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 the nth roots of unity complex numbers maths

If $\displaystyle \omega $ is fifth root of unity, then $\displaystyle \log _2 \mid 1+\omega +\omega ^{2}+\omega ^{3}-\omega ^{-1}\mid $ is equal to

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

$1+w+w^{2}+...w^{3}-\dfrac{1}{w}$
$=\dfrac{w+w^{2}+w^{3}+...w^{4}-1}{w}$


$=\dfrac{1}{w}[\dfrac{w(1-w^{4})}{1-w}-1]$

$=\dfrac{1}{w}[\dfrac{w-w^{5}}{1-w}-1]$

$=\dfrac{1}{w}[\dfrac{w-1}{1-w}-1]$

$=\dfrac{1}{w}[-2]$

Now
$|\dfrac{-2}{w}|$

$=\dfrac{|-2|}{|w|}$

$=\dfrac{2}{1}$

$=2$

$=log _{2}(2)$

$=1$
Hence, option 'A' is correct.

Multiple choice the nth roots of unity complex numbers maths

If $w$ be complex $n^{th}$ root of unity and $r$ is an integer not divisible by $n$, then the sum of the $r$th powers of the nth roots of unity is

  1. $0$
  2. $1$
  3. $w$
  4. $n$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We have, for $n^{th}$ root of unity denoted by $\alpha i$
$1 + \alpha _{1}+\alpha _{2}+\alpha _{3}---------\alpha _{n-1}=\dfrac{1(1-(\alpha)^{2})}{1-\alpha}$                    [where $\alpha _{i}= e^{i \dfrac{2 \pi k}{n}}$ Thus, $\alpha _{1}=e^{i\dfrac{2 \pi}{n}}$ $\alpha _{1}=e^{i\dfrac{4 \pi}{n}}$---------------form a G.P. with ratio $\alpha = e^{i 2 \pi /2}$]
for power r, expression changes to,
$1+\alpha _{1}^{r}+\alpha _{2}^{r}+\alpha _{3}^{r}---------\alpha _{n-1}^{r}=\dfrac{1-(\alpha^{r})^{2}}{}$ [Given $r \neq kn$ $\Rightarrow \alpha^{r} \neq 1$]
$=\dfrac{1-(\alpha^{n})^{r}}{1-\alpha^{r}}$
But $\alpha^{2}=1$
Thus $\dfrac{1-1}{1-\alpha^{r}}=0$
Multiple choice the nth roots of unity complex numbers maths

If $1,\alpha,\alpha^ 2......\alpha^{n}$ are the $n^{th}$ roots of unity then $^nC _1+ ^nC _2.\alpha + ^nC _3.\alpha^2 ........+^nC _n.\alpha^{n}$ is equal to

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

$^nC _1 + ^nC _2.\alpha + ^nC _3.\alpha^2 + ......... + ^nC _n.\alpha^{n1}$

$\Rightarrow \displaystyle \frac{1}{\alpha} [^n C _1. \alpha + ^n C _2 . \alpha^2 + ....... + ^n C _n. \alpha^n]$

$\Rightarrow \displaystyle \frac{1}{\alpha} [^nC _0.1+^n C _1. \alpha + ^n C _2 . \alpha^2 + ....... + ^n C _n. \alpha^n -1] \quad \dots (^nC _0=1)$

$\Rightarrow \displaystyle \frac{1}{\alpha} [(1 + \alpha)^n - 1]$

$^nC _1 + ^nC _2.\alpha + ^nC _3.\alpha^2 + ......... + ^nC _n.\alpha^{n1}=\dfrac{1}{\alpha} [(1 + \alpha)^n - 1]$

Multiple choice the nth roots of unity complex numbers maths

In the multiplicative group of $n^{th}$ roots of unity the inverse of ${ \omega  }^{ k },\left( k<n \right) $ is

  1. ${ \omega }^{ { 1 }/{ k } }$
  2. ${ \omega }^{ -1 }$
  3. ${ \omega }^{ n-k }$
  4. ${ \omega }^{ { n }/{ k } }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since $\omega $ is the $n^{th}$ root of unity, hence $\omega^{n}=1$.
Now inverse of $\omega ^{k}$ where $k<n$ is
$=\dfrac{1}{\omega ^{k}}$
$=\dfrac{\omega^{n}}{\omega^{k}}$
$=\omega^{n-k}$.
Hence inverse of $\omega^{k}$
$={\omega^{-k}}$
$=\omega^{n-k}$. 

