Question

Important Question for jee mains.
mathamatics :- Practice set 1​

Answer 1

$\large\underline{\sf{Solution-}}$

We know,

From cube roots of unity,

$\rm :\longmapsto\: \omega \: = \dfrac{ - 1 + \sqrt{3} i}{2}$

and

$\rm :\longmapsto\:1 + \omega \: + {\omega }^{2} = 0$

and

$\rm :\longmapsto\: {\omega }^{3} = 1$

and

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:1 + {\omega }^{n} + {\omega }^{2n} -\begin{cases} &\sf{3 \: if \: n \: is \: multiple \: of \: 3} \\ &\sf{0 \: if \: n \: is \: not \: multiple \: of \: 3} \end{cases}\end{gathered}\end{gathered}$

Let's solve the problem now!!

It is given that

$\rm :\longmapsto\:z = - 2 + 2 \sqrt{3}i$

$\rm :\longmapsto\:z = 2(- 1 + \sqrt{3}i)$

$\rm :\longmapsto\:z = 2 \times 2\omega$

$\bf\implies \:z = 4\omega$

Now,

Consider,

$\rm :\longmapsto\: {z}^{2n} + {2}^{2n} {z}^{n} + {2}^{4n}$

On substituting the value of z, evaluated above, we get

$\rm \: = \: {(4\omega )}^{2n} + {2}^{2n} {(4\omega )}^{n} + {2}^{4n}$

$\rm \:  =  \:  \: {16}^{n} {\omega }^{2n} + {4}^{n} {4}^{n} {\omega }^{n} + {16}^{n}$

$\rm \:  =  \:  \: {16}^{n} {\omega }^{2n} + {16}^{n} {\omega }^{n} + {16}^{n}$

$\rm \:  =  \:  \: {16}^{n} ({\omega }^{2n} + {\omega }^{n} + 1)$

$\rm \:  =  \:  \: {4}^{2n} \times 3 \: if \: n \: is \: multiple \: of \: 3$

Hence,

$\boxed{\rm \:{z}^{2n} +{2}^{2n} {z}^{n} +{2}^{4n}  ={4}^{2n} \times 3 \: if \: n \: is \: multiple \: of \: 3}$

Therefore,

• Option (c) is correct