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 congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

$\begin{array} { l } { \text { If } z _ { 1 } \text { and } z _ { 2 } \text { are complex numbers, then } \left| z _ { 1 } + z _ { 2 } \right| ^ { 2 } = \left| z _ { 1 } \right| ^ { 2 } + \left| z _ { 2 } \right| ^ { 2 } \text { if and only if } z _ { 1 } \overline { z } _ { 2 } \text { is } } \ { \text { purely imaginary. } } \end{array}$

  1. True

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

|z1 + z2|^2 = (z1 + z2)(conjugate(z1) + conjugate(z2)) = |z1|^2 + |z2|^2 + z1*conjugate(z2) + conjugate(z1)*z2. For this to equal |z1|^2 + |z2|^2, the cross terms must sum to zero: z1*conjugate(z2) + conjugate(z1*conjugate(z2)) = 0. This means 2*Re(z1*conjugate(z2)) = 0, so z1*conjugate(z2) must be purely imaginary.

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $P$ and $Q$ are represented by complex numbers $z _{1}$ and $z _{2}$ such that $\left| \dfrac { 1 }{ { z } _{ 1 } } +\dfrac { 1 }{ { z } _{ 2 } }  \right| =\left| \dfrac { 1 }{ { z } _{ 1 } } -\dfrac { 1 }{ { z } _{ 2 } }  \right| $ then the circumference of $\triangleOPQ(O is origin)$ is

  1. $\dfrac{{ z } _{ 1 } -{ z } _{ 2 } }{2}$
  2. $\dfrac{{ z } _{ 1 } +{ z } _{ 2 } }{2}$
  3. $\dfrac{{ z } _{ 1 } +{ z } _{ 2 } }{3}$
  4. ${ z } _{ 1 } +{ z } _{ 2 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $\left|z\right| <\sqrt{2} -1$, then $\left|z^2 + 2 z  cos  \alpha \right|$ is

  1. less than 1
  2. $\sqrt{2} + 1$
  3. $\sqrt{2} -1$

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

$\left| z \right| <\sqrt { 2 } -1\ \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| \le \left| { z }^{ 2 } \right|+ \left| 2z\cos { \alpha  }  \right| \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| { z } _{ 1 }+{ z } _{ 2 } \right| \le \left| { z } _{ 1 } \right| +\left| { z } _{ 2 } \right|  \right} \ \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| \le |z|(|z|+2) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| \cos { \alpha  }  \right| \le 1\quad  \right} \ \Rightarrow \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| <(\sqrt{2}-1){ \left( \sqrt { 2 } +1 \right)  }<1\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| z \right| <\sqrt { 2 } -1\quad  \right} \ \therefore \quad \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| <1\ $
Hence, option 'A' is correct.

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If z be a complex number for which $|2z  cos  \theta + z^2| = 1$, then the minimum value of |z|
 is ......................

  1. $\sqrt{3} -1$
  2. $\sqrt{3} +1$
  3. $\sqrt{2} -1$
  4. $\sqrt{2} +1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$|z^{2}+2zcos\theta|$
$=|z(z+2cos\theta)|$
$=|z|.|z+2cos\theta|$
$=1$
Now 
$|z|=1$ and 
$|z+2cos\theta|=1$
Now 
$|z+2cos\theta|\leq |z|+|2cos\theta|$
Considering 
$|z+2\cos\theta|=|z|+|2cos\theta|=1$
Hence
$|z|=|2cos\theta|\pm1$
Considering $z=|2cos\theta|-1$ we get the minimum value at multiples of $\theta=45^{0}$
Hence
$z=\sqrt{2}-1$.

