Tag: complex numbers

Questions Related to complex numbers

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

the nth roots of unity complex numbers maths

  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

Show answer
Correct Option: A
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
$=\dfrac{3^{n}+1}{4}$

$(1-\omega +\omega^{2})(1-\omega^{2}+\omega^{4})(1-\omega^{4}+\omega^{8})......$to 2n factors =

the nth roots of unity complex numbers maths

  1. 2

  2. $2^{2n}$

  3. 2n

  4. none of these

Show answer
Correct Option: B
Explanation:
Given,

$(1−ω+ω^2)(1−ω^2+ω^4)(1−ω^4+ω^8)$...... to $2n$

we have,

$1+w+w^2=0$

and $w^3=1$

$\Rightarrow (-w-w)(-w^2-w^2)(-w-w)....................2n$

$=(-2w)(-2w^2)(-2w).............2n$

here, we can see, 2 consecutive terms are same and are even, so we get,

$=(4w^3)(4w^3)(4w^3).........2n$

$=4 \times 4 \times .........2n$

$=2^2.........2n$

for $2n$ terms, we get,

$=2^{2n}$

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

the nth roots of unity complex numbers maths

  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 } } $

Show answer
Correct Option: B
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
$S(1-\alpha)=\dfrac{\alpha^{n}-1}{\alpha-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{1-1}{\alpha-1}-n$
$S(1-\alpha)=-n$
$S=-\dfrac{n}{1-\alpha}$

if $\displaystyle\ z _{\gamma }=\cos \frac{2\gamma \pi}{5}+i\sin \frac{2\gamma \pi}{5}=0$, $\displaystyle\ \gamma = 0,1,2,3,4.....$ then $\displaystyle\ z _{1}z _{2}z _{3}z _{4}z _{5}$ is equal to

the nth roots of unity complex numbers maths

  1. $\displaystyle\ -1$

  2. $\displaystyle\ 0$

  3. $\displaystyle\ 1$

  4. $none\ of\ these$

Show answer
Correct Option: C
Explanation:

Given, $z _\gamma = \cos{\frac{2 \pi \gamma}{5}} + i \sin{\frac{2 \pi \gamma}{5}}$ where $\gamma = 1, 2, 3, 4, 5.$
Thus, $z _\gamma = e^(i \frac{2 \pi \gamma}{5})$
Thus, $z _1z _2z _3z _4z _5 = e^(\frac{2 \pi}{5} + \frac{4 \pi}{5} + \frac{6 \pi}{5} + \frac{8 \pi}{5} + \frac{10 \pi}{5}) $
$= e^{6 \pi}$
$= \cos(6 \pi) + i \sin(6 \pi)$
$= 1$

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

the nth roots of unity complex numbers maths

  1. $\displaystyle\ 1$

  2. $\displaystyle\ 0$

  3. $\displaystyle\ i$

  4. None of these

Show answer
Correct Option: B
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 $i^2=-1$
$\therefore z _1^2+z _2^2+z _3^2+z _4^2=1+1+i^2+i^2=2-2=0$

If $a = cos \dfrac{2\pi}{7}+i  sin\dfrac{2\pi}{7}$, then find the quadratic equation whose roots are $a = a + a^2 + a^4$ and $\beta = a^3 + a^5 + a^6$.

the nth roots of unity complex numbers maths

  1. $x^2 + x - 1=0$

  2. $x^2 + x - 2=0$

  3. $x^2 + x + 1=0$

  4. $x^2 + x + 2=0$

Show answer
Correct Option: D
Explanation:

$a = cos (2\pi/7)+i  sin(2\pi/7)$
$\Longrightarrow a^7 = [cos(2\pi/7)+i  sin(2\pi/7)]^7$
$= cos 2\pi + i sin 2\pi = 1$          (1)
$S = \alpha + \beta = (a + a^2 + a^4) + (a^3 + a^5 + a^6)$
$= a +a^2 +a^3 +a^4 + a^5 +a^6 = \frac{a(1-a^6)}{1-a}$
$= \frac{a-a^7}{1-a} = \frac{a-1}{1-a} = -1$                           (2)
$P= \alpha \beta = (a+a^2+a^4)(a^3+a^5+a^6)$
$= a^4+a^6 +a^7+a^5+a^7+a^8+a^7+a^9+a^{10}$
$= a^4+a^6+1+a^5+1+a+1+a^2+a^3$         [From Eq. (1)] 
$= 3+(a+ a^2+ a^3+ a^4+ a^5 + a^6)$ 
$= 3+S = 3-1=2$              [From Eq. (2)]
Therefore, the required equation is 
$x^2 -Sx + P = 0$
$\Longrightarrow x^2 + x + 2=0$


Ans: D

If $\displaystyle z=\cos \frac{8\pi }{11}+i\sin\frac{8\pi }{11},$ then Real $\displaystyle \left ( z+z^{2}+z^{3}+z^{4}+z^{5} \right )$ is

the nth roots of unity complex numbers maths

  1. $\displaystyle -\frac{1}{2}$

  2. 0

  3. $\displaystyle \frac{1}{2}$

  4. none

Show answer
Correct Option: A
Explanation:

Real (z) $\displaystyle =\frac{z+\bar{z}}{2}=\frac{1}{2}\left ( z+\frac{1}{z} \right )$


$\displaystyle \because z\bar{z}=\cos ^{2}\frac{8\pi }{11}+\sin ^{2}\frac{8\pi }{11}=1\ \ \therefore \bar{z}=\frac{1}{z}$

$\displaystyle \because$ E=Real part of $\displaystyle \left ( z+z^{2}+z^{3}+z^{4}+z^{5}+\frac{1}{z}+\frac{1}{z^{2}}+\frac{1}{z^{3}}+\frac{1}{z^{4}}+\frac{1}{z^{5}} \right )$

Now $\displaystyle z^{11}= \cos 8\pi +i\sin 8\pi = 1$

$\displaystyle \therefore \frac{1}{z^{4}}= \frac{z^{7}}{z^{11}}= z^{7}$ etc.
$\displaystyle \therefore E= \frac{1}{2}\left [z+z^{2}+z^{3}+z^{4}+z^{5}+z^{10}+z^{9}+z^{8}+z^{7}+z^{6} \right ]$

Add and subtract $\displaystyle z^{11}.$
$\displaystyle \therefore E= \frac{1}{2}$ [sum of G.P. of 11 terms-$\displaystyle z^{11}$]

$\displaystyle = \frac{1}{2}\left [ \frac{z\left ( 1-z^{11} \right )}{1-z}-z^{11} \right ]= \frac{1}{2}\left ( 0-1 \right )= -\frac{1}{2}$ by (I)

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

the nth roots of unity complex numbers maths

  1. $1$

  2. $0$

  3. $-1$

  4. $2$

Show answer
Correct Option: A
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.

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

the nth roots of unity complex numbers maths

  1. $0$

  2. $1$

  3. $w$

  4. $n$

Show answer
Correct Option: A
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 $\dfrac{1-1}{1-\alpha^{r}}=0$

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

the nth roots of unity complex numbers maths

  1. $\displaystyle \frac{1}{\alpha}$

  2. $\displaystyle \frac{1}{\alpha} (2^n -1)$

  3. $\alpha$

  4. $\displaystyle \frac{1}{\alpha} \left[ (1+\alpha)^n - 1\right]$

Show answer
Correct Option: D
Explanation:

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

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

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

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

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