Theory of Equations 1 Question 11
12. Let $\alpha$ and $\beta$ be two roots of the equation $x^{2}+2 x+2=0$, then $\alpha^{15}+\beta^{15}$ is equal to
(2019 Main, 9 Jan I)
(a) 256
$t$ (b) 512
(c) -256
(d) -512
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Answer:
Correct Answer: 12. (c)
Solution:
- We have, $x^{2}+2 x+2=0$
$\Rightarrow x=\frac{-2 \pm \sqrt{4-8}}{2} \quad\left[\because\right.$ roots of $a x^{2}+b x+c=0$ are given by $\left.x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\right]$
$\Rightarrow \quad x=-1 \pm i$
Let $\alpha=-1+i$ and $\beta=-1-i$.
Then, $\alpha^{15}+\beta^{15}=(-1+i)^{15}+(-1-i)^{15}$
$ \begin{aligned} &=-\left[(1-i)^{15}+(1+i)^{15}\right] \\ &=- \sqrt{2} \frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}+\sqrt{2} \frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}} \\ &=-\sqrt{2} \cos \frac{\pi}{4}-i \sin \frac{\pi}{4} \\ &+\sqrt{2} \cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \\ &=-(\sqrt{2})^{15} \cos \frac{15 \pi}{4}-i \sin \frac{15 \pi}{4}+\cos \frac{15 \pi}{4}+i \sin \frac{15 \pi}{4} \end{aligned} $
[using De’ Moivre’s theorem $\left.(\cos \theta \pm i \sin \theta)^{n}=\cos n \theta \pm i \sin n \theta, n \in Z\right]$
$=-(\sqrt{2})^{15} 2 \cos \frac{15 \pi}{4}=-(\sqrt{2})^{15} 2 \times \frac{1}{\sqrt{2}}$
$ \because \cos \frac{15 \pi}{4}=\cos 4 \pi-\frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} $
$=-(\sqrt{2})^{16}=-2^{8}=-256$.
Alternate Method
$ \begin{aligned} \alpha^{15}+\beta^{15} & =(-1+i)^{15}+(-1-i)^{15} \\ & =-\left[(1-i)^{15}+(1+i)^{15}\right] \\ & =-\frac{(1-i)^{16}}{1-i}+\frac{(1+i)^{16}}{1+i} \\ & =-\frac{\left[(1-i)^{2}\right]^{8}}{1-i}+\frac{\left[(1+i)^{2}\right]^{8}}{1+i} \end{aligned} $
$ \begin{aligned} & =-\frac{\left[1+i^{2}-2 i\right]^{8}}{1-i}+\frac{\left[1+i^{2}+2 i\right]^{8}}{1+i} \\ & =-\frac{(-2 i)^{8}}{1-i}+\frac{(2 i)^{8}}{1+i} \\ & =-2^{8} \frac{1}{1-i}+\frac{1}{1+i} \quad\left[\because i^{4 n}=1, n \in Z\right] \\ & =-256 \frac{2}{1-(i)^{2}}=-256 \frac{2}{2}=-256 \end{aligned} $