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles


$|\mathrm{z} _{1}-\mathrm{z} _{2}|=$

  1. $\geq||z _{1}|-|z _{2}||$
  2. $\leq|z _{1}|-|z _{2}|$
  3. $=|\mathrm{z} _{1}|-|\mathrm{z} _{2}|$
  4. $\geq|\mathrm{z} _{1}|-|\mathrm{z} _{2}|$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $ argz _1=\theta _1  \quad argz _2=\theta _2$
we know that
$|z _{1}-z _{2}|^{2}=|z _{1}|^{2}+|z _{2}|^{2}-2|z _{1}||z _{2}|cos(\theta _{1}-\theta _{2})$


now, $+1\geq cos(\theta _{1}-\theta _{2})\geq -1$


$-2|z _{1}||z _{2}|cos(\theta _{1}-\theta _{2})\geq -2|z _{1}||z _{2}|$


$\therefore |z _{1}|^{2}+|z _{2}|^{2}-2|z _{1}||z _{2}|cos(\theta _{1}-\theta _{2})\geq |z _{1}|^{2}+|z _{2}|^{2}-2|z _{1}||z _{2}|$


$\therefore |z _{1}-z _{2}|^{2}\geq (|z _{1}|-|z _{2}|)^{2}\Rightarrow |z _{1}-z _{2}|\geq ||z _{1}|-|z _{2}||$


Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles


 lf $|\mathrm{z} _{1}|=2,\ |\mathrm{z} _{2}|=3$, then $|\mathrm{z} _{1}+\mathrm{z} _{2}+5+12\mathrm{i}|$ is less than or equal to

  1. $8$
  2. $18$
  3. $10$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that $|z _{1}+z _{2}+ z _3 |\leq |z _{1}|+|z _{2}| + | z _3| $
$|z _{1}|+|z _{2}|=5$
$z _3 = 5+12 i $

$|z _3 | = 13 $
$\therefore 18\geq |z _{1}+z _{2}+5+12i|$
Hence, option B is correct

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles


 Let $z _{1}=24+7i$ and $z _{2}$ be complex number whose magnitude is unity, then

  1. Maximum value of $|z _{1}+z _{2}|$ is 26
  2. Maximum value of $|z _{1}+z _{2}|$ is 31
  3. Minimum value of $|z _{1}+z _{2}|$ is 24
  4. Minimum value of $|z _{1}+z _{2}|$ is 19
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$ |z _1 | =  25 $

$| z _2 | = 1$

We have, 
$\left| |z _1| - |z _2| \right| \leq |z _1+z _2 | \leq \left | |z _1| + |z _2| \right |$
$\Rightarrow 24 \leq |z _1+z _2| \leq 26$
Hence, options A and C are correct 

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

The complex number $z$ satisfies the condition $\left|\displaystyle {z}-\frac{25}{z}\right|=24$. Then the maximum distance from the origin to the point '$z$' in the argand plane is

  1. 20

  2. 25

  3. 30

  4. 35
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$|z| = |z - \frac{25}{z} + \frac{25}{z}| \leq 24 + \frac{25}{|z|}$
$\therefore$ $|z|^2 - 24|z| - 25 \leq 0.$
$\therefore$ $(|z| - 25)(|z| + 1) \leq 0., \Rightarrow |z| \leq 25.$
Hence maximum distance of z from origin is 25.

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

$z _0$ is a root of the equation $z^n cos \theta _o+z^{n-1} cos\theta _1+....+z cos\theta _{n-1}+cos\theta _n=2$, where $\theta, \epsilon R$, then

  1. $|z _0| > 1$
  2. $|z _0| > \dfrac {1}{2}$
  3. $|z _0| > \dfrac {1}{4}$
  4. $|z _0| > \dfrac {3}{2}$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

