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Step-by-Step Solution
Step 1: Express the base in polar form
Observe that
$1-\sqrt{3}\,i = 2\left(\frac{1}{2} - \frac{\sqrt{3}}{2}\,i\right).$
We recognize
$\frac{1}{2} - \frac{\sqrt{3}}{2}\,i = \cos\left(\frac{\pi}{3}\right)\;-\;i\,\sin\left(\frac{\pi}{3}\right).$
Hence,
$$
1 - \sqrt{3}\,i \;=\;2\bigl(\cos(\tfrac{\pi}{3}) - i\,\sin(\tfrac{\pi}{3})\bigr).
$$
Step 2: Raise the expression to the 200th power
When we raise $2\left(\cos\left(\frac{\pi}{3}\right)-i\,\sin\left(\frac{\pi}{3}\right)\right)$ to the 200th power, use De Moivre’s theorem:
$$
\bigl(r(\cos\theta - i\,\sin\theta)\bigr)^n \;=\; r^n\bigl(\cos(n\theta) - i\,\sin(n\theta)\bigr).
$$
Here, $r=2$, $\theta = \tfrac{\pi}{3}$, and $n=200.$ Therefore
$$
(1 - \sqrt{3}\,i)^{200}
\;=\;
\bigl(2e^{-i\pi/3}\bigr)^{200}
\;=\;
2^{200}\,\Bigl(\cos\bigl(\tfrac{200\pi}{3}\bigr)-i\,\sin\bigl(\tfrac{200\pi}{3}\bigr)\Bigr).
$$
Step 3: Simplify the trigonometric part
Note that
$$
\frac{200\pi}{3} \;=\; 66\pi + \frac{2\pi}{3}.
$$
Since $\cos(66\pi+\theta)=\cos(\theta)$ and $\sin(66\pi+\theta)=\sin(\theta)$ (because $66\pi$ is a multiple of $2\pi$), we get
$$
\cos\bigl(\tfrac{200\pi}{3}\bigr)
= \cos\bigl(\tfrac{2\pi}{3}\bigr),
\quad
\sin\bigl(\tfrac{200\pi}{3}\bigr)
= \sin\bigl(\tfrac{2\pi}{3}\bigr).
$$
Thus,
$$
\cos\left(\tfrac{2\pi}{3}\right) = -\frac{1}{2},
\quad
\sin\left(\tfrac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}.
$$
So
$$
(1 - \sqrt{3}\,i)^{200}
= 2^{200}\left(-\tfrac12 - i\,\tfrac{\sqrt{3}}{2}\right)
= 2^{199}\bigl(-1 - \sqrt{3}\,i\bigr).
$$
Step 4: Identify p and q
We have
$$
(1 - \sqrt{3}\,i)^{200} = 2^{199} (p + i q).
$$
Comparing with the simplified form
$2^{199} (-1 - \sqrt{3}\,i)$,
we get:
$$
p = -1
\quad\text{and}\quad
q = -\sqrt{3}.
$$
Step 5: Form the expressions involving p and q
We are told that
$p + q + q^2
\quad\text{and}\quad
p - q + q^2
$
are roots of the desired quadratic equation. Let us calculate these two values:
For $p - q + q^2$
$$
=\,(-1) - \bigl(-\sqrt{3}\bigr) + \bigl(-\sqrt{3}\bigr)^2
=\, -1 + \sqrt{3} + 3
=\, 2 + \sqrt{3}.
$$
For $p + q + q^2$
$$
=\,(-1) + \bigl(-\sqrt{3}\bigr) + \bigl(-\sqrt{3}\bigr)^2
=\, -1 - \sqrt{3} + 3
=\, 2 - \sqrt{3}.
$$
Denote
$$
\alpha = 2 + \sqrt{3}, \quad \beta = 2 - \sqrt{3}.
$$
Step 6: Derive the quadratic equation with these roots
If $\alpha$ and $\beta$ are roots of the quadratic, then the sum and product of roots are
$$
\alpha + \beta = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4,
$$
$$
\alpha \beta = \bigl(2 + \sqrt{3}\bigr)\bigl(2 - \sqrt{3}\bigr) = 4 - 3 = 1.
$$
A quadratic with roots $\alpha$ and $\beta$ is
$$
x^2 \;-\; (\alpha + \beta)\,x \;+\; \alpha\beta \;=\; 0.
$$
Substituting the values:
$$
x^2 \;-\; 4x \;+\; 1 \;=\; 0.
$$
Answer
The required quadratic equation is
$$
x^2 - 4x + 1 = 0.
$$
