Tag: maths

Questions Related to maths

Given $z$ is a complex number with modulus $1$. Then the equation $\dfrac{(1+ia)}{(1-ia)}$ = $z$ has

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

  1. all roots real and distinct

  2. two real and one imaginary

  3. three roots real and one imaginary

  4. one root real and three imaginary

Show answer
Correct Option: A
Explanation:

$\displaystyle { \left( \frac { 1+ia }{ 1-ia }  \right)  }^{ 4 }=z\quad $         ...(1)
$\displaystyle & \quad \left| z \right| =1$
$\displaystyle z=cisA=\cos { A } +i\sin { A } $
Substitute $z$  in equation (1)
$\displaystyle { \left( \frac { 1+ia }{ 1-ia }  \right)  }={ cisA }^{ \frac { 1 }{ 4 }  }=cis\frac { 2k\pi +A }{ 4 } $       ...{De Moivre's Theorem}
where $ k=0,1,2,3$

Let $\displaystyle B=\frac { 2k\pi +A }{ 4 } $

$\displaystyle \Longrightarrow ia=\frac { -1+cisB }{ 1+cisB } =\frac { \sin { \frac { B }{ 2 } \left( i\cos { \frac { B }{ 2 }  } -\sin { \frac { B }{ 2 }  }  \right)  }  }{ \cos { \frac { B }{ 2 } \left( \cos { \frac { B }{ 2 }  } +i\sin { \frac { B }{ 2 }  }  \right)  }  } $

$\displaystyle \Longrightarrow ia=\frac { i\sin { \frac { B }{ 2 } \left( \cos { \frac { B }{ 2 }  } +i\sin { \frac { B }{ 2 }  }  \right)  }  }{ \cos { \frac { B }{ 2 } \left( \cos { \frac { B }{ 2 }  } +i\sin { \frac { B }{ 2 }  }  \right)  }  } $

$\displaystyle \Longrightarrow a=\tan { \frac { B }{ 2 }  } $

Therefore roots are real and distinct.

Ans: A

If n is a natural number$ \ge$ 2, such that $z^n = (z+ 1)^n$, then 

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

  1. roots of equation lie on a straight line parallel to the y-axis

  2. roots of equation lie on a straight line parallel to the x-axis

  3. sum of the real parts of the roots is -[(n-1)/2]

  4. none of these

Show answer
Correct Option: A,C
Explanation:

$\displaystyle { z }^{ n }={ \left( z+1 \right)  }^{ n }$   where, $n\ge 2$     ...(1)

$\displaystyle \Rightarrow { \left( \frac { z+1 }{ z }  \right)  }^{ n }=1=\cos { 0 } +i\sin { 0 } $

$\displaystyle \Rightarrow \frac { z+1 }{ z } ={ \left( \cos { 0 } +i\sin { 0 }  \right)  }^{ \frac { 1 }{ n }  }$

$\displaystyle \Rightarrow \frac { z+1 }{ z } =\cos { \frac { 2k\Pi  }{ n }  } +i\sin { \frac { 2k\Pi  }{ n }  } $      ...{De Moivre's Theorem}

$\displaystyle \Rightarrow z=\frac { -1 }{ 1-\cos { \frac { 2k\Pi  }{ n }  } -i\sin { \frac { 2k\Pi  }{ n }  }  } \quad =\frac { -1 }{ 2\sin { \frac { k\Pi  }{ n } \left( \sin { \frac { k\Pi  }{ n } -i } \cos { \frac { k\Pi  }{ n }  }  \right)  }  } =\frac { -1\left( \sin { \frac { k\Pi  }{ n } +i } \cos { \frac { k\Pi  }{ n }  }  \right)  }{ 2\sin { \frac { k\Pi  }{ n }  }  } $

$\displaystyle \Rightarrow z=\frac { -\left( 1+i\cot { \frac { k\Pi  }{ n }  }  \right)  }{ 2 } $

Where$ k=1,2,3,....,n-1 $     ..{Since at k=0, z is not defined}

$\because \quad Re\left( z \right) $ is constant.
Therfore roots of ${ z }^{ n }={ \left( z+1 \right)  }^{ n }$ lie on straight line parellel to y-axis.

$\displaystyle \because \quad z=\frac { -\left( 1+i\cot { \frac { k\Pi  }{ n }  }  \right)  }{ 2 } $ and $k=1,2,3,....,(n-1)$

Sum of $\displaystyle Re\left( z \right) $= $-\frac { \left( n-1 \right)  }{ 2 } $.

