Question

The equation $x^2+2x+2=0$ has roots $\alpha$ and $\beta$. Then value of ${\alpha }^{15}+{\beta }^{15}$ is

• A. $512$
• B. $256$
• C. $-512$
• D. $-256$

Answer 1

The correct option is D

$-256$

Explanation for the correct option:

Step 1: Find the roots of the given equation.

A quadratic equation $x^2+2x+2=0$ is given.

Compute the roots using the quadratic formula.

$$\begin{gathered} x=\frac{-2\pm \sqrt{2^2-4·1·2}}{2·1} \\ \Rightarrow x=\frac{-2\pm \sqrt{4-8}}{2} \\ \Rightarrow x=\frac{-2\pm \sqrt{-4}}{2} \\ \Rightarrow x=\frac{-2\pm 2i}{2} \\ \Rightarrow x=\left\{-1-i,-1+i\right\} \end{gathered}$$

The roots of the quadratic equation $x^2+2x+2=0$ are $-1-i$ and $-1+i$.

Step 2: Find the value of the given expression.

It is given that, the roots of $x^2+2x+2=0$ are $\alpha ,\beta$.

Therefore, $\alpha =-1-i$ and $\beta =-1+i$.

According to Euler's representation of the complex number, a complex number $z=r(\cos \theta +i\sin \theta )$ can be represented as $z=r·e^{i\theta }$.

Where, $e^{i\theta }=\cos \theta +i\sin \theta$.

We can rewrite $\alpha$ as follows:

$$\begin{gathered} \alpha =\sqrt{2}\left(-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\right) \\ \Rightarrow \alpha =\sqrt{2}\left[\cos \left(-\frac{3\mathrm{\pi }}{4}\right)+i\sin \left(-\frac{3\mathrm{\pi }}{4}\right)\right] \end{gathered}$$

Use Euler's representation.

$\alpha =\sqrt{2}{e}^{i\left(\frac{-3\mathrm{\pi }}{4}\right)}$

Similarly, $\beta =\sqrt{2}{e}^{i\left(\frac{3\mathrm{\pi }}{4}\right)}$

Now, evaluate the given expression as follows:

$$\begin{gathered} {\alpha }^{15}+{\beta }^{15}={\left[\sqrt{2}{e}^{i\left(\frac{-3\mathrm{\pi }}{4}\right)}\right]}^{15}+{\left[\sqrt{2}{e}^{i\left(\frac{3\mathrm{\pi }}{4}\right)}\right]}^{15} \\ \Rightarrow {\alpha }^{15}+{\beta }^{15}={\left(\sqrt{2}\right)}^{15}\left[e^{i\left(\frac{-45\mathrm{\pi }}{4}\right)}+e^{i\left(\frac{45\mathrm{\pi }}{4}\right)}\right] \\ \Rightarrow {\alpha }^{15}+{\beta }^{15}={\left(\sqrt{2}\right)}^{15}\left[\cos \left(\frac{-45\mathrm{\pi }}{4}\right)+i\sin \left(\frac{-45\mathrm{\pi }}{4}\right)+\cos \left(\frac{45\mathrm{\pi }}{4}\right)+i\sin \left(\frac{45\mathrm{\pi }}{4}\right)\right] \\ \Rightarrow {\alpha }^{15}+{\beta }^{15}={\left(\sqrt{2}\right)}^{15}\left[\cos \left(\frac{-45\mathrm{\pi }}{4}\right)-i\sin \left(\frac{45\mathrm{\pi }}{4}\right)+\cos \left(\frac{45\mathrm{\pi }}{4}\right)+i\sin \left(\frac{45\mathrm{\pi }}{4}\right)\right] \\ \Rightarrow {\alpha }^{15}+{\beta }^{15}={\left(\sqrt{2}\right)}^{15}\left[2\cos \left(\frac{45\mathrm{\pi }}{4}\right)\right] \\ \Rightarrow {\alpha }^{15}+{\beta }^{15}={\left(\sqrt{2}\right)}^{15}\left[-\frac{2}{\sqrt{2}}\right] \\ \Rightarrow {\alpha }^{15}+{\beta }^{15}=-{\left(2\right)}^8 \\ \Rightarrow {\alpha }^{15}+{\beta }^{15}=-256 \end{gathered}$$

Therefore, the value of ${\alpha }^{15}+{\beta }^{15}$ is $-256$.

Hence, the option $D$ is correct.