Multiple choice the nth roots of unity complex numbers maths

If $\omega, \omega^2, \omega^3, ........ \omega^{n - 1}$ are nth roots of unity then $(1- \omega) (1- \omega^2) ....... (1 - \omega^{n  -1})$ equals:

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

Since $1, \omega, \omega^2, \omega^3, ......... \omega^{n - 1}$ are nth roots of unity, therefore, we have the identity
$(x - 1)(x - \omega)(x - \omega^2) ...... (x - \omega^{n - 1})$
$= x^n - 1$
or $(x - \omega)(x - \omega^2) ...... (x - \omega^{n - 1}) = \displaystyle \frac{x^n - 1}{x - 1}$
$= x^{n - 1} + x^{n - 2} + ...... + x + 1$
Putting x = 1 on both sides, we get $(1 - \omega) (1 - \omega^2) (1 - \omega^{n - 1}) = n$

Multiple choice the nth roots of unity complex numbers maths

$1 , z _1, z _2, z _3, ..., z _{n-1}$ are the $n$th roots of unity, then the value of $\displaystyle\frac{1}{(3-z _1)} +\displaystyle\frac{1}{(3-z _2)} + ... +\displaystyle\frac{1}{(3-z _{n-1})}$ is equal to  

  1. $\displaystyle \frac { n{ 3 }^{ n-1 } }{ { 3 }^{ n }-1 } -\displaystyle \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. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If $\alpha _1, \alpha _2, ... , \alpha _m$ are the roots of polynomial equation
$\quad f(x) = a _0x^m + a _1x^{m-1} + ... + a _{m-1}x + a _m = 0$
Then,
$\quad f(x) = a _0(x-\alpha _1) ... (x-\alpha _m).$


and $\quad \displaystyle\frac{f'(x)}{f(x)} = \displaystyle\frac{1}{x-\alpha _1}+...+\displaystyle\frac{1}{x-\alpha _m}$

The equation in question is ${ x }^{ n }-1=0$

 $f(x)={ x }^{ n }-1=(x-1)(x-{ z } _{ 1 })...(x-{ z } _{ n-1 })$


Thus, $\dfrac { f'(x) }{ f(x) } =\dfrac { n{ x }^{ n-1 } }{ { x }^{ n }-1 } =\dfrac { 1 }{ x-1 } +\dfrac { 1 }{ x-{ z } _{ 1 }  } +...+\dfrac { 1 }{ x-{ z } _{ n-1 } } $

Substituting $x=3$:

$\dfrac { 1 }{ 3-1 } +\dfrac { 1 }{ 3-{ z } _{ 1 }  } +...+\dfrac { 1 }{ 3-{ z } _{ n-1 } } =\dfrac { n{ 3 }^{ n-1 } }{ { 3 }^{ n }-1 } $

Hence, $\dfrac { 1 }{ 3-{ z } _{ 1 }  } +...+\dfrac { 1 }{ 3-{ z } _{ n-1 } } =\dfrac { n{ 3 }^{ n-1 } }{ { 3 }^{ n }-1 } - \dfrac { 1 }{ 2 }$

Hence,  (A) is correct.

Multiple choice the nth roots of unity complex numbers maths

If $1,\omega,\omega^{2},...,\omega^{n-1}$ are $n^{th}$ roots of unity, then the value of $(5-\omega)(5-\omega^{2})...(5-\omega^{n-1})=$

  1. $\displaystyle \frac{5^{n}-2}{4}$
  2. $\displaystyle \frac{5^{n}+2}{4}$
  3. $\displaystyle \frac{5^{n}+1}{4}$
  4. $\displaystyle \frac{5^{n}-1}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

${ x }^{ n }-1=0$ has n roots (of unity).
Thus, ${ x }^{ n }-1=(x-1)(x-\omega )(x-{ \omega  }^{ 2 })...(x-{ \omega  }^{ n-1 })$.
Substitute $x=5$: 
${ 5 }^{ n }-1=(5-1)(5-\omega )(5-{ \omega  }^{ 2 })...(5-{ \omega  }^{ n-1 })$
=> $(5-\omega )(5-{ \omega  }^{ 2 })...(5-{ \omega  }^{ n-1 })=\dfrac { { 5 }^{ n }-1 }{ 4 } $
Hence, option D is correct.