Ans: A,C

For positive integers $\displaystyle n _{1}$ and $\displaystyle n _{2}$ the value of the expression $\displaystyle (1+i)^{n _{1}}+(1+i^{3})^{n _{1}}+(1+i^{5})^{n _{2}}+(1+i^{2})^{n _{2}}$ where
$\displaystyle i= \sqrt{-1}$ is a real number iff

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

  1. $\displaystyle n _{1}= n _{2}$

  2. $\displaystyle n _{2}= n _{2}-1$

  3. $\displaystyle n _{1}= n _{2}+1$

  4. $\displaystyle \forall n _{1}$ and $\displaystyle n _{2}$

Show answer
Correct Option: D
Explanation:

$(1+i)^{n _{1}}  = $ $  ^{n _{1}}C _{0} + ^{n _{1}}C _{1} i +^{n _{1}}C _{2} i^2 + ......+^{n _{1}}C _{n _{1}} i^{n _{1}}$ --------(1)


$(1+i^3)^{n _{1}}  =(1-i)^{n _{1}}=  $ $ ^{n _{1}}C _{0} - ^{n _{1}}C _{1} i +^{n _{1}}C _{2} i^2 - ......+^{n _{1}}C _{n _{1}} i^{n _{1}}$--------(2)

$(1+i^5)^{n _{2}}  =(1+i)^{n _{2}}=  $ $ ^{n _{2}}C _{0} + ^{n _{2}}C _{1} i +^{n _{2}}C _{2} i^2 + ......+^{n _{2}}C _{n _{2}} i^{n _{2}}$--------(3)

$(1+i^7)^{n _{2}}  =(1-i)^{n _{2}}=  $ $ ^{n _{2}}C _{0} - ^{n _{2}}C _{1} i +^{n _{2}}C _{2} i^2 - ......+^{n _{2}}C _{n _{2}} i^{n _{2}}$--------(4)

Adding (1),(2),(3) and (4),

$(1+i)^{n _{1}} +(1+i^3)^{n _{1}} +(1+i^5)^{n _{2}} +(1+i^7)^{n _{2}}$ 
$= 2(^{n _{1}}C _{0} +^{n _{1}}C _{2}i^2 +^{n _{1}} C _{4}i^4 +...........)+2(^{n _{2}}C _{0} +^{n _{2}}C _{2}i^2 +^{n _{2}} C _4 i^4+...........)$

$ = 2(^{n _{1}}C _{0} -^{n _{1}}C _{2} +^{n _{1}} C _{4} +...........)+2(^{n _{2}}C _{0} -^{n _{2}}C _{2} +^{n _{2}} C _4 +...........)$
$\Rightarrow$  As there are only even powers of $i$, the expression is real for all  positive integers $n _{1}$ and $n _{2}$

If ${ x }^{ 6 }={ \left( 4-3i \right)  }^{ 5 }$, then the product of all of its roots is (where $\displaystyle \theta =-\tan ^{ -1 }{ \frac { 3 }{ 4 }  } $)

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

  1. ${ 5 }^{ 5 }\left( \cos { 5\theta  } +i\sin { 5\theta  }  \right) $

  2. $-{ 5 }^{ 5 }\left( \cos { 5\theta  } +i\sin { 5\theta  }  \right) $

  3. ${ 5 }^{ 5 }\left( \cos { 5\theta  } -i\sin { 5\theta  }  \right) $

  4. $-{ 5 }^{ 5 }\left( \cos { 5\theta  } -i\sin { 5\theta  }  \right) $

Show answer
Correct Option: B
Explanation:

$\displaystyle { x }^{ 6 }={ \left( 4-3i \right)  }^{ 5 }\Rightarrow { x }^{ 6 }={ 5 }^{ 6 }\left( \frac { 4 }{ 5 } -\frac { 3i }{ 5 }  \right) ={ 5 }^{ 5 }{ \left( \cos { \theta  } +i\sin { \theta  }  \right)  }^{ 5 }$

where $\displaystyle \theta =-\tan ^{ -1 }{ \frac { 3 }{ 4 }  } ={ 5 }^{ 5 }\left( \cos { 5\theta  } +i\sin { 5\theta  }  \right) $
$\displaystyle x={ 5 }^{ 5/6 }{ \left( \cos { 5\theta  } +i\sin { 5\theta  }  \right)  }^{ 1/6 }={ 5 }^{ 5/6 }\left[ \cos { \left( \frac { 2k\pi +5\theta  }{ 6 }  \right)  } +i\sin { \left( \frac { 2k\pi +5\theta  }{ 6 }  \right)  }  \right] $
${ x } _{ 1 }{ x } _{ 2 }{ x } _{ 3 }...{ x } _{ 6 }={ 5 }^{ 5 }\left( \cos { \left( 5\pi +5\theta  \right)  } +i\sin { \left( 5\pi +5\theta  \right)  }  \right) \ ={ 5 }^{ 5 }\left( -\cos { 5\theta  } -i\sin { 5\theta  }  \right) =-{ 5 }^{ 5 }\left( \cos { 5\theta  } +i\sin { 5\theta  }  \right) $

