The expression you've provided is $(1 + \sqrt{3}i)^6$.

To solve this, let's approach it by expressing $1 + \sqrt{3}i$ in polar form. This will allow us to use De Moivre's theorem to find the sixth power of the complex number.

Step-by-Step Solution

1. Express $1 + \sqrt{3}i$ in Polar Form:

   • A complex number $a + bi$ can be converted to polar form $r(\cos \theta + i \sin \theta)$, where $r = \sqrt{a^2 + b^2}$ and $\theta = \tan^{-1}(b/a)$.
   • Here, $a = 1$ and $b = \sqrt{3}$.
   • Calculate $r$: $r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$
   • Calculate $\theta$: $\theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$
   • So, $1 + \sqrt{3}i$ in polar form is: $2 \left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right)$

2. Apply De Moivre's Theorem:

   • To find $(1 + \sqrt{3}i)^6$, raise the polar form to the power of 6.
   • $\left(2 \left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right)\right)^6 = 2^6 \left(\cos \left(6 \cdot \frac{\pi}{3}\right) + i \sin \left(6 \cdot \frac{\pi}{3}\right)\right)$
   • Simplify $2^6$: $2^6 = 64$
   • Simplify the angle: $6 \cdot \frac{\pi}{3} = 2\pi$
   • Since $\cos(2\pi) = 1$ and $\sin(2\pi) = 0$, we have: $64 (\cos 2\pi + i \sin 2\pi) = 64 (1 + 0i) = 64$

Final Answer

$(1 + \sqrt{3}i)^6 = 64$

Would you like further details or have any questions?

Here are some related questions to consider:

1. How does De Moivre's theorem apply to complex number powers?
2. What is the significance of converting complex numbers to polar form?
3. How can complex numbers in polar form simplify complex arithmetic?
4. What are the applications of De Moivre's theorem in trigonometry?
5. How can powers of complex numbers be visualized on the complex plane?

Tip: When dealing with powers of complex numbers, converting to polar form is often the easiest method, especially when the exponent is large.