Mathematics · Quantitative Aptitude

Polynomial and Quadratic Equations

239 Questions

Polynomial and quadratic equations involve finding the roots of expressions with varying degrees. The focus is on the relationship between coefficients and roots, symmetric functions, and solving higher degree polynomials. This topic is a core component of algebra in competitive testing.

Roots of quadratic equationsSymmetric root functionsCubic polynomialsRoot approximation methodsSum and product of roots

Polynomial and Quadratic Equations Questions

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If $o<\alpha<\beta<\gamma<\dfrac {\pi}{2}$, then the equation $\dfrac {1}{x-\sin \alpha}+\dfrac {1}{x-\sin\beta}+\dfrac {1}{x-\sin \gamma}=0$ has

  1. Imaginary roots

  2. Real and equal roots

  3. Real and unequal roots

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

$\begin{array}{l} \frac { 1 }{ { x-\sin  \alpha  } } +\frac { 1 }{ { x-\sin  \beta  } } +\frac { 1 }{ { x-\sin  \gamma  } } =0 \ 0<\alpha <\beta <\gamma <\frac { \pi  }{ 2 }  \ let\, \alpha ={ 30^{ 0 } },\beta ={ 45^{ 0 } },\gamma ={ 60^{ 0 } } \ \Rightarrow \frac { 1 }{ { x-\frac { 1 }{ 2 }  } } =\frac { 1 }{ { x-2\sqrt { 2 }  } } =\frac { 1 }{ { x-\sqrt { \frac { 3 }{ 2 }  }  } } =0 \end{array}$

hence roots are real and unequal

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The value of $'a'$ for which the equation ${ x }^{ 3 }+ax+1=0$ and ${ x }^{ 4 }+a{ x }^{ 2 }+1=0$, have a common root is

  1. $a=2$
  2. $a=-2$
  3. $a=0$

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

Consider the following equation
$x^{4}+ax^{2}+1=0$
$x^{3}+ax+1=0$
Subtracting equation (ii) from (i), we get 
$x^{4}-x^{3}+a(x^{2}-x)=0$
$x^{3}(x-1)+ax(x-1)=0$
$(x-1)(x^{3}+ax)=0$
$x(x-1)(x^{2}+a)=0$
Hence, we get $x=0$ $x=1$ and $x^{2}=-a$
Now out of the above two, $x=0$ is not a root of the following two equations.
We do not know the nature of '$a$'. 

Hence, we cannot determine that $x^{2}=-a$ will have real or imaginary roots.
Hence, we get $x=1$ as a common root for the above two equations.
Now for both the equations to have $x=1$ as a common root, 
$f(1)=0$
$1+a+1=0$
$a=-2$
Similarly substituting in the second equation, we get $a=-2$.
Hence, the required value of $a$ is $-2$.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

lf the difference of the roots of the equation $x^{2}-bx+c=0$ is equal to the differecne of the roots of the equation ${x}^{2}-{c}x+b=0$ and $b\neq c$, then $b+c=$

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

Let $(\alpha, \beta )$ and $ (\gamma, \delta )$ be the roots of the first equation and second equation respectively. 

Then, for the first equation
$ \alpha +\beta =b$ and $ \alpha \beta =c$
Now $ (\alpha +\beta )^{ 2 }={ b }^{ 2 }$
$ \Rightarrow (\alpha -\beta )^{ 2 }+4\alpha \beta ={ b }^{ 2 }$
$ \Rightarrow (\alpha -\beta )^{ 2 }={ b }^{ 2 }-4\alpha \beta $
$ \Rightarrow |\alpha -\beta |=\sqrt { { b }^{ 2 }-4c } $
Similarly for the second equation
$ |\gamma -\delta |=\sqrt { { c }^{ 2 }-4b } $
As per the given condition,
$ \sqrt { { b }^{ 2 }-4c } =\sqrt { { c }^{ 2 }-4b } $
$\Rightarrow { b }^{ 2 }-4c={ c }^{ 2 }-4b$
$\Rightarrow { b }^{ 2 }-{ c }^{ 2 }=-4(b-c)$
$ \Rightarrow (b+c)(b-c)=-4(b-c)$
Therefore, option D is correct.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Let $\displaystyle a _{1}, a _{2},a _{3},a _{4},a _{5} \, \varepsilon \, R$ denote a rearrangement of equation $\displaystyle p _{1}x^{5}+p _{2}x^{3}+p _{3}x^{2}+p _{4}x+p _{5}=0$ then, equation  $\displaystyle a _{1}x^{4}+a _{2}x^{3}+a _{3}x^{2}+a _{4}x +a _{5}=0$ has 

