Tag: complex numbers

Questions Related to complex numbers

lf $a=\displaystyle \cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}, \alpha=a+a^{2}+a^{4}$ and $\beta=a^{3}+a^{5}+a^{6}$, then $\alpha, \beta$ are the roots of the equation

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

  1. $x^{2}+x+1=0$

  2. $x^{2}+x+2=0$

  3. $x^{2}+2x+2=0$

  4. $x^{2}+2x+3=0$

Show answer
Correct Option: B
Explanation:

$a={ e }^{ i2\pi /7 }\ a^ 7=1\ \alpha +\beta $

$=a+a^ 2+a^ 3+a^ 4+a^ 5+a^ 6\ =\dfrac{a(a^ 6-1)}{(a-1)}\ =\dfrac{(a^ 7-a)}{(a-1)}\ =\dfrac{(1-a)}{(a-1)}$
$=-1$
$\alpha \beta =(a+a^ 2+a^ 4)(a^ 3+a^ 5+a^ 6)\ =(a^ 4+a^ 6+a^ 7+a^ 5+a^ 7+a^ 8+a^ 7+a^ 9+a^ {10})\ =(a^ 4+a^ 6+1+a^ 5+1+a+1+a^ 2+a^ 3)\ =(3+a+a^ 2+a^ 3+a^ 4+a^ 5+a^ 6)\ =(3-1)$
$=2 $
The equation can be written as
 $x^ 2-(\alpha +\beta )x +\alpha \beta  =x^ 2+x+2$
Hence, option B is correct.

If the expression $z^5 =32$ can be factorised into linear and quadratic factors over real coefficients as $(z^5 - 32)=(z - 2) (z^2-pz+4)(z^2-qz+4)$, where p > q, then the value of $p^2-  2q$
finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $8$

  2. $4$

  3. $-4$

  4. $-8$

Show answer
Correct Option: A
Explanation:

Given,

$ { z }^{ 5 }=32 $ can be factorized as
$(z-2)({ z }^{ 2 }-pz+4)({ z }^{ 2 }-qz+4)$ where $ p>q.$
 To find the value of $ { p }^{ 2 }-2q$
 Solution,
${ z }^{ 5 }=32\ { z }^{ 5 }-32=0$
$ { z }^{ 5 }-{ 2 }^{ 5 }=0$
$ z=2$ 
Will be one of the factor of given equation.
$ \quad \quad \quad \quad \quad \quad \quad { z }^{ 4 }+{ 2z }^{ 3 }+{ 4z }^{ 2 }+8z+16\ \therefore \quad z-2\sqrt { { z }^{ 5 }-32\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  } \ \quad \quad \quad \cfrac { \pm { z }^{ 5 }\mp { 2z }^{ 4 } }{ { \quad \quad \quad \quad \quad \quad 2z }^{ 4 }-32 } \ \quad \quad \quad \quad \quad \quad \cfrac { \pm { 2z }^{ 4 }\mp 4{ z }^{ 3 } }{ \quad \quad \quad \quad \quad \quad \quad 4{ z }^{ 3 }-32 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cfrac { \pm 4{ z }^{ 3 }\mp 8{ z }^{ 2 } }{ \quad \quad \quad \quad \quad \quad \quad 8{ z }^{ 2 }-32 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cfrac { \pm 8{ z }^{ 2 }\mp 16z }{ \quad \quad \quad \quad \quad \quad \quad 16z-32 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cfrac { \pm 16z\mp 32 }{ 0 } \ (z-2)({ z }^{ 4 }+2{ z }^{ 3 }+4{ z }^{ 2 }+8z+16)\longrightarrow (1)\ { z }^{ 5 }-32=(z-2)({ z }^{ 2 }-pz+4)({ z }^{ 2 }-qz+4)\quad \ { z }^{ 5 }-32=(z-2)({ z }^{ 4 }+(p+q)z^{ 3 }+(8+pq){ z }^{ 2 }+16-(4p+4q)z)\longrightarrow (2)$
On comparing $ (1)& (2)$ we get
$ p=-2-q\ 8+(-2-q)q=4\ -2q-{ q }^{ 2 }=-4\ { q }^{ 2 }+2q-4=0\ d=4+16=20\ q=\cfrac { -2\pm 2\sqrt { 5 }  }{ 5 } \ q=-1\pm \sqrt { 5 } \ { \parallel  }^{ rly }p=-1\pm \sqrt { 5 } $
But$\quad p>q\therefore p=-1+\sqrt { 5 } & q=-1-\sqrt { 5 } $
value of ${ p }^{ 2 }-2q={ \left( -1+\sqrt { 5 }  \right)  }^{ 2 }+2{ \left( -1-\sqrt { 5 }  \right)  }\ =1+5-2\sqrt { 5 } +2+2\sqrt { 5 } \ { p }^{ 2 }-2q=8\ $