Multiple choice the nth roots of unity complex numbers maths
Let $z$ be any complex number. To factorise the expression of the form ($z^n- 1$), we consider the equation $z^n = 1$. This equation is solved using De moiver's theorem. Let $1, \alpha _1, \alpha _2 .......\alpha _{n-1}$ be the roots of this equation, then $z^n-1=(z-1)(z-\alpha _1)(z-\alpha _2).......(z-\alpha _{n-1})$. This method can be generalised to factorize any expression of the form $z^n-k^n$.
For example, $z^7+1=\displaystyle \Pi _{m=0}^6\left (z-CiS\left (\dfrac {2m\pi}{7}+\dfrac {\pi}{7}\right )\right )$
This can be further simplified as
$z^7+1(z+1)(z^2-2zcos\dfrac {\pi}{7}+1)(z^2-2z cos\dfrac {3\pi}{7}+1)(z^2-2z cos \dfrac {5\pi}{7}+1)$ .......$(i)$
These factorisations are useful in proving different trigonometric identities e.g. in equation $(i)$ if we put $z = i,$ then equation $(i)$ becomes
$(1-i)=(i+1)(-2i cos \dfrac {\pi}{7})(-2i cos \dfrac {3\pi}{7})(-2i cos \dfrac {5\pi}{7})$
i.e, $cos \dfrac {\pi}{7}cos \dfrac {3\pi}{7}cos \dfrac {5\pi}{7}=-\dfrac {1}{8}$


By using the factorisation for $z^5+1$, the value of $4 sin\dfrac {\pi}{10} cos \dfrac {\pi}{5}$ comes out to be :

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

$z^{5}+1=0$ Implies 

$z=-1,(cos\dfrac{\pi}{5}+isin\dfrac{\pi}{5}),(cos\dfrac{-\pi}{5}+isin\dfrac{-\pi}{5}),(cos\dfrac{3\pi}{5}+isin\dfrac{3\pi}{5}),(cos\dfrac{-3\pi}{5}+isin\dfrac{-3\pi}{5})$
Now 
$4sin\dfrac{\pi}{10}.cos\dfrac{\pi}{5}$
$=4sin\dfrac{\pi}{10}sin\dfrac{3\pi}{10}$

$=2[cos\dfrac{\pi}{5}-cos\dfrac{2\pi}{5}]$

$=2[cos\dfrac{\pi}{5}+cos\dfrac{3\pi}{5}]$

$=cos\dfrac{\pi}{5}+cos\dfrac{\pi}{5}+cos\dfrac{3\pi}{5}+cos\dfrac{3\pi}{5}$

$=cos\dfrac{\pi}{5}+cos\dfrac{3\pi}{5}+cos\dfrac{-3\pi}{5}+cos\dfrac{-\pi}{5}$

$=\sum Re z _{i}$

$=|z|$
$=1$

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

If $A =\begin{bmatrix} 1&9  & -7\ i & \omega^n & 8\ 1 & 6 &\omega^{2n} \end{bmatrix}$ where $i= \sqrt{-1} $ and $\omega$ is complex cube root of unity, then tr(A) will be 

  1. $1, \,if \,n = 3k,\, k \in\, N$
  2. $3, \,if \,n = 3k,\, k \in\, N$
  3. $0,\, if \,n\neq \,3k,\, k \epsilon \in N$
  4. $-1,\, if \,n\neq \,3k, \,k \epsilon \in N$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$tr(A)=1+\omega^n+\omega^{2n}$
if $n=3k$ i.e Mutiple of $3$.
$\Rightarrow tr(A)=1+\omega^{3k}+\omega^{6k}=1+1+1=3$
if $n\neq 3k$ i.e not a multiple of $3$.
then $n=3k+1$ or $n=3k+2$
$\Rightarrow tr(A)=1+\omega^{3k+1}+\omega^{2(3k+1)}=1+\omega+\omega^2=0$
Hence, options B and C.

