If $$(\sqrt{3} + i)^{100} = 2^{99}(p + iq)$$, then $$p$$ and $$q$$ are roots of the equation:

We have the complex number $$\sqrt{3}+i$$. First we express it in polar (modulus-argument) form, because raising a complex number to a high power is easiest when the number is written as $$r(\cos\theta+i\sin\theta)$$.

The modulus is found from the formula $$r=\sqrt{(\text{Re})^{2}+(\text{Im})^{2}}$$. Here $$\text{Re}=\sqrt{3}$$ and $$\text{Im}=1$$, so

$$r=\sqrt{(\sqrt{3})^{2}+1^{2}}=\sqrt{3+1}=2.$$

For the argument we use $$\theta=\tan^{-1}\left(\dfrac{\text{Im}}{\text{Re}}\right)$$. Hence

$$\theta=\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)=\tan^{-1}(\tan 30^{\circ})=30^{\circ}=\dfrac{\pi}{6}.$$

So we can write

$$\sqrt{3}+i=2\bigl(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\bigr).$$

Now we are asked to evaluate $$(\sqrt{3}+i)^{100}$$. We will use De Moivre’s theorem, which states:

$$\bigl(r(\cos\theta+i\sin\theta)\bigr)^{n}=r^{\,n}\bigl(\cos n\theta+i\sin n\theta\bigr).$$

Applying the theorem with $$r=2,\;\theta=\dfrac{\pi}{6},\;n=100$$ gives

$$\bigl(\sqrt{3}+i\bigr)^{100}=2^{100}\bigl(\cos(100\cdot\tfrac{\pi}{6})+i\sin(100\cdot\tfrac{\pi}{6})\bigr).$$

The problem statement rewrites this power in the form $$2^{99}(p+iq)$$. To match that form we divide the expression we have by $$2^{99}$$:

$$\bigl(\sqrt{3}+i\bigr)^{100}=2^{99}\Bigl[2\bigl(\cos\tfrac{100\pi}{6}+i\sin\tfrac{100\pi}{6}\bigr)\Bigr]=2^{99}\bigl(p+iq\bigr).$$

Comparing the two brackets, we see directly that

$$p+iq=2\bigl(\cos\tfrac{100\pi}{6}+i\sin\tfrac{100\pi}{6}\bigr).$$

Now we simplify the angle. First compute

$$\dfrac{100\pi}{6}=\dfrac{50\pi}{3}.$$

Next, reduce $$\dfrac{50\pi}{3}$$ to an angle between $$0$$ and $$2\pi$$. Write

$$\dfrac{50\pi}{3}=16\pi+\dfrac{2\pi}{3}.$$

Because multiples of $$2\pi$$ do not change sine or cosine, we subtract $$16\pi=8(2\pi)$$ and keep only the remainder $$\dfrac{2\pi}{3}$$. Therefore

$$\cos\dfrac{50\pi}{3}=\cos\dfrac{2\pi}{3},\qquad\sin\dfrac{50\pi}{3}=\sin\dfrac{2\pi}{3}.$$

We know the exact trigonometric values:

$$\cos\dfrac{2\pi}{3}=-\dfrac12,\qquad\sin\dfrac{2\pi}{3}=\dfrac{\sqrt{3}}{2}.$$

Substituting these into the expression for $$p+iq$$ gives

$$p+iq=2\bigl(-\dfrac12+i\dfrac{\sqrt{3}}{2}\bigr)=-1+i\sqrt{3}.$$

Hence we read off

$$p=-1,\qquad q=\sqrt{3}.$$

We now form the quadratic equation whose roots are exactly $$p$$ and $$q$$. For a quadratic with roots $$\alpha$$ and $$\beta$$, the standard form is

$$x^{2}-(\alpha+\beta)x+\alpha\beta=0.$$

Here $$\alpha=p=-1$$ and $$\beta=q=\sqrt{3}$$, so

Sum of roots $$\alpha+\beta=(-1)+\sqrt{3}=\sqrt{3}-1,$$

Product of roots $$\alpha\beta=(-1)(\sqrt{3})=-\sqrt{3}.$$

Putting these into the standard form gives the required quadratic:

$$x^{2}-(\sqrt{3}-1)x-\sqrt{3}=0.$$

This matches Option D in the list given. Hence, the correct answer is Option 4.