  1. at least two real roots

  2. all four real roots

  3. only imaginary roots

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$a _{1}x^{4}+a _{2}x^{3}+a _{3}x^{2}+a _{4}x+a _{5}=0$
for $ x=1 $
$a _{1}+a _{2}+a _{3}+a _{4}+a _{5}=0$
for the given set of equation,
sum of  $p _{1}+p _{2}+p _{3}+p _{4}+p _{5}=0$ 
So, $a _{1}+a _{2}+a _{3}+a _{4}+a _{5}\epsilon R $  for any set of arrangement
Hence at least two real roots.
Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If the sum of two roots of the equation $x^{4}+px^{3}+qx^{2}+rx+8=0$ is equal to the sum of the other two, then $p^{3}+8r=$

  1. $p^2 - 4pq$
  2. $2pq$
  3. $p^2 - pq$
  4. $4pq$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the roots of the equation be $a,b,c,d$. 
As per the question,
$a+b = c+d$ 
From the theory of polynomials, 
$a+b+c+d = -p$
$ \Rightarrow a+b=c+d= \displaystyle \frac{-p}{2} $

Also,
$ab+ac+ad+bd+bc+cd  = q $
$ \Rightarrow (a+b)(c+d) +ab+cd  =q $
$ \Rightarrow ab+cd = q - \displaystyle \frac{p^2}{4} $

Also, 
$abc+abd+bcd+adc = -r $
$ \Rightarrow ab(c+d) +cd(a+b) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} (ab+cd) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} ( q - \displaystyle \frac{p^2}{4} ) = -r $
$ \Rightarrow -4pq + p^3 = -8r $
$ \Rightarrow p^3 + 8r = 4pq $

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

lf one root of the equation $ax^{2}+bx+c=0$ is the square of the other, then

  1. $b^{2}+ac^{2}+a^{2}c=3abc$
  2. $b^{3}+ac^{2}+a^{2}c=3abc$
  3. $b^{2}+ac^{2}+a^{2}c+3abc=0$
  4. $b^{3}+ac^{2}+a^{2}c+3abc=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given equation $a{ x }^{ 2 }+bx+c$
Given that, one root of the equation is square of another.
So, lets assume $\alpha$ , ${ \alpha  }^{ 2 }$ are roots of the given equation
We know that,
Sum of roots $=$ $\alpha +{ \alpha  }^{ 2 }=\dfrac { -b }{ a }$ 
Product of roots $=$ $ \alpha \times { \alpha  }^{ 2 }=\dfrac { c }{ a }$
$\alpha (1+\alpha )=\dfrac { -b }{ a } \longrightarrow 1  $
${ \alpha  }^{ 3 }=\dfrac { c }{ a } \longrightarrow 2 $
Cubing equation (1) on both sides and substitute the value from equation (2).
${ \alpha  }^{ 3 }{ (1+\alpha ) }^{ 3 }=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ { \alpha  }^{ 3 }({ \alpha  }^{ 3 }+1+3{ \alpha  }(1+\alpha ))=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ \dfrac { c }{ a } \left (\dfrac { c }{ a } +1+3\left (\dfrac { -b }{ a } \right)\right)=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ \dfrac { ({ c }^{ 2 }+ac-3bc) }{ { a }^{ 2 } } =\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ a({ c }^{ 2 }+ac-3bc)=-{ b }^{ 3 }\ { b }^{ 3 }+a{ c }^{ 2 }+{ a }^{ 2 }c=3abc $

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

The equation $\sqrt{x+4}$- $\sqrt{x-3}$+ 1=0 has:

  1. no root

  2. one real root

  3. one real root and one imaginary root

  4. two imaginary roots

  5. two real roots
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Longrightarrow \sqrt { x+4 } -\sqrt { x-3 } +1=0\ \Longrightarrow \sqrt { x+4 } +1=\sqrt { x-3 } \ \Longrightarrow x+4+1+2\sqrt { x+4 } =x-3\ \Longrightarrow 2\sqrt { x+4 } =-8\ \Longrightarrow x+4=16\ \therefore x=12$

But x = 12 will not satisfy given equation.
$\therefore$ No roots for given equation.

