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

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

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

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

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

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

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

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

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

$\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

Dividing and multiplying by $\sqrt{13}$
$=\sqrt{13}[(\dfrac{5}{13}-\dfrac{12i}{13})^{\dfrac{1}{2}}+(-\dfrac{5}{13}-\dfrac{12i}{13})^{\dfrac{1}{2}}]$
$=\sqrt{13}[e^{i\dfrac{-\theta}{2}}+e^{i\dfrac{\theta-\pi}{2}}]$
$=\sqrt{13}[cos\dfrac{\theta}{2}-isin\dfrac{\theta}{2}+sin\dfrac{\theta}{2}-icos\frac{\theta}{2}]$
$=\sqrt{13}[cos\dfrac{\theta}{2}+sin\dfrac{\theta}{2}-i(sin\dfrac{\theta}{2}+cos\dfrac{\theta}{2})]$
$=z$
Here $\theta=sin^{-1}(\dfrac{12}{13})$
Hence
$|Re(z)|=|Im(z)|$
Hence argument of $Z$ is in the form of $\dfrac{2n-1(\pi)}{4}$ $n\epsilon::Integers$

We have $\displaystyle { \left( \frac { 1+i }{ \sqrt { 2 }  }  \right)  }^{ 8 }+{ \left( \frac { 1-i }{ \sqrt { 2 }  }  \right)  }^{ 8 }$

${ z }^{ 10 }-{ z }^{ 5 }+1=0$

Let ${ z }^{ 5 }=w$
$\Rightarrow { w }^{ 2 }-w+1=0$

$\Rightarrow w=\frac { 1\pm \sqrt { 3 }  }{ 2 } =\cos { \frac { \Pi  }{ 3 }  } \pm \sin { \frac { \Pi  }{ 3 }  } =cis\left( \pm \frac { \Pi  }{ 3 }  \right) $


$\Rightarrow { z }^{ 5 }=cis\left( \pm \frac { \Pi  }{ 3 }  \right) $

Case 1:
${ z }^{ 5 }=cis\left( \frac { \Pi  }{ 3 }  \right) $

$\Rightarrow z={ \left( cis\left( \frac { \Pi  }{ 3 }  \right)  \right)  }^{ \frac { 1 }{ 5 }  }=cis\left( \frac { 2k\Pi +\Pi  }{ 15 }  \right) \ $        ...{De Moivre's Theorem}

Where k=0,1,2,3,4.
Therefore number of solutions are 5.

Case 2:
${ z }^{ 5 }=cis\left( -\frac { \Pi  }{ 3 }  \right) $

$\Rightarrow z={ \left( cis\left( -\frac { \Pi  }{ 3 }  \right)  \right)  }^{ \frac { 1 }{ 5 }  }=cis\left( \frac { 2k\Pi -\Pi  }{ 15 }  \right) $       ...{De Moivre's Theorem}

Where k=0,1,2,3,4.
Therefore number of solutions are 5.

From case 1 & case 2 total number of solutions of equation ${ z }^{ 10 }-{ z }^{ 5 }+1=0$ are 10.

Ans: D

$z=1+\cos { \frac { 2\pi  }{ 3 }  } +i\sin { \frac { 2\pi  }{ 3 }  } =2\cos ^{ 2 }{ \frac { \pi  }{ 3 }  } +2i\sin { \frac { \pi  }{ 3 } \cos { \frac { \pi  }{ 3 }  }  } $

$\displaystyle \Rightarrow z=2\cos { \frac { \pi  }{ 3 }  } \left( \cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  }  \right) =\cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  } $

$\displaystyle \Rightarrow { z }^{ 5 }={ \left( \cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  }  \right)  }^{ 5 }=\cos { \frac { 5\pi  }{ 3 }  } +i\sin { \frac { 5\pi  }{ 3 }  } $        ...{De Moivre's Theorem}
 
$\displaystyle \Rightarrow { z }^{ 5 }=\frac { 1-i\sqrt { 3 }  }{ 2 } $

$\displaystyle \therefore \quad Re\left( { z }^{ 5 } \right) =\frac { 1 }{ 2 } \quad & \quad Im\left( { z }^{ 5 } \right) =\frac { -\sqrt { 3 }  }{ 2 } $
Hence, option B is correct.

We know, $\displaystyle \alpha = \cos \theta +i\sin \theta , \beta = \cos \theta -i\sin \theta $
$\displaystyle \alpha ^{n}= \cos n\theta +i\sin n\theta ,$
$\displaystyle \beta ^{n}= \cos n\theta -i\sin n\theta $
$\displaystyle S= 2\cos n\theta , P= 1 \therefore x^{2}-Sx+P= 0$
or $\displaystyle x^{2}-2\cos n\theta x+1= 0$ is the required equation.