If $C _{o},C _{1},C _{2}...C _{n}$ are the Binomial coefficient in the expansion of $\left ( 1+x \right )^{n}$ then which is not correct

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

  1. $C _{0}-C _{2}+C _{4}-C _{6}+...=2\tfrac{n}{2}\cos \frac{n\pi }{4}$

  2. $C _{1}-C _{3}+C _{5}+...=2\tfrac{n}{2}\sin \frac{n\pi }{4}$

  3. $C _{1}+C _{5}+C _{9}+C _{13}+...=\tfrac{1}{2}\left ( 2^{n-1}+2\tfrac{n}{2}\sin \frac{n\pi }{4} \right )$

  4. None of these

Show answer
Correct Option: D
Explanation:

By Binomial theorem
${ \left( 1+x \right)  }^{ n }={ C } _{ 0 }+{ C } _{ 1 }x+{ C } _{ 2 }{ x }^{ 2 }+{ C } _{ 3 }{ x }^{ \ 3 }+{ C } _{ 4 }{ x }^{ 4 }+...$      ...(1)

Substitute $x=i$ in equation (1), we get 
$\Rightarrow { \left( 1+i \right)  }^{ n }={ C } _{ 0 }+{ C } _{ 1 }i-{ C } _{ 2 }-{ C } _{ 3 }i+{ C } _{ 4 }+.....$       ...(2)

Substitute $x=-i$ in equation (1), we get
${ \left( 1-i \right)  }^{ n }={ C } _{ 0 }-{ C } _{ 1 }i-{ C } _{ 2 }+{ C } _{ 3 }i+{ C } _{ 4 }+.....$      ...(3)

Adding (2) & (3), we have
$\Rightarrow 2\left( { C } _{ 0 }{ -C } _{ 2 }{ +C } _{ 4 }+.... \right) ={ \left( 1+i \right)  }^{ n }+{ \left( 1-i \right)  }^{ n }$

$\Rightarrow { C } _{ 0 }{ -C } _{ 2 }{ +C } _{ 4 }+....=\dfrac { { \left( 1+i \right)  }^{ n }+{ \left( 1-i \right)  }^{ n } }{ 2 } ={ 2 }^{ \dfrac { n }{ 2 }  }\left( \dfrac { { \left( \cos { \dfrac { \pi  }{ 4 }  } +i\sin { \dfrac { \pi  }{ 4 }  }  \right)  }^{ n }+{ \left( \cos { \dfrac { \pi  }{ 4 }  } -i\sin { \dfrac { \pi  }{ 4 }  }  \right)  }^{ n } }{ 2 }  \right) $

$\Rightarrow { C } _{ 0 }{ -C } _{ 2 }{ +C } _{ 4 }+....={ 2 }^{ \dfrac { n }{ 2 }  }\left( \dfrac { \cos { \dfrac { n\pi  }{ 4 }  } +i\sin { \dfrac { n\pi  }{ 4 } +\cos { \dfrac { n\pi  }{ 4 }  } -i\sin { \dfrac { n\pi  }{ 4 }  }  }  }{ 2 }  \right) $

$\Rightarrow { C } _{ 0 }{ -C } _{ 2 }{ +C } _{ 4 }+....={ 2 }^{ \dfrac { n }{ 2 }  }\cos { \dfrac { n\pi  }{ 4 }  } $

Subtracting (2) & (3), we have
$\Rightarrow 2i\left( { C } _{ 1 }{ -C } _{ 3 }{ +C } _{ 5 }+.... \right) ={ \left( 1+i \right)  }^{ n }-{ \left( 1-i \right)  }^{ n }$