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

If $z _{1}$ is a root of the equation $a^{n} _{0}z^{n}+a _{1}z^{n-1}+....+a _{n-1^{z}}+a _{n}=3$, where $|a _{i}|<2$ for $i=0,1,....,n.$ Then,

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

According to question,

${l} { a _{ 0 } }{ z^{ n } }+{ a _{ 1 } }{ z^{ n-1 } }+.............+{ a _{ n-1 } }z+{ a _{ n } }=3 \ \Rightarrow \left| { { a _{ 0 } }{ z^{ n } }+{ a _{ 1 } }{ z^{ n-1 } }+.............+{ a _{ n-1 } }z+{ a _{ n } } } \right| =\left| 3 \right|  \ \Rightarrow \left| { { a _{ 0 } } } \right| \, { \left| z \right| ^{ n } }+\left| { { a _{ 1 } } } \right| \, { \left| z \right| ^{ n-1 } }\, +..........+\left| { { a _{ n-1 } } } \right| \, \left| z \right| \, +\left| { { a _{ n } } } \right| \ge 3 \ \Rightarrow 2\, ({ \left| z \right| ^{ n } }+{ \left| z \right| ^{ n-1 } }+...........\left| z \right| +1)\, \, >\, 3 \ \Rightarrow (1+\left| z \right| +{ \left| z \right| ^{ 2 } }+...........+{ \left| z \right| ^{ n } })\, \, >\, \frac { 3 }{ 2 }  \ \Rightarrow \frac { { \, \, \, \, 1-{ { \left| z \right|  }^{ n+1 } } } }{ { 1-\left| z \right|  } } \, \, >\, \frac { 3 }{ 2 }  \ \Rightarrow 2-2{ \left| z \right| ^{ n+1 } }\, >\, 3-3\left| z \right|  \ \Rightarrow 2{ \left| z \right| ^{ n+1 } }<3\left| z \right| \, -1 \ \Rightarrow 3\, \left| z \right| -1\, >0 \ \, \, \, \, \, \therefore \, \, \, \left| z \right| \, >\, \frac { 1 }{ 3 }  \ so\, the\, correct\, option\, is\, \, A$

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

If $p, q, r, s, t$ are the roots of the equation $x^5-1 = 0$, then $p^{ 10 }+q^{ 10 }+{ r }^{ 10 }+{ s }^{ 10 }+t^{ 10 }=$

  1. $0$
  2. $1$
  3. $3$
  4. $5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$p, q, r, s, t$ are all fifth root of unity
$\Rightarrow p^5=q^5= r^5= s^5= t^5=1$   
$ \Rightarrow p^{10}=q^{10}= r^{10}= s^{10}= t^{10}=1$
Hence, the required sum is $5$

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

If $1,{ \alpha  } _{ 1 },{ \alpha  } _{ 2 },{ \alpha  } _{ 3 }$ and $\alpha _4$ be the roots of $x^5-1=0$, then $\displaystyle \frac { \omega -{ \alpha  } _{ 1 } }{ { \omega  }^{ 2 }-{ \alpha  } _{ 1 } } .\frac { \omega -{ \alpha  } _{ 2 } }{ { \omega  }^{ 2 }-{ \alpha  } _{ 2 } } .\frac { \omega -{ \alpha  } _{ 3 } }{ { \omega  }^{ 2 }-{ \alpha  } _{ 3 } } .\frac { \omega -{ \alpha  } _{ 4 } }{ { \omega  }^{ 2 }-{ \alpha  } _{ 4 } } =$ 

  1. $1$
  2. $\omega$
  3. ${ \omega }^{ 2 }$

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

Since $1,{ \alpha  } _{ 1 },{ \alpha  } _{ 2 },{ \alpha  } _{ 3 },{ \alpha  } _{ 4 }$ are roots of the equation ${ x }^{ 5 }-1=0$

