Convert the complex number z = sqrt(3) + i to polar form and calculate z5 using De Moivre's Theorem.

Given $z = \sqrt{3} + i$.
The rectangular form is $z = x + yi$, where $x = \sqrt{3}$ and $y = 1$.

Calculate the modulus $r$:

$$r = |z| = \sqrt{x^2 + y^2} = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$$

Calculate the argument $\theta$:
Since $x > 0$ and $y > 0$, $\theta$ is in the first quadrant.

$$\tan \theta = \frac{y}{x} = \frac{1}{\sqrt{3}}$$ $$\theta = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$$

So, $z$ in polar form is $z = 2\left(\cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right)\right)$.

De Moivre's Theorem states that for a complex number $z = r(\cos \theta + i \sin \theta)$, its $n$-th power is $z^n = r^n(\cos(n\theta) + i \sin(n\theta))$.
For $n=5$:

$$z^5 = 2^5\left(\cos\left(5 \times \frac{\pi}{6}\right) + i \sin\left(5 \times \frac{\pi}{6}\right)\right)$$ $$z^5 = 32\left(\cos\left(\frac{5\pi}{6}\right) + i \sin\left(\frac{5\pi}{6}\right)\right)$$

The angle $\frac{5\pi}{6}$ is in the second quadrant.

$$\cos\left(\frac{5\pi}{6}\right) = -\cos\left(\pi - \frac{5\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$ $$\sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$

$$z^5 = 32\left(-\frac{\sqrt{3}}{2} + i \frac{1}{2}\right)$$ $$z^5 = 32 \times \left(-\frac{\sqrt{3}}{2}\right) + 32 \times \left(\frac{1}{2}\right)i$$ $$z^5 = -16\sqrt{3} + 16i$$

The calculated value for $z^5$ is $-16\sqrt{3} + 16i$.
The options provided in the image are:
A. 2
B. $3+i$
C. $3-i$
D. 4
None of the given options match the calculated result.

The final answer is $\boxed{-16\sqrt{3} + 16i}$.