$z^n cos\theta _0+z^{n-1} cos\theta _1+.....+z cos\theta _{n-1}+cos\theta _n=2$

or $2=|z _0^n cos\theta _0+z _0^{n-1} cos\theta _1+....+z _0 cos\theta _{n-1}+cos\theta _n|$

or $2\leq |z _0|^n |cos\theta _0|+|z|^{n-1}|cos\theta _1|+....+|z _0||cos\theta _{n-1}|+|cos\theta _n|$

or $2\leq |z _0|^n+|z _0|^{n-1}+|z _0|^{n-2}+.....+|z _0|+1$

which is clearly satisfied for $|z _0| \geq 1$. If $|z _0| < 1$, then

$2 < 1+|z _0|+|z _0|^2+.....+|z|^n+....\infty$

$\Rightarrow 2 < \dfrac {1}{1-|z _0|}$

$\Rightarrow |z _0| > \dfrac {1}{2}$

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $z _{1},\ z _{2}--,\ z _{n}$ are complex numbers such that $|z _{i}|<\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{d}\lambda _{i}>0$ for $i=1,2,---n$ and $\lambda _{1}+\lambda _{2}+--+\lambda _{n}=1$ then $|\lambda _{1}z _{1}+\lambda _{2}z _{2}+--+\lambda _{n}\mathrm{z} _{1}|?$

  1. $=1$
  2. $<1$
  3. $>1$
  4. $=n$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given:

$\lambda _i>0$  and  $\lambda _1+\lambda _2+...+\lambda _n=1$
$\therefore 0<\lambda _i<1$
Also,  $|z _i|<1$
$\therefore \lambda _1|z _1|+\lambda _2|z _2|+.....+\lambda _n|z _n|<1$                ......( 1 )

$\therefore |\lambda _1z _1+\lambda _2z _2+.....+\lambda _nz _n|$
$\leq |\lambda _1z _1|+|\lambda _2z _2|+.....+|\lambda _nz _n|$
$\leq \lambda _1|z _1|+\lambda _2|z _2|+.....+\lambda _n|z _n|$
$<1$

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

If $\omega$ is a cube root of unity and $x+ y + z = a, x + \omega y + \omega^2 z = b, x + \omega^2 y + \omega z = c$, then $x = $ ............ 

  1. $ \dfrac{a+b+ c}{3}$
  2. $ \dfrac{a + \omega^2 b + \omega c}{3}$
  3. $\dfrac{a + \omega b + \omega^2 c}{3}$
  4. $\dfrac{a+b+ c \omega}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $\omega$ is a cube root of unity and $x+ y + z = a, x + \omega y + \omega^2 z = b, x + \omega^2 y + \omega z = c$.

Adding all the given equations gives

$3x+y(1+\omega+\omega^2)+z(1+\omega+\omega^2)=a+b+c$

$\Rightarrow 3x=a+b+c$     $\because 1+\omega+\omega^2=0$

$\therefore x=\displaystyle\frac{a+b+c}{3}$

Hence, option A.

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

If $\displaystyle \omega$ is cube root of unity and $\displaystyle x + y + z = a$, $\displaystyle x + \omega y + \omega^{2} z = b$, $\displaystyle x + \omega^{2} y + \omega z = b$ then which of the following is not correct?

  1. $\displaystyle x = \frac{a + b + c}{3}$
  2. $\displaystyle y = \frac{a + b \omega^{2} + \omega c}{3}$
  3. $\displaystyle x = \frac{a + b \omega + \omega^{2} c}{3}$

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

Given $x+y+z=a $...(i)

$\displaystyle x+\omega y+\omega ^{2}z= b$ ...(ii)

$\displaystyle x+\omega ^{2}y+\omega z= c$ ...(iii)

By adding (i), (ii) and (iii), we get

$\displaystyle x= \frac{a+b+c}{3}$

Hence option A is correct.

Again $\displaystyle \left ( i \right )+\left ( ii \right )\times \omega ^{2}+\left ( iii \right )\times \omega $, we get

$\displaystyle 3y= a\omega ^{3}+b\omega ^{2}+c\omega$

$\displaystyle y= \frac{a+b\omega ^{2}+c\omega }{3}$

Hence, option B is correct.

Similarly, $\displaystyle \left ( i \right )+\left ( ii \right )\times \omega +\left ( iii \right )\times \omega ^{2}$ we get

$\displaystyle z= \frac{a+b\omega +c\omega ^{2}}{3}$

Hence, option C is correct

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.