Key Concepts and Formulas

• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Polar Form of Complex Numbers: A complex number $z = x + iy$ can be represented in polar form as $z = r(\cos \theta + i \sin \theta)$, where $r = |z| = \sqrt{x^2 + y^2}$ and $\theta = \arg(z) = \tan^{-1}(y/x)$, adjusted for the correct quadrant.
• De Moivre's Theorem: For any complex number $z = r(\cos \theta + i \sin \theta)$ and any integer $n$, $z^n = r^n (\cos(n\theta) + i \sin(n\theta))$.

Step-by-Step Solution

Step 1: Finding the Roots of the Quadratic Equation

We need to find the roots of the given quadratic equation $x^2 + 2x + 2 = 0$. This is done using the quadratic formula to determine the values of $\alpha$ and $\beta$.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 2$, and $c = 2$. Substituting these values, we get: $x = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)}$ $x = \frac{-2 \pm \sqrt{4 - 8}}{2}$ $x = \frac{-2 \pm \sqrt{-4}}{2}$ $x = \frac{-2 \pm 2i}{2}$ $x = -1 \pm i$ So the roots are $\alpha = -1 + i$ and $\beta = -1 - i$.

Step 2: Converting Roots to Polar Form

To apply De Moivre's Theorem, we convert the roots $\alpha$ and $\beta$ to polar form, $r(\cos \theta + i \sin \theta)$. This simplifies the process of raising these complex numbers to a high power.

For $\alpha = -1 + i$: The modulus is $r = \sqrt{(-1)^2 + (1)^2} = \sqrt{2}$. The argument $\theta$ is such that $\tan \theta = \frac{1}{-1} = -1$. Since the complex number is in the second quadrant, $\theta = \frac{3\pi}{4}$. Therefore, $\alpha = \sqrt{2} \left( \cos \left(\frac{3\pi}{4}\right) + i \sin \left(\frac{3\pi}{4}\right) \right)$.

For $\beta = -1 - i$: The modulus is $r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$. The argument $\theta$ is such that $\tan \theta = \frac{-1}{-1} = 1$. Since the complex number is in the third quadrant, $\theta = -\frac{3\pi}{4}$. Therefore, $\beta = \sqrt{2} \left( \cos \left(-\frac{3\pi}{4}\right) + i \sin \left(-\frac{3\pi}{4}\right) \right) = \sqrt{2} \left( \cos \left(\frac{3\pi}{4}\right) - i \sin \left(\frac{3\pi}{4}\right) \right)$.

Step 3: Applying De Moivre's Theorem

Now, we apply De Moivre's Theorem to find $\alpha^{15}$ and $\beta^{15}$. This transforms the problem into manipulating trigonometric functions.

$\alpha^{15} = \left(\sqrt{2} \left( \cos \left(\frac{3\pi}{4}\right) + i \sin \left(\frac{3\pi}{4}\right) \right)\right)^{15}$ $\alpha^{15} = (\sqrt{2})^{15} \left( \cos \left(\frac{45\pi}{4}\right) + i \sin \left(\frac{45\pi}{4}\right) \right)$ $\alpha^{15} = 2^{15/2} \left( \cos \left(\frac{45\pi}{4}\right) + i \sin \left(\frac{45\pi}{4}\right) \right)$

$\beta^{15} = \left(\sqrt{2} \left( \cos \left(-\frac{3\pi}{4}\right) + i \sin \left(-\frac{3\pi}{4}\right) \right)\right)^{15}$ $\beta^{15} = (\sqrt{2})^{15} \left( \cos \left(-\frac{45\pi}{4}\right) + i \sin \left(-\frac{45\pi}{4}\right) \right)$ $\beta^{15} = 2^{15/2} \left( \cos \left(-\frac{45\pi}{4}\right) + i \sin \left(-\frac{45\pi}{4}\right) \right) = 2^{15/2} \left( \cos \left(\frac{45\pi}{4}\right) - i \sin \left(\frac{45\pi}{4}\right) \right)$

Step 4: Calculating $\alpha^{15} + \beta^{15}$

We sum the expressions for $\alpha^{15}$ and $\beta^{15}$. The imaginary terms will cancel out because $\alpha$ and $\beta$ are complex conjugates.

$\alpha^{15} + \beta^{15} = 2^{15/2} \left( \cos \left(\frac{45\pi}{4}\right) + i \sin \left(\frac{45\pi}{4}\right) \right) + 2^{15/2} \left( \cos \left(\frac{45\pi}{4}\right) - i \sin \left(\frac{45\pi}{4}\right) \right)$ $\alpha^{15} + \beta^{15} = 2^{15/2} \cdot 2 \cos \left(\frac{45\pi}{4}\right)$

Step 5: Simplifying the Trigonometric Expression and Final Calculation

We simplify the angle in the cosine function and calculate the final numerical value.

We have $\frac{45\pi}{4} = \frac{44\pi}{4} + \frac{\pi}{4} = 11\pi + \frac{\pi}{4}$. Since $\cos(11\pi + \frac{\pi}{4}) = \cos(\pi + \frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$, $\alpha^{15} + \beta^{15} = 2^{15/2} \cdot 2 \cdot \left(-\frac{1}{\sqrt{2}}\right) = 2^{15/2} \cdot 2^1 \cdot (-2^{-1/2}) = -2^{\frac{15}{2} + 1 - \frac{1}{2}} = -2^{\frac{16}{2}} = -2^8 = -256$

Common Mistakes & Tips

• Quadrant Awareness: When finding the argument of a complex number, always consider the quadrant to ensure the correct angle.
• Angle Reduction: Simplify angles by adding or subtracting multiples of $2\pi$ to find a coterminal angle within a standard range.
• Conjugate Simplification: Remember the shortcut: If $\alpha$ and $\beta$ are conjugates, $\alpha^n + \beta^n = 2r^n \cos(n\theta)$.

Summary

By finding the roots of the quadratic equation, converting them to polar form, applying De Moivre's theorem, and using trigonometric identities to simplify the result, we found that $\alpha^{15} + \beta^{15} = -256$.

The final answer is \boxed{-256}, which corresponds to option (A).