$\Rightarrow { C } _{ 1 }{ -C } _{ 3 }{ +C } _{ 5 }+....=\dfrac { { \left( 1+i \right)  }^{ n }-{ \left( 1-i \right)  }^{ n } }{ 2i } ={ 2 }^{ \dfrac { n }{ 2 }  }\left( \dfrac { { \left( \cos { \dfrac { \pi  }{ 4 }  } +i\sin { \dfrac { \pi  }{ 4 }  }  \right)  }^{ n }-{ \left( \cos { \dfrac { \pi  }{ 4 }  } -i\sin { \dfrac { \pi  }{ 4 }  }  \right)  }^{ n } }{ 2i }  \right) $

$\Rightarrow { C } _{ 1 }{ -C } _{ 3 }{ +C } _{ 5 }+....={ 2 }^{ \dfrac { n }{ 2 }  }\left( \dfrac { \cos { \dfrac { n\pi  }{ 4 }  } +i\sin { \dfrac { n\pi  }{ 4 } -\cos { \dfrac { n\pi  }{ 4 }  } +i\sin { \dfrac { n\pi  }{ 4 }  }  }  }{ 2i }  \right) $

$\Rightarrow { C } _{ 1 }{ -C } _{ 3 }{ +C } _{ 5 }+.....={ 2 }^{ \dfrac { n }{ 2 }  }\sin { \dfrac { n\pi  }{ 4 }  } $        ...(4)

Substitute $x=1$ in equation (1), we get
$\Rightarrow { C } _{ 0 }+{ C } _{ 1 }{ +C } _{ 2 }+{ C } _{ 3 }+{ C } _{ 4 }+.....\quad ={ 2 }^{ n }$          ...(5)

Substitute $x=-1$ in equation (1), we have
$\ \Rightarrow { C } _{ 0 }{ -C } _{ 1 }{ +C } _{ 2 }-{ C } _{ 3 }+{ C } _{ 4 }+.....\quad =0$       ...(6)

Subtracting (5) & (6), we get
$\Rightarrow 2\left( { C } _{ 1 }+{ C } _{ 3 }+{ C } _{ 5 }+..... \right) ={ 2 }^{ n }$
$\Rightarrow { C } _{ 1 }+{ C } _{ 3 }+{ C } _{ 5 }+.....\quad ={ 2 }^{ n-1 }$       ...(7)

Adding (4) & (7), we get
$\Rightarrow 2\left( { C } _{ 1 }+{ C } _{ 5 }+{ C } _{ 9 }+..... \right) ={ 2 }^{ n-1 }+{ 2 }^{ \dfrac { n }{ 2 }  }\sin { \dfrac { n\pi  }{ 4 }  } $

$\Rightarrow { C } _{ 1 }+{ C } _{ 5 }+{ C } _{ 9 }+.....=\dfrac { 1 }{ 2 } \left( { 2 }^{ n-1 }+{ 2 }^{ \dfrac { n }{ 2 }  }\sin { \dfrac { n\pi  }{ 4 }  }  \right) $
Hence, option 'D' is correct.

If $ x+\dfrac{1}{x}=2\cos \theta \   and \ y+\dfrac{1}{y}=2\cos \phi$  then which of the following is not correct?

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

  1. $\displaystyle \frac{x}{y} +\frac{y}{x}=2\cos \left ( \theta -\phi \right )$

  2. $x^{m}y^{n}=\cos \left ( m\theta +n\phi \right )+i\sin \left ( m\theta +n\phi \right )$

  3. $x^{m}y^{n}+x^{-m}y^{-n}=2\cos \left ( m\theta +n\phi \right )$

  4. None of these

Show answer
Correct Option: C
Explanation:

$x+\dfrac { 1 }{ x } =2\cos { \theta  } \quad & \quad y+\dfrac { 1 }{ y } =2\cos { \phi  } \ $

$\Rightarrow x=\cos { \theta  } +i\sin { \theta  } =cis\theta \ \quad & \quad y=\cos { \phi  } +i\sin { \phi  } =cis\phi \ $

$\dfrac { x }{ y } +\dfrac { y }{ x } =\dfrac { cis\theta  }{ cis\phi  } +\dfrac { cis\phi  }{ cis\theta  } =cis\left( \theta -\phi  \right) +cis\left( -\theta +\phi  \right) $

$\therefore \quad \dfrac { x }{ y } +\dfrac { y }{ x } =2\cos { \left( \theta -\phi  \right)  } $

${ x }^{ m }{ y }^{ n }={ \left( cis\theta  \right)  }^{ m }{ \left( cis\phi  \right)  }^{ n }=\left( cism\theta  \right) \left( cisn\phi  \right) $          ...{ De Moivre's Theorem}