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

If $z + \frac{1}{z} = 2\cos {6^0}$, then ${z^{1000}} + \frac{1}{{{z^{1000}}}} + 1$ is equal to 

  1. 0

  2. 1

  3. -1

  4. 2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$z+\cfrac{1}{z}=2\cos {6}^{o}$

$z=\cos ]theta+i\sin ]theta$
$z+\cfrac{1}{z}=\cos \theta +i\sin\theta +\cos \theta -i\sin\theta $
$2\cos{6}^{o}=2\cos \theta $
$\theta ={6}^{o}$
${z}^{1000}+\cfrac{1}{{z}^{1000}}=\cos(1000\theta )+i\sin(1000\theta )+\cos (1000\theta )-i\sin (1000\theta )+1$
$=2\cos(1000\theta )+1$
$=2\cos({6000}^{o} )+1$
$=2\cos(5760+240)+1$
$2\cos({240}^{o})+1=2\cos({120}^{o})+1$
$=-2\cos 60+1-1+1=0$
$\therefore$ ${z}^{1000}+\cfrac{1}{{z}^{1000}}+1=0$

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

For positive integers $n _1, n _2, $ the value of the expression $(1 + i)^{n _1} + (1 + i^3)^{n _1} + (1 + i^5)^{n _2} + (1 + i^7)^{n _2}$, where $i = \sqrt{-1}$ is a 

  1. real

  2. complex number
  3. $0$
  4. $i$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Simplify the terms: (1+i)^n1 + (1-i)^n1 + (1+i)^n2 + (1-i)^n2. Since (1+i) and (1-i) are conjugates, their sum to any power n is real. Thus, the entire expression is real.

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

If $\dfrac { z+2i }{ z-2i } $ is purely imaginary then $\left| z \right| $ is 

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

$=\dfrac{z+2i}{z-2i}\times \dfrac{z+2i}{z+2i}$

$=\dfrac{z^2+4z-4}{z^2-4}$

Let $z=x+iy$

$=\dfrac{(x+iy)^2+4(x+iy)-4}{(x+iy)^2-4}$

$=\dfrac{x^2-y^2+2ixy-4+4x+4iy}{(x+iy)^2-4}$

$=\dfrac{(x^2-y^2+4x-4)+i(2xy+4y)}{(x+iy)^2-4}$

z is purely imaginary. So, Re(z)=0

$\dfrac{x^2-y^2+4x-4}{(x+iy)^2-4}=0$

$\Rightarrow x^2-y^2+4x-4=0$

$\therefore x^2-y^2+4x=4$

The above equation represents the hyperbola on x-axis, with $(1,0)$

$\therefore z=1$
Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

The complex number z having least positive argument which satisfies the condition $|z - 25i| \le 15$   is:

  1. $25i$
  2. $12+5i$
  3. $16+12i$
  4. $12+16i$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Solution:

$|z-25 i| \leq 15$

Let $z= r(cos \theta + i \, sin \theta)$

$\theta$ must be minimum

$| r \, cos \theta +i ( r\, sin \theta-25)|\leq 15$

$|\sqrt{r^2cos^2\theta+r^2 sin^2 \theta+ 625- 50 r \, sin \theta} \,|\leq 15$

square both side

$r^2 (cos^2 \theta+ sin^2 \theta)+625 - 50 r \, sin \theta \leq 225$

$r^250 r \, sin \theta \leq - 400$

$f(r)=\dfrac {400+r^2}{50 \, r}\leq sin \theta $

Find maximum value of $f(r)=\dfrac {400+r^2}{50\, r}$

$f'(r)=\dfrac {100 r^2- 50(400+r^2)}{2500 r^2}=0$

$50 r^2- 50 \times 400=0$

$r= 20$

$f(r=20)=\dfrac {800}{1000}\leq sin \theta $

$\dfrac {4}{5}\leq  sin \theta $

Least value of $sin \theta $ is $4/5$

$ tan \, \theta = 4/3 \,\,\,\,\,\,\,\,\,\, cos \theta  = 3/5$

$z= 20(3/5+4/5 \,i)$

$z= 12+16\, i$

D is correct.
Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

Suppose z and $\omega$ are two complex numbers such that $|z| \leq 1, |\omega| \leq 1$, and $|z+i\omega|=|z-i\omega|=2$.

Which of the following is true about $|z|$ and $|\omega|$?

  1. $|z|=|\omega|=\frac {1}{2}$
  2. $|z|=\frac {1}{2}, |\omega|=\frac {3}{4}$
  3. $|z|=|\omega|=\frac {3}{4}$
  4. $|z|=|\omega|=1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have,


$|z+i\omega|\le |z|+|i\omega|\le |z|+|i||\omega|\le 2$

$|z-i\omega|\le |z|+|-i\omega|\le |z|+|i||-\omega|=|z|+|i||\omega|\le 2$

Also,

$|z+i\omega|=|z-i\omega|$

and

$ |z+i\omega|=2$ 

Hence, 

$|z|  = |\omega| = 1 $