Question

If z = (√3/2 + i/2)⁵ + (√3/2 - i/2)⁵, then?
[A] Im(z) < 0
[B] Re(z) > 0
[C] Im(z) = 0
[D] Im(z) > 0

Answer 1

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

Given that,

$\rm \: z = {\bigg(\dfrac{ \sqrt{3} }{2} + \dfrac{i}{2} \bigg) }^{5} + {\bigg(\dfrac{ \sqrt{3} }{2} - \dfrac{i}{2} \bigg) }^{5}$

can be rewritten as

$\rm \: z = {\bigg(\dfrac{ \sqrt{3} + i}{2} \bigg) }^{5} + {\bigg(\dfrac{ \sqrt{3} - i}{2}\bigg) }^{5} - - - - (1)$

We know,

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

So,

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

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

$\rm\implies \:\dfrac{i + \sqrt{3} }{2} \: = - \: i\omega - - - (2)$

Also, we know that

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

$\rm \: i\omega^{2} = \dfrac{ - i - {i}^{2} \sqrt{3} }{2}$

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

$\rm\implies \:\dfrac{\sqrt{3} - i }{2} \: = \: i\omega^{2} - - - (3)$

So, on substituting equation (2) and (3) in (1), we get

$\rm \: z = {\bigg(\dfrac{ \sqrt{3} }{2} + \dfrac{i}{2} \bigg) }^{5} + {\bigg(\dfrac{ \sqrt{3} }{2} - \dfrac{i}{2} \bigg) }^{5}$

$\rm \: = \: {( - i\omega )}^{5} + {(i {\omega }^{2}) }^{5}$

$\rm \: = \: - {i}^{5} {\omega }^{5} + \:{i}^{5} {\omega }^{10}$

$\rm \: = \: - \: {i}^{5}( {\omega }^{5} - {\omega }^{10})$

$\rm \: = \: {i}^{4} \: i \: ( {\omega }^{3} \: {\omega }^{2} - {( {\omega }^{3}) }^{3} \: \omega )$

We know,

$\boxed{\tt{ \: \: {i}^{4} = 1 \: }}$

$\boxed{\tt{ \: \: {\omega }^{3} \: = \: 1 \: \: }}$

So, using this we get

$\rm \: = \: - \: i \: ( {\omega }^{2} - \omega )$

On substituting the values, we get

$\rm \: = - \:\sqrt{3}$

$\rm \: = \: -\sqrt{3} + 0i$

$\rm\implies \:Re(z) < 0 \: Im(z) = 0$

So, Option [C] is correct.

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Additional Information :-

Cube Roots of unity

$\rm \: x = {\bigg(1 \bigg) }^{\dfrac{1}{3} }$

On cubing both sides, we get

$\rm \: {x}^{3} = 1$

$\rm \: {x}^{3} - 1 = 0$

$\rm \: (x - 1)( {x}^{2} + x + 1) = 0$

$\rm\implies \:x = 1$

or

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

$\rm \: x = \dfrac{ - 1 \: \pm \: \sqrt{ {1}^{2} - 4 \times 1 \times 1} }{2 \times 1}$

$\rm \: x = \dfrac{ - 1 \: \pm \: \sqrt{ 1 - 4} }{2 }$

$\rm \: x = \dfrac{ - 1 \: \pm \: \sqrt{ - 3} }{2 }$

$\rm \: x = \dfrac{ - 1 \: \pm \: i\sqrt{3} }{2 }$

$\rm \: x = \dfrac{ - 1 \: + \: i\sqrt{3} }{2 } \: \: or \: \: \dfrac{ - 1 \: - \: i\sqrt{3} }{2 }$

$\rm\implies \:x = \omega \: \: or \: \: {\omega }^{2}$

where,

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

and

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