Ans: C

As we know that,
$\dfrac { \sqrt { 3 }  }{ 2 } +\dfrac { i }{ 2 } =\cos { \dfrac { \pi  }{ 6 }  } +i\sin { \dfrac { \pi  }{ 6 }  } $

and $\dfrac { \sqrt { 3 }  }{ 2 } -\dfrac { i }{ 2 } =\cos { \dfrac { \pi  }{ 6 }  } -i\sin { \dfrac { \pi  }{ 6 }  } $

$z=\left( \dfrac { \sqrt { 3 }  }{ 2 } +\dfrac { i }{ 2 }  \right) ^{ 2009 }+\left( \dfrac { \sqrt { 3 }  }{ 2 } -\dfrac { i }{ 2 }  \right) ^{ 2009 }$

$\Rightarrow z=\left( \cos { \dfrac { \pi  }{ 6 }  } +i\sin { \dfrac { \pi  }{ 6 }  }  \right) ^{ 2009 }+\left( \cos { \dfrac { \pi  }{ 6 }  } -i\sin { \dfrac { \pi  }{ 6 }  }  \right) ^{ 2009 }$         ......{ De Moivre's Theorem}

$\Rightarrow z=\cos { \dfrac { 2009\pi  }{ 6 }  } +i\sin { \dfrac { 2009\pi  }{ 6 }  } +\cos { \dfrac { 2009\pi  }{ 6 }  } -i\sin { \dfrac { 2009\pi  }{ 6 }  } $

$\Rightarrow z=2\cos { \dfrac { 2009\pi  }{ 6 }  } $

$z=\dfrac{\sqrt{3}+i}{2}=\cos \dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}=cis\dfrac{\pi}{6}$

$z={ \left( \sin { \frac { \pi  }{ 8 } +i } \cos { \frac { \pi  }{ 8 }  }  \right)  }^{ 8 }={ \left[ i\left( \cos { \frac { \pi  }{ 8 } -i\sin { \frac { \pi  }{ 8 }  }  }  \right)  \right]  }]^8$
     ...{$\because \quad { i }^{ 8 }=1$}
$\Rightarrow z=\cos { \pi -i\sin { \pi  }  } =-1$        ...{De Moivre's Theorem}
Hence, option 'A' is correct.

By Binomial Theorem
${ \left( 1+z \right)  }^{ n }={ C } _{ 0 }+{ C } _{ 1 }z+{ C } _{ 2 }{ z }^{ 2 }+{ C } _{ 3 }{ z }^{ \ 3 }+....+{ C } _{ n }{ z }^{ n }=\sum _{ r=0 }^{ n }{ { C } _{ r }{ z }^{ r } } $      ...(1)

Substituting $z=\cos { 2\theta  } +i\sin { 2\theta  }  $ in eq. (1), we get

${ \left( 1+\cos { 2\theta  } +i\sin { 2\theta  }  \right)  }^{ n }=\sum _{ r=0 }^{ n }{ { C } _{ r }\left( \cos { 2\theta  } +i\sin { 2\theta  }  \right) ^{ r } } $

$\Rightarrow \sum _{ r=0 }^{ n }{ { C } _{ r }\left( \cos { 2r\theta  } +i\sin { 2r\theta  }  \right)  }$      ...{De Moivre's Theorem}

$\Rightarrow { \left[ 2\cos { \theta  } \left( \cos { \theta  } +i\sin { \theta  }  \right)  \right]  }^{ n }=\sum _{ r=0 }^{ n }{ { C } _{ r }\cos { 2r\theta  }  } +i\sum _{ r=0 }^{ n }{ { C } _{ r }\sin { 2r\theta  }  } $

$\Rightarrow \sum _{ r=0 }^{ n }{ { C } _{ r }\cos { 2r\theta  }  } +i\sum _{ r=0 }^{ n }{ { C } _{ r }\sin { 2r\theta  }  } ={ 2 }^{ n }\cos ^{ n }{ \theta  } { \left( \cos { n\theta  } +i\sin { n\theta  }  \right)  }$         ...{De Moivre's Theorem}

$\Rightarrow \sum _{ r=0 }^{ n }{ { C } _{ r }\cos { 2r\theta  }  } +i\left( \sum _{ r=0 }^{ n }{ { C } _{ r }\sin { 2r\theta  }  }  \right) ={ 2 }^{ n }\cos ^{ n }{ \theta  } \cos { n\theta  } +i\left( { 2 }^{ n }\cos ^{ n }{ \theta  } \sin { n\theta  }  \right) $

On comparing real and Imaginary parts, we get
$\sum _{ r=0 }^{ n }{ { C } _{ r }\cos { 2r\theta  }  } ={ 2 }^{ n }\cos ^{ n }{ \theta  } \cos { n\theta  } \quad & \quad \sum _{ r=0 }^{ n }{ { C } _{ r }\sin { 2r\theta  }  } ={ 2 }^{ n }\cos ^{ n }{ \theta  } \sin { n\theta  } $
Hence, option 'A' and 'C' are correct.

