple choice the nth roots of unity complex numbers maths

The value of the expression $1+(2-\omega )+(2-{ \omega  }^{ 2 })+2+(3-\omega )+(3-{ \omega  }^{ 2 })+..........+(n-1)(n-\omega )(n-{ \omega  }^{ 2 })$ where $\omega $ is an imaginary cube root of unity is-

  1. ${ (\frac { n(n+1) }{ 2 } ) }^{ 2 }$
  2. ${ (\frac { n(n+1) }{ 2 } ) }^{ 2 }-n$
  3. ${ (\frac { n(n+1) }{ 2 } ) }^{ 2 }+n$

  4. None of above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The general term is (k-1)(k-w)(k-w^2) = (k-1)(k^2 + k + 1) = k^3 - 1. Summing this from k=2 to n gives the result (n(n+1)/2)^2.

Multiple choice the nth roots of unity complex numbers maths

If 1,${ a } _{ 1 }{ a } _{ 2,........, }{ a } _{ n-1 }$ are the ${ n }^{ th }$ roots of unity, then $\left( 1-{ a } _{ 1 } \right) \left( 1-{ a } _{ 2 } \right) ....\left( 1-{ a } _{ n-1 } \right) $ is equal to

  1. n

  2. 0

  3. 1

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

The roots of unity are solutions to z^n - 1 = 0. The expression (1-a1)(1-a2)...(1-an-1) is the evaluation of (z^n - 1)/(z - 1) at z=1, which equals n.

Multiple choice the nth roots of unity complex numbers maths

Let the four roots of unity be $z _1, z _2, z _3$, and $z _4$, respectively.
Statement 1: $z _1^2+z _2^2+z _3^2+z _4^2=0$
Statement 2: $z _1+z _2+z _3+z _4=0$.

  1. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

  2. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

  3. Statement 1 is true and Statement 2 is false.

  4. Statement 1 is false and Statement 2 is true.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$x^{4}=1$
$x^{2}=\pm1$
$x^{2}=1$ and $x^{2}=-1$
$x=\pm1$ and $x=\pm i$
Hence the four roots are
$1,-1,i,-i$.
Now
$z _{1}=1=-z _{2}$
$z _{3}=i=-z _{4}$
Hence
$z _{1}^{2}+z _{2}^{2}+z _{3}^{2}+z _{4}^{2}$
$=1+1+(i)^{2}+(-i)^{2}$
$=2-2$
$=0$ ...(i)
And also
$z _{1}+z _{2}+z _{3}+z _{4}$
$=1-1+i-i$
$=0$ ...(ii)
Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

Multiple choice the nth roots of unity complex numbers maths

If $\alpha$ is the n$^{th}$ root of unity, then $1+2\alpha+3\alpha^2+.... $ to $n$ terms equal to

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

$S=1+2\alpha+3\alpha^{2}+...n\alpha^{n-1}$
$\alpha S=\:\:\alpha+2\alpha^{2}+3\alpha^{3}+...(n-1)\alpha^{n-1}+n\alpha^{n}$
$S(1-\alpha)=1+\alpha+\alpha^{2}+\alpha^{3}+...\alpha^{n-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{1-\alpha^{n}}{1-\alpha}-n\alpha^{n}$
Now $\alpha^{n}=1$ since it is the $n^{th}$ root of unity.
Therefore,
$S(1-\alpha)=-n$
$S=\dfrac{-n}{1-\alpha}$

Multiple choice the nth roots of unity complex numbers maths

If n is an odd positive integer and $ I,\alpha _{1},\alpha _{2},....\alpha _{n-1}$ are the $n,n^{th}$ roots of unity, then $\left ( 3+\alpha ^{1} \right )\left ( 3+\alpha ^{2} \right )....\left ( 3+\alpha ^{n-1} \right )$ equals

  1. $\displaystyle \frac{3^{n}+1}{4}$
  2. $\displaystyle \frac{3^{n}-1}{2}$
  3. $\displaystyle \frac{3^{n}-1}{4}$

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

$x^{n}-1=(x-1)(x-\alpha _{1})(x-\alpha _{2})(x-\alpha _{3})...(x-\alpha _{n-1})$
$\dfrac{x^{n}-1}{x-1}=(x-\alpha _{1})(x-\alpha _{2})(x-\alpha _{3})...(x-\alpha _{n-1})$
$(x-\alpha _{1})(x-\alpha _{2})(x-\alpha _{3})...(x-\alpha _{n-1})=1+x+x^{2}+...x^{n-1}$
Substituting  $x=-3$.
$(3+\alpha _{1})(3+\alpha _{2})(3+\alpha _{3})...(3+\alpha _{n-1})=1-3+3^{2}+...(-1)^{n-1}3^{n-1}$
Therefore $(3+\alpha _{1})(3+\alpha _{2})(3+\alpha _{3})...(3+\alpha _{n-1})$
$=\dfrac{1-(-3)^{n}}{1-(-3)}$
Now, $n$ is odd, therefore
$=\dfrac{3^{n}+1}{4}$

Multiple choice the nth roots of unity complex numbers maths

If $\alpha$ is the $n^{th}$ root of unity, then $1+2\alpha+3\alpha^{2}+...$ to $n$ terms is equal to

  1. $\displaystyle -\frac { n }{ { \left( 1-\alpha \right) }^{ 2 } } $
  2. $\displaystyle -\frac { n }{ { \left( 1-\alpha \right) }} $
  3. $\displaystyle -\frac { 2n }{ { \left( 1-\alpha \right) } } $
  4. $\displaystyle -\frac { 2n }{ { \left( 1-\alpha \right) }^{ 2 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$S=1+2\alpha+3\alpha^{2}+....n\alpha^{n-1}$
$S\alpha=\alpha+2\alpha^{2}+3\alpha^{3}...(n-1)\alpha^{n-1}+n\alpha^{n}$
$S(1-\alpha)=1+\alpha+\alpha^{2}+...\alpha^{n-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{\alpha^{n}-1}{\alpha-1}-n\alpha^{n}$
Since $\alpha$ is the nth root of unity, hence $\alpha^{n}=1$
Thus
$S(1-\alpha)=\dfrac{\alpha^{n}-1}{\alpha-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{1-1}{\alpha-1}-n$
$S(1-\alpha)=-n$
$S=-\dfrac{n}{1-\alpha}$

Multiple choice the nth roots of unity complex numbers maths

If the fourth roots of unity are $\displaystyle\ z _{1},z _{2},z _{3},z _{4}$ then $\displaystyle\ z _{1}^{2}+z _{2}^{2}+z _{3}^{2}+z _{4}^{2}$ is equal to

  1. $\displaystyle\ 1$
  2. $\displaystyle\ 0$
  3. $\displaystyle\ i$

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

If $z$ be the fourth roots of unity then, $z^4=1$
$\Rightarrow (z^4-1)=0\Rightarrow (z^2-1)(z^2+1)=0$
$\Rightarrow z=\pm 1, \pm i,$ where $i^2=-1$
$\therefore z _1^2+z _2^2+z _3^2+z _4^2=1+1+i^2+i^2=2-2=0$