Suppose A is a complex number and $ n \in N, $ such that $A^{n} = (A + 1)^{n} =1, $ then the least value of $n$ is

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

  1. $3$

  2. $6$

  3. $9$

  4. $12$

Show answer
Correct Option: B
Explanation:

$Since\quad { z }^{ n }=1\ \Rightarrow \quad { \left| z \right|  }^{ n }=1\ \quad \quad \quad \left| z \right| =1\ similarly,\ { \left( z+1 \right)  }^{ n }=1\ \Rightarrow \quad { \left| z+1 \right|  }^{ n }=1\ \left| z+1 \right| =1\ Let\quad z=a+ib\ \left| z \right| =\left| z+1 \right| \quad \Rightarrow \quad { a }^{ 2 }+{ b }^{ 2 }={ \left( a+1 \right)  }^{ 2 }+{ b }^{ 2 }\ { a }^{ 2 }+{ b }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2a+1\ \Rightarrow \quad 2a+1=0\ \therefore \quad a=\frac { -1 }{ 2 } \ putting\quad the\quad value\quad of\quad a\quad in\quad eq.\ \Rightarrow \quad { \left( \frac { -1 }{ 2 }  \right)  }^{ 2 }+{ b }^{ 2 }=1\ \Rightarrow \quad { b }^{ 2 }=\frac { 3 }{ 4 } \ \Rightarrow \quad b=\pm \frac { \sqrt { 3 }  }{ 2 } \ Now,\quad z+1=\quad \frac { 1 }{ 2 } \pm \frac { \sqrt { 3 }  }{ 2 } \ \Rightarrow \quad z+1={ e }^{ \pm \frac { zni }{ 3 }  }\ { \left( z+1 \right)  }^{ n }=\quad { e }^{ \pm \frac { zni }{ 3 }  }\ For\quad { \left( z+1 \right)  }^{ n }\quad to\quad be\quad 1\quad cos\quad \pm \frac { zn }{ 3 } =1\quad and\quad sin\quad \pm \frac { zn }{ 3 } =0\ This\quad can\quad only\quad happen\quad if\quad \pm \frac { zn }{ 3 } =2ak\quad for\quad integer\quad k.\ Solving\quad for\quad n,\quad we\quad get:\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \pm \frac { zn }{ 3 } =2ak\quad \Rightarrow \quad n=6k\ \quad \quad \quad \quad \quad \quad \quad k=\frac { 6 }{ n } \ least\quad value=\quad 6\ $

If $1,$$\alpha _{1},\alpha _{2,} \alpha _{3},\alpha _{4}$ be the  roots of $z^{5}-1=0$ and $\omega $ be an imaginary cube root of unity, 


then  $ \displaystyle \left ( \frac{\omega -\alpha _{1}}{\omega ^{2}-\alpha _{1}} \right )\left ( \frac{\omega -\alpha _{2}}{\omega ^{2}-\alpha _{2}} \right )\left ( \frac{\omega -\alpha _{3}}{\omega ^{2}-\alpha _{3}} \right )\left ( \frac{\omega -\alpha _{4}}{\omega ^{2}-\alpha _{4}} \right )$ is ?

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

  1. $\omega $

  2. $\omega^{2}$

  3. $1$

  4. $2$

Show answer
Correct Option: A
Explanation:

We have,
$z^{5}-1=\left ( z-1 \right )\left ( z-\alpha _{1} \right )\left ( z-\alpha _{2} \right )\left ( z-\alpha 3 \right )\left ( z-\alpha _{4} \right )$
putting z=w,
$\Rightarrow w^{5}-1=\left ( w-1 \right )\left ( w-\alpha _{1} \right )\left ( w-\alpha _{2} \right )\left ( w-\alpha _{3} \right )\left ( w-\alpha _{4} \right )..............(1)$
Similarly putting $z=w^{2}$
$\left ( w^{2} \right )^{5}-1=w^{10}-1=\left ( w^{2}-1 \right )\left ( w^{2}-\alpha _{1} \right )\left ( w^{2}-\alpha _{2}\right )\left ( w^{2} -\alpha _{3}\right )\left ( w^{2}-\alpha _{4} \right ).........(2)$
So the required expression reduces to 
$ \displaystyle =\frac{\left ( w^{5} -1\right )/\left ( w-1 \right )}{\left ( w^{10}-1 \right )/\left ( w^{2} -1\right )}$
$=\frac{\left ( w^{2} -1\right )/\left ( w-1 \right )}{\left ( w-1 \right )/\left ( w^{2} -1\right )}$
$\left [ \because w^{3}=1,w^{5} =w^{3}w^{2}=w^{2}\right ]$
$=\frac{\left ( w^{2}-1 \right )^{2}}{\left ( w-1 \right )^{2}}$
$=\frac{\left ( w-1 \right )^{2}\left ( w+1 \right )^{2}}{\left ( w-1 \right )^{2}}$
$=1+2w+w^{2}$
$=w\left [ \because 1+w^{2} =-w\right ]$

