Let `p,q in R` and `(1-sqrt3i)^(200)=2^(199) (p=iq),i=sqrt(-1)` then `p+q+q^2` and `p-q+q^2` are roots of the equation

Options

A. `x^2 – 4x + 1 = 0`
B. `x^2 + 4x + 1 = 0`
C. `x^2 – 4x – 1 = 0`
D. `x^2 + 4x – 1 = 0`

Answer

To solve the given problem step by step, we will start from the equation provided and simplify it to find the values of \( p \) and \( q \).

### Step 1: Rewrite the given expression
We start with the expression:
\[ (1 - \sqrt{3}i)^{200} = 2^{199}(p + iq) \]
To simplify \( 1 - \sqrt{3}i \), we can express it in polar form.

### Step 2: Convert to polar form
The modulus \( r \) of \( 1 - \sqrt{3}i \) is calculated as:
\[ r = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]
The argument \( \theta \) is given by:
\[ \theta = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3} \]
Thus, we can express \( 1 - \sqrt{3}i \) in polar form as:
\[ 1 - \sqrt{3}i = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right) \]

### Step 3: Raise to the power of 200
Using De Moivre's theorem:
\[ (1 - \sqrt{3}i)^{200} = (2)^{200} \left(\cos\left(-\frac{200\pi}{3}\right) + i\sin\left(-\frac{200\pi}{3}\right)\right) \]
Calculating \( -\frac{200\pi}{3} \):
\[ -\frac{200\pi}{3} = -66\pi + \frac{2\pi}{3} = -66\pi + 2\pi/3 \]
Since \( -66\pi \) is a multiple of \( 2\pi \), we can simplify:
\[ \cos\left(-\frac{200\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}, \quad \sin\left(-\frac{200\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Thus:
\[ (1 - \sqrt{3}i)^{200} = 2^{200}\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 2^{199}(-1 + i\sqrt{3}) \]

### Step 4: Compare coefficients
From the original equation:
\[ 2^{199}(p + iq) = 2^{199}(-1 + i\sqrt{3}) \]
By comparing real and imaginary parts, we find:
\[ p = -1, \quad q = \sqrt{3} \]

### Step 5: Calculate \( p + q + q^2 \) and \( p - q + q^2 \)
Now we calculate:
1. \( p + q + q^2 \):
\[ p + q + q^2 = -1 + \sqrt{3} + (\sqrt{3})^2 = -1 + \sqrt{3} + 3 = 2 + \sqrt{3} \]
2. \( p - q + q^2 \):
\[ p - q + q^2 = -1 - \sqrt{3} + 3 = 2 - \sqrt{3} \]

### Step 6: Form the equation with the roots
Let the roots be \( r_1 = 2 + \sqrt{3} \) and \( r_2 = 2 - \sqrt{3} \).

- Sum of the roots:
\[ r_1 + r_2 = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \]
- Product of the roots:
\[ r_1 \cdot r_2 = (2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \]

### Step 7: Write the quadratic equation
The quadratic equation with these roots is:
\[ x^2 - (sum \, of \, roots)x + (product \, of \, roots) = 0 \]
Thus:
\[ x^2 - 4x + 1 = 0 \]

### Final Answer
The required equation is:
\[ \boxed{x^2 - 4x + 1 = 0} \]