Q19 / 108 · NTA Abhyas (JEE Main) 2020 · Mathematics · Complex Number

Let $α$ and $β$ be the roots of the equation $x^{2} + 2 x + 2 = 0,$ then $α^{15} + β^{15}$ is equal to

A
$- 512$
B
$128$
C
$512$
Correct answer
D
$- 256$

Solution

Let the two roots of

x^{2} + 2 x + 2 = 0

be

α

and

β

\Rightarrow α, β = \frac{- 2 \pm \sqrt{4 - 8}}{2} = - 1 \pm i

\Rightarrow α, β = \sqrt{2} (\frac{- 1}{\sqrt{2}} \pm i \frac{1}{\sqrt{2}})

= \sqrt{2} (c o s \frac{3 π}{4} \pm i s i n \frac{3 π}{4})

\Rightarrow α^{15} + β^{15} = {[\sqrt{2} (c o s \frac{3 π}{4} + i s i n \frac{3 π}{4})]}^{15} + {[\sqrt{2} (c o s \frac{3 π}{4} - i s i n \frac{3 π}{4})]}^{15}

= 2^{\frac{15}{2}} [(c o s \frac{45 π}{4} + i s i n \frac{45 π}{4}) + (c o s \frac{45 π}{4} - i s i n \frac{45 π}{4})]

= {2.2}^{\frac{15}{2}} . c o s \frac{45 π}{4} = 2^{\frac{17}{2}} . c o s \frac{5 π}{4} = - 2^{\frac{17}{2}} . \frac{1}{\sqrt{2}}

= - 256

Hence, the correct answer is option: 'D'.

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