If $iz^4 + 1 = 0$, then z can take the value

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $\displaystyle \frac{1 + i}{\sqrt 2}$

  2. $\cos \displaystyle \frac{\pi}{8} + i \sin \frac{\pi}{8}$

  3. $\displaystyle \frac{1}{4 i}$

  4. $i$

Show answer
Correct Option: B
Explanation:
$iz^4 + 1 = 0 \Rightarrow z^4 = - \displaystyle \frac{1}{i} = \frac{i^2}{i} = i$
Let $z^4 = \cos  \displaystyle \frac{\pi}{2} + i  \sin  \frac{\pi}{2}$
$\therefore z = \displaystyle \left [ \cos  \frac{\pi}{2} + i  \sin  \frac{\pi}{2} \right ]^{1/4}$
U\sin g De-Moivre's theorem,
$(\cos   \theta + i  \sin  \theta)^n = \cos   n \theta + i  \sin   n \theta, n  \varepsilon I$
Hence, $z = \cos  \displaystyle \frac{\pi}{8} + i  \sin  \frac{\pi}{8}$

The product of the values of $\displaystyle{\left[ {\cos {\pi  \over 3} + i\sin {\pi  \over 3}} \right]^{{3 \over 4}}}$ is

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $-1$

  2. $1$

  3. $i$

  4. $-i$

Show answer
Correct Option: B
Explanation:

Given: $\displaystyle{\left[ {\cos {\pi  \over 3} + i\sin {\pi  \over 3}} \right]^{{3 \over 4}}}$


$=[e^{i(\pi/3)}]^{(3/4)}=e^{i\pi(1/3)(3/4)}=e^{4\pi i}=cos4\pi+isin{4\pi}=1-0i=1$ 

Number of integral values of n for which the quantity ${n+i}^{4}$ where ${i}^{2}=-1$, is an integer is

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $1$

  2. $2$

  3. $3$

  4. Infinite

Show answer
Correct Option: D
Explanation:

Simply $n+i^{4}$ = $n+1$ which is integer for every integral value of n .

So there are infinite values of n .

De Moivre's theorem

$(\cos\theta +i\sin \theta )=\cos n\theta $ if n is an integer and $\cos n\theta +i \sin n\theta $ is one of the values of $(\cos\theta +i\sin\theta )^{n}$, if n is a fraction.

Corollary : The q values of ($(\cos\theta +i\sin\theta )^{\frac{1}{q}}$ are obtained from

cos $\frac{2n\pi +\theta }{q}+i\sin\frac{2n\pi +\theta }{q}$ by putting n = 0, 1, 2, ..., (q - 1).


de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. Both are correct

  2. Only first statement is true.

  3. Only second ststement is true

  4. None 

Show answer
Correct Option: A

If $a = {\mathop{\rm cis}\nolimits} \alpha ,b = cis\beta ,c = cis\gamma $ then $\dfrac{{{a^3}{b^3}}}{{{c^2}}} = $

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $cis(3\alpha + 3\beta + 2\gamma )$

  2. $cis(3\alpha + 3\beta - 2\gamma )$

  3. $cis( - 3\alpha - 3\beta + 2\gamma )$

  4. $cis(3\alpha - 3\beta + 2\gamma )$

Show answer
Correct Option: B
Explanation:

$a=cis\alpha=Cos\alpha+iSin\alpha= \ e^{i\alpha}$

$b=cis\beta=Cos\beta+iSin\beta= \ e^{i\beta}$      

$a=cis\gamma=Cos\gamma+iSin\gamma= \ e^{i\gamma}$

$\therefore \dfrac{a^3b^3}{c^2}=\dfrac{({e^{i\alpha}})^3({e^{i\beta}})^3}{({e^{i\gamma}})^2}$

$=\dfrac{e^{3i\alpha}e^{3i\beta}} {e^{2i\gamma}}=\ e^{i(3\alpha+3\beta-2\gamma)}$

$=Cos(3\alpha+3\beta-2\gamma)+iSin(3\alpha+3\beta-2\gamma)$

$=cis(3\alpha+3\beta-2\gamma)$

If $a=\cos { \left( \cfrac { 8\pi  }{ 11 }  \right)  } +i\sin { \left( \cfrac { 8\pi  }{ 11 }  \right)  } $, then $Re(a+{a}^{2}+{a}^{3}+{a}^{4}+{a}^{5})=$

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $0$

  2. $-\cfrac{1}{2}$

  3. $\cfrac{1}{2}$

  4. $1$

Show answer
Correct Option: B