Tag: complex numbers

Changing the above expression to Eular's form, we get
$e^{i\theta}e^{2i\theta}e^{3i\theta}...e^{in\theta})=1$
$e^{i(\theta+2\theta+3\theta+...n\theta}=1$
$e^{i\cfrac{n(n+1)}{2}\theta}=e^{2m\pi}$
Therefore, simplifying we get
$\dfrac{n(n+1)}{2}\theta=2m\pi$
$\theta=\dfrac{4m\pi}{n(n+1)}$

Let
$z=(cos\theta+isin\theta)^{3}$
Taking the fifth root, we get
$z^{\dfrac{1}{5}}=(cos\theta+isin\theta)^{\dfrac{3}{5}}=x$
Now there will be $5$, corresponding values of $x$.
Product if all the $5$ values will be
$x^{5}$
$=(cos\theta+isin\theta)^{3}$
$=cos3\theta+isin3\theta$ ... using De-Moivre's rule.
Now consider $x^{n}=1$
Hence if $n$ is odd, the nth roots of unity will be
$1,a _{1},\overline{a _{1}},a _{2},\overline{a _{2}}....$
Now $a _{1}.\overline{a _{1}}=|a _{1}|^{2}=1$
$a _{2}.\overline{a _{2}}=|a _{2}|^{2}=1$
:
:
Hence one root will be one, and the rest $n-1$ roots will occur in pair with its conjugate.
Hence product will be $1$.
Substituting, $n=5$, we get the roots as
$1,a _{1},\overline{a _{1}},a _{2},\overline{a _{2}}$
Hence product if all the roots will be $1$. And sum of all the roots will be $0$.
Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

$cis\left( \theta  \right) =\cos { \theta  } +i\sin { \theta  } $
De Moivre's Theorem for fractional power:
${ \left( cis\theta  \right)  }^{ \frac { 1 }{ n }  }=cis\left( \frac { 2k\Pi +\theta  }{ n }  \right) $

${ \left( x-3 \right)  }^{ 3 }+1=0$
$\Longrightarrow x=3+{ \left( cis\left( \Pi  \right)  \right)  }^{ \frac { 1 }{ 3 }  }$
$x=3+{ \left( cis\left( \frac { 2k\Pi +\Pi  }{ 3 }  \right)  \right)  }$      ...{De Moivre's Theorem}
Where, $k=0,1,2$
for  $k=0$,
$x _{ 1 }=3+cis\left( \frac { \Pi  }{ 3 }  \right)$ 

for $k=1$,
$x _{ 2 }=3+cis\left( \Pi  \right) $

for $k=2,$
$x _{ 3 }=3+cis\left( \frac { 5\Pi  }{ 3 }  \right) $

$\Longrightarrow { x } _{ 1 }+{ x } _{ 3 }=6+cis\left( \frac { \Pi  }{ 3 }  \right) +cis\left( \frac { 5\Pi  }{ 3 }  \right) \ \Longrightarrow { x } _{ 1 }+{ x } _{ 3 }=7$
 
Ans: D

$\displaystyle { \left( 2+z \right)  }^{ 6 }{ +\left( 2-z \right)  }^{ 6 }=0\quad \quad & \quad w=\frac { 2+z }{ 2-z } $

$\displaystyle \Rightarrow { \left( \frac { 2+z }{ 2-z }  \right)  }^{ 6 }=-1\ \Rightarrow { w }^{ 6 }=-1$

$\displaystyle { \therefore \quad w=\left( -1 \right)  }^{ \frac { 1 }{ 6 }  }$

$\displaystyle \because \quad \frac { 2+z }{ 2-z } =w\ \Rightarrow 2\left( w-1 \right) =z\left( w+1 \right) $

$\displaystyle \therefore \quad z=\frac { 2\left( w-1 \right)  }{ w+1 } $

$\displaystyle { \because \quad w=\left( -1 \right)  }^{ \frac { 1 }{ 6 }  }$

$\displaystyle w={ \left( \cos { \pi  } +i\sin { \pi  }  \right)  }^{ \frac { 1 }{ 6 }  }=\cos { \left( \frac { 2p\pi +\pi  }{ 6 }  \right) +i } \sin { \left( \frac { 2p\pi +\pi  }{ 6 }  \right)  } $       ..{De Moivre's Theorem}

Where$ p=0,1,2,3,4,5.$

$\displaystyle \Rightarrow w={ e }^{ i\frac { \left( 2p+1 \right) \pi  }{ 6 }  }$
Hence, option 'D' is correct.

$\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

$\displaystyle { z }^{ n }={ \left( z+1 \right)  }^{ n }$   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

$(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)

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

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.