The equation you have provided asks to prove:

$(-1 + \sqrt{3}i)^{3n} + (-1 - \sqrt{3}i)^{3n} = 2^{3n+1}$

where $n$ is any integer.

Step-by-step solution:

1. Recognize the terms:

   • The complex numbers $-1 + \sqrt{3}i$ and $-1 - \sqrt{3}i$ are complex conjugates. These numbers can be written in polar form, which helps in understanding their powers.
   • These complex numbers resemble the cube roots of unity, specifically $e^{2\pi i / 3}$ and $e^{-2\pi i / 3}$. This is because $-1 + \sqrt{3}i = 2e^{i2\pi/3}$ and $-1 - \sqrt{3}i = 2e^{-i2\pi/3}$, with a modulus of 2 and argument $\pm 120^\circ$ (or $\pm \frac{2\pi}{3}$).

2. Use De Moivre's Theorem:

   • De Moivre's Theorem states that for any complex number $r(\cos \theta + i \sin \theta)$, its powers can be expressed as: $(r (\cos \theta + i \sin \theta))^n = r^n (\cos(n\theta) + i \sin(n\theta))$
   • Let’s apply this to the terms $-1 + \sqrt{3}i$ and $-1 - \sqrt{3}i$.

3. Simplify both terms:

   • First, consider $(-1 + \sqrt{3}i)$ in polar form: $-1 + \sqrt{3}i = 2\left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right)$ Similarly, $(-1 - \sqrt{3}i)$ becomes: $-1 - \sqrt{3}i = 2\left(\cos \frac{-2\pi}{3} + i \sin \frac{-2\pi}{3}\right)$
   • Now raise both to the power of $3n$: $(-1 + \sqrt{3}i)^{3n} = 2^{3n} \left( \cos(3n \cdot \frac{2\pi}{3}) + i \sin(3n \cdot \frac{2\pi}{3}) \right)$ $(-1 - \sqrt{3}i)^{3n} = 2^{3n} \left( \cos(3n \cdot \frac{-2\pi}{3}) + i \sin(3n \cdot \frac{-2\pi}{3}) \right)$

4. Add the two terms:

   • Since the sine terms will cancel each other out, you will only be left with the cosine terms: $(-1 + \sqrt{3}i)^{3n} + (-1 - \sqrt{3}i)^{3n} = 2^{3n} \left( \cos(3n \cdot \frac{2\pi}{3}) + \cos(3n \cdot \frac{-2\pi}{3}) \right)$
   • Using the fact that $\cos \theta = \cos(-\theta)$, this simplifies to: $= 2^{3n+1} \cos \left( 3n \cdot \frac{2\pi}{3} \right)$

5. Final cosine simplification:

   • When $3n$ is an integer multiple of $\frac{2\pi}{3}$, the cosine will cycle between 1 and -1, depending on $n$. For integer values of $n$, the cosine term becomes $1$, leading to: $(-1 + \sqrt{3}i)^{3n} + (-1 - \sqrt{3}i)^{3n} = 2^{3n+1}$

Thus, we have proved that:

$(-1 + \sqrt{3}i)^{3n} + (-1 - \sqrt{3}i)^{3n} = 2^{3n+1}$

Let me know if you'd like further details or have questions!

Related Questions:

1. How does De Moivre's theorem help with complex number powers?
2. What are cube roots of unity and how do they relate to this equation?
3. Can you express $\cos(2\pi/3)$ and $\sin(2\pi/3)$ geometrically?
4. How can polar form simplify calculations involving powers of complex numbers?
5. What is the role of the modulus in the polar representation of complex numbers?

Tip:

Understanding complex numbers in both rectangular and polar forms is crucial when dealing with powers and roots. Always try converting between these forms for simplification.