Thus ${ x }^{ 5 }-1=\left( x-1 \right) \left( x-{ \alpha  } _{ 1 } \right) \left( x-{ \alpha  } _{ 2 } \right) \left( x-{ \alpha  } _{ 3 } \right) \left( x-{ \alpha  } _{ 4 } \right)$
$ \Rightarrow \displaystyle\frac { { x }^{ 5 }-1 }{ \left( x-1 \right)  } =\left( x-{ \alpha  } _{ 1 } \right) \left( x-{ \alpha  } _{ 2 } \right) \left( x-{ \alpha  } _{ 3 } \right) \left( x-{ \alpha  } _{ 4 } \right) $........1
Putting $x=w$ in 1 we get,
$\Rightarrow \displaystyle\frac { { 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) $
$\Rightarrow \displaystyle\frac { { w }^{ 2 }-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) $............2
Putting $x={ w }^{ 2 }$ in 1 we get
$\Rightarrow \displaystyle\frac { { 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) $
$\Rightarrow \displaystyle\frac { { w }-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) $..........3
Dividing 2 by 3 we get, 
$\Rightarrow \displaystyle\frac { \left( { w }-{ \alpha  } _{ 1 } \right) \left( { w }-{ \alpha  } _{ 2 } \right) \left( { w }-{ \alpha  } _{ 3 } \right) \left( { w }-{ \alpha  } _{ 4 } \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)  } =\frac { { \left( { w }^{ 2 }-1 \right)  }^{ 2 } }{ { \left( { w }-1 \right)  }^{ 2 } } $
$=\displaystyle\frac { { w }^{ 4 }+1-2{ w }^{ 2 } }{ { w }^{ 2 }+1-2w } =\frac { { w }+1-2{ w }^{ 2 } }{ { w }^{ 2 }+1-2w } =\frac { -{ w }^{ 2 }-2{ w }^{ 2 } }{ -w-2w } =w$

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

The maximum number of real root of the equation $\displaystyle x^{2n} - 1 = 0$ is

  1. $\displaystyle 2$
  2. $\displaystyle 3$
  3. $\displaystyle n$
  4. $\displaystyle 2n$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x^{2n}=1$
Now if  $n$ is odd we have
$x^{n}=\pm1 $
$x^{n}=1$ and $x^{n}=-1$
$x^{n}=-1$
$x=-1$
Now if $n$ is odd
$x^{n}-1=0$
$(x-1)(1+x+x^{2}+..x^{n-1})=0$
Hence $x=1$ and  the equation $1+x+x^{2}+..x^{n-1}=0$ gives $nth$  roots of unity.
Hence at most $2$ real roots.
Similarly if $n$  is even.
Then
$x^{n}=\pm1 $
$x^{n}=1$ and $x^{n}=-1$
Now 
$x^{n}=-1$ will given imaginary roots.
$x^{n}=1$ can be further simplified in to
$x^{\frac{n}{2}}=\pm1 $ and so on.
Hence we will get remaining pairs of imaginary roots ans two real roots $1$ and  $-1$  at the end.
Hence at-most $2$ real roots.

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 root of $x^6=1$ then $\displaystyle \frac{\alpha ^5+\alpha ^3+\alpha +1}{\alpha ^2+1}=$

  1. -$\alpha ^2$

  2. 0
  3. $\alpha ^2$
  4. $\alpha $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\alpha$ is non real root of $x^6 = 1$

one possible complex value of 
$\alpha = \cos (2 \pi / 6) + i \sin (2\pi / 6)$
$\alpha = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i$
$\alpha^2 = \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i \right) \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i \right) = \dfrac{1}{4} + \dfrac{\sqrt{3}}{4} i + \dfrac{\sqrt{3}}{4} i - \dfrac{3}{4} = \dfrac{-1}{2} + \dfrac{\sqrt{3}}{2} i$
$\alpha^3 = \left(\dfrac{-1}{2} + \dfrac{\sqrt{3}}{2} i \right) \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i \right) = \dfrac{-1}{4} = \dfrac{\sqrt{3}}{4} i + \dfrac{\sqrt{3}}{4} - \dfrac{3}{4}= -1$
$\dfrac{\alpha^5 + \alpha^3 + \alpha + 1}{\alpha^2 + 1} = \alpha^3 + \dfrac{\alpha + 1}{\alpha^2 + 1}$
$= -1 + \dfrac{\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i + 1 \right)}{\left(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i + 1 \right)}$
$= \dfrac{\dfrac{1}{2} - \dfrac{\sqrt{3}}{2} i - 1 + \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i + 1}{(\dfrac{-1}{2} + \dfrac{\sqrt{3}}{2} i + 1)}$
$= \dfrac{1}{\alpha^2 + 1} = \dfrac{\alpha^2}{\alpha^4 + \alpha^2}$
$= \alpha^4 = \alpha^2 . \alpha^2 = \left(\dfrac{\sqrt{3}}{2} i - \dfrac{1}{2} \right) \left(\dfrac{\sqrt{3}}{2} i - \dfrac{1}{2} \right) = -\dfrac{\sqrt{3}}{2} i - \dfrac{1}{2}$
$= \dfrac{\alpha^2}{\alpha^4 + \alpha^2} = \dfrac{\alpha^2}{\left(\dfrac{-1}{2} + \dfrac{\sqrt{3}}{2} i - \dfrac{1}{2} - \dfrac{\sqrt{3}}{2} i\right)} = - \alpha^2$
Considering option C as $- \alpha^2$ , it is correct