$\therefore \quad { x }^{ m }{ y }^{ n }=cis\left( m\theta +n\phi  \right) =\cos { \left( m\theta +n\phi  \right)  } +i\sin { \left( m\theta +n\phi  \right)  } $

${ x }^{ -m }{ y }^{ -n }={ \left( cis\theta  \right)  }^{ -m }{ \left( cis\phi  \right)  }^{ -n }=\left( cis\left( -m\theta  \right)  \right) \left( cis\left( -n\phi  \right)  \right) \ \therefore \quad { x }^{ -m }{ y }^{ -n }=cis\left( -m\theta -n\phi  \right) =\cos { \left( m\theta +n\phi  \right)  } -i\sin { \left( m\theta +n\phi  \right)  } \ $

$\therefore \quad { x }^{ m }{ y }^{ n }+{ x }^{ -m }{ y }^{ -n }=2\cos { \left( m\theta +n\phi  \right)  } $


Let $\mathrm{z}=\cos\theta+\mathrm{i}\sin\theta$. Then the value of $\displaystyle \sum _{\mathrm{m}=1}^{15}{\rm Im}(\mathrm{z}^{2\mathrm{m}-1})$ at $\theta =2^{\mathrm{o}}$ is

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

  1. $\displaystyle \frac{1}{\sin 2^{\mathrm{o}}}$

  2. $\displaystyle \frac{1}{3\sin 2^{\mathrm{o}}}$

  3. $\displaystyle \frac{1}{2\sin 2^{\mathrm{o}}}$

  4. $\displaystyle \frac{1}{4\sin 2^{\mathrm{o}}}$

Show answer
Correct Option: D
Explanation:

$\mathrm{z}=\cos\theta+\mathrm{i}\sin\theta$
$\Rightarrow z^{2m-1} = \cos(2m-1)\theta+i\sin (2m-1)\theta$


Let $X = \displaystyle \sum _{\mathrm{m}=1}^{15}{\rm Im}(\mathrm{z}^{2\mathrm{m}-1})$

$\therefore \mathrm{X}=\sin\theta+\sin 3\theta+\ldots+\sin 29\theta$


$ 2(sin\theta)X=2\sin\theta\sin\theta + 2\sin\theta\sin3\theta...........+2\sin\theta\sin29\theta$
$\Rightarrow  2(\sin\theta)\mathrm{X}=1-\cos 2\theta+\cos 2\theta-\cos 4\theta+\ldots+\cos 28\theta-\cos 30\theta$

$\displaystyle \therefore  \mathrm{X}=\frac{1-\cos 30\theta}{2\sin\theta}=\frac{1}{4\sin 2^{\mathrm{o}}}$

The given table shows the average temperature of some cities of India in a particular month.

City Average maximum temperature Average minimum temperature
New Delhi $39^oC$ $27^o$
Mumbai $36^oC$ $19^oC$
Ahemdabad $37^oC$ $19^oC$
Chennai $42^oC$ $26^oC$
Kolkata $38^oC$ $24^oC$

Which city has the highest difference its average maximum temperature and average minimum temperature?

maths operations adding and subtracting numbers using place value addition & subtraction mental additions and subtractions

  1. Kolkata

  2. Mumbai

  3. Chennai

  4. Ahemdabad

Show answer
Correct Option: D
Explanation:
Highest difference of arg maximum temperature & arg minimum temperature.

New Delhi = $39^{\circ}-27^{\circ}=12^{\circ}C$

Mumbai = $36^{\circ}-19^{\circ}=17^{\circ}C$

Ahmadabad = $37^{\circ}-19^{\circ}=18^{\circ}C$

Chennai = $42^{\circ}-26^{\circ}=16^{\circ}C$

Kolkata = $38^{\circ}-24^{\circ}=14^{\circ}C$

$\therefore $ Highest difference = $18^{\circ}C$ in city Ahmadabad.

Jenny has $4$ red roses and $5$ yellow roses. How many roses does Jenny have in all

maths operations adding and subtracting numbers using place value addition & subtraction mental additions and subtractions

  1. $7$

  2. $4$

  3. $5$

  4. $9$

Show answer
Correct Option: D
Explanation:

$Total >roses =4+5=9$

Shraddha needs $6$ containers which can hold $15600$ oil. Find the capacity of each container.

maths operations adding and subtracting numbers using place value addition & subtraction mental additions and subtractions

  1. $2500$

  2. $2600$

  3. $3200$

  4. $152100$

Show answer
Correct Option: B
Explanation:

$6$ containers can hold $15600$ amount of oil

capacity of each container is $\dfrac{15600}{6}=2600$