Using De-Moivre's Theorem, the given expression
$\displaystyle = \frac{\left ( \cos \theta -i\sin \theta  \right )^{14}\left ( \cos \theta +i\sin \theta  \right )^{-15}}{\left ( \cos \theta +i\sin \theta  \right )^{48}\left ( \cos \theta +i\sin \theta  \right )^{-30}}$
$\displaystyle=\frac{\left ( e^{i\theta } \right )^{-29}}{\left ( e^{i\theta } \right )^{18}}=\left ( e^{i\theta } \right )^{-47}$
$\displaystyle=\left ( \cos \theta +i\sin \theta  \right )-^{47}=\cos 47\theta -\sin47\theta.$ 

Ans: B

As we know that,
$\dfrac { \sqrt { 3 }  }{ 2 } +\dfrac { i }{ 2 } =\cos { \dfrac { \pi  }{ 6 }  } +i\sin { \dfrac { \pi  }{ 6 }  } $
and $\dfrac { \sqrt { 3 }  }{ 2 } -\dfrac { i }{ 2 } =\cos { \dfrac { \pi  }{ 6 }  } -i\sin { \dfrac { \pi  }{ 6 }  } $

$z=\left( \dfrac { \sqrt { 3 }  }{ 2 } +\dfrac { i }{ 2 }  \right) ^{ 5 }+\left( \dfrac { \sqrt { 3 }  }{ 2 } -\dfrac { i }{ 2 }  \right) ^{ 5 }$

$\Rightarrow z=\left( \cos { \dfrac { \pi  }{ 6 }  } +i\sin { \dfrac { \pi  }{ 6 }  }  \right) ^{ 5 }+\left( \cos { \dfrac { \pi  }{ 6 }  } -i\sin { \dfrac { \pi  }{ 6 }  }  \right) ^{ 5 }$         ......{ De Moivre's Theorem}

$\Rightarrow z=\cos { \dfrac { 5\pi  }{ 6 }  } +i\sin { \dfrac { 5\pi  }{ 6 }  } +\cos { \dfrac { 5\pi  }{ 6 }  } -i\sin { \dfrac { 5\pi  }{ 6 }  } $

$\Rightarrow z=-\dfrac { \sqrt { 3 }  }{ 2 } $
Therefore, $Im(z)=0$

Ans: B

$z=(e^{3-\tfrac{i\pi}{4}})^{3}$
$=(e^{3}.e^{-\tfrac{i\pi}{4}})^{3}$
$=e^{9}.e^{-\tfrac{3i\pi}{4}}$
$=|z|e^{i\arg(z)}$ 
By comparing RHS and LHS we get
$|z|=e^{9}$ and $\arg(z)=\dfrac{-3\pi}{4}$.

$\displaystyle\alpha =\cos { \left( \frac { 8\pi  }{ 11 }  \right)  } +i\sin { \left( \frac { 8\pi  }{ 11 }  \right)  } $

${ u }^{ 2 }-2u+2=0$
$\Longrightarrow \quad u=1\pm i$
So,$\alpha =1+i\quad and\quad \beta =1-i$
Now given that,$x=\cot { \theta  } -1$
so,$\displaystyle \frac { { (x+\alpha ) }^{ n }-{ (x+\beta ) }^{ n } }{ \alpha -\beta  } =\frac { { (\cot { \theta  } -1+1+i) }^{ n }-{ (\cot { \theta  } -1 }+1-i)^{ n } }{ 2i } \ \ $
$=\displaystyle \frac { { (\cot { \theta  } +i) }^{ n }-(\cot { \theta  } -i)^{ n } }{ 2i } =\frac { { (\cos { \theta  } +i\sin { \theta  } ) }^{ n }-{ (\cos { \theta  } -\sin { \theta  } ) }^{ n } }{ ({ \sin { \theta  }  })^{ n }(2i) } \ \ $
$=\displaystyle \frac { { e }^{ (in\theta ) }-{ e }^{ -(in\theta ) } }{ ({ \sin { \theta ) }  }^{ n }2i } \ \ $
=$\displaystyle \frac { (\cos { (n\theta ) } +i\sin { (n\theta )) } -(\cos { (n\theta ) } -i\sin { (n\theta )) }  }{ ({ \sin { \theta ) }  }^{ n }2i } =\frac { \sin { (n\theta ) }  }{ { (\sin { \theta ) }  }^{ n } } \ \ $

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.

$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)  } $

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