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

The roots of the equation  $z^{5}+z^{4}+z^{3}+z^{2}+z+1=0$   are given by

  1. $-1$
  2. $\displaystyle -\frac{1}{2}+\frac{i\sqrt{3}}{2}$
  3. $\displaystyle \frac{1}{2}+\frac{i\sqrt{3}}{2}$
  4. $\displaystyle \frac{-1-i\sqrt{3}}{2}$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

$\displaystyle { z }^{ 5 }+{ z }^{ 4 }+{ z }^{ 3 }+{ z }^{ 2 }+z+1=0\ \Rightarrow \frac { { z }^{ 6 }-1 }{ z-1 } =0\ \Rightarrow z\neq 1\quad &amp; \quad { z }^{ 6 }=1=\cos { 0 } +i\sin { 0 } \ \Rightarrow z={ \left( \cos { 0 } +i\sin { 0 }  \right)  }^{ \frac { 1 }{ 6 }  }=\cos { \frac { 2k\pi  }{ 6 }  } +i\sin { \frac { 2k\pi  }{ 6 }  } \ \Rightarrow z=\cos { \frac { k\pi  }{ 3 }  } +i\sin { \frac { k\pi  }{ 3 }  } $
where $k=0,1,2,3,4,5$.

For $k=0$,
$z=1$ but $z\neq 1$

For $k=1$,
$\displaystyle z=\cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  } =\frac { 1+i\sqrt { 3 }  }{ 2 } $

For  $k=2$,
$\displaystyle z=\cos { \frac { 2\pi  }{ 3 }  } +i\sin { \frac { 2\pi  }{ 3 }  } =\frac { -1+i\sqrt { 3 }  }{ 2 } $

For  $k=3$,
$\displaystyle z=\cos { \frac { 3\pi  }{ 3 }  } +i\sin { \frac { 3\pi  }{ 3 }  } =-1$

For  $k=4$,
$\displaystyle z=\cos { \frac { 4\pi  }{ 3 }  } +i\sin { \frac { 4\pi  }{ 3 }  } =\frac { -1-i\sqrt { 3 }  }{ 2 } $

For $k=4$,
$\displaystyle z=\cos { \frac { 5\pi  }{ 3 }  } +i\sin { \frac { 5\pi  }{ 3 }  } =\frac { 1-i\sqrt { 3 }  }{ 2 } $

Hence, all the options A,B,C and D are correct.

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

If $\alpha,\ \beta,\ \gamma$ and $\Delta $ are the roots of the equation $x^{4}-1=0$, then the value of $\displaystyle \frac{a\alpha+b\beta+c\gamma+d\Delta}{a\gamma+b\Delta +c\alpha+d\beta}+\frac{a\gamma+b\Delta +c\alpha+d\beta}{a\alpha+b\beta+c\gamma+d\Delta }$ is

  1. $ 3\beta$
  2. $0$
  3. $ 2\gamma$
  4. $-2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Clearly,
$\alpha = e^{i0} = 1$
$\beta = e \frac{i2\pi}{4} = i$
$\gamma = e \frac{i4\pi}{4} = -1$
$\delta = e \frac{i6\pi}{4} = -i$
So, $\dfrac {a\alpha+b\beta+c\gamma+d\delta}{ a\gamma+b\delta+c\alpha+d\beta}=\dfrac {a+bi-c-di}{- a-bi+c+di}$
$=-1$
Similarly second expression is nothing but reciprocal of first
$=\frac{1}{-1}=-1$
Ans $ = -1-1 = -2$

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

The number of roots of the equation $z^{15}=1$ satisfying $|\arg(z)|<\pi/2$ is

  1. 6

  2. 7

  3. 8

  4. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
For the nth root of unity of complex number $z$ i.e. $z^n = 1$, there are 'n' total roots.
In the present case, $n=15$, thus, we have 15 roots.

$|arg(z)|<\cfrac {\pi}{2}$ 
$\Rightarrow -\cfrac {\pi}{2} < arg(z) < \cfrac {\pi}{2}$

Each root is at equal angular distance i.e. $\dfrac{2\pi}{15}$      ...(because $\dfrac{2\pi}{n}$).

$\therefore$ between $-\cfrac {\pi}{2}$ and $\cfrac {\pi}{2}$, their will be 7 roots (imcluding 1).
Hence, the correct option is B.