Question

Let $p , q \in R$ and $(1-\sqrt{3})^{200}=2^{199}(p+i q), t=\sqrt{-1}$ Then $p + q + q ^2$ and $p - q + q ^2$ are roots of the equation

• A. $x^2-4 x+1=0$
• B. $x^2+4 x+1=0$
• C. $x^2-4 x-1=0$
• D. $x^2+4 x-1=0$

Answer 1

The Correct Option is A

The correct answer is (A) : $x^2-4 x+1=0$
$(1-\sqrt{3}i{)}^{200}=2^{199}(p+iq)$
$2^{200}{\left(\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}\right)}^{200}=2^{199}(p+iq)$
$2\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)=p+iq$
$p=-1,q=-\sqrt{3}$
$\alpha =p+q+q^2=2-\sqrt{3}$
$\beta =p-q+q^2=2+\sqrt{3}$
$\alpha +\beta =4$
$\alpha \cdot \beta =1$
equation $x^2-4x+1=0$

Answer 2

We are given: \[ (1 - \sqrt{3}i)^{200} = 2^{199}(p + iq). \] Convert \( 1 - \sqrt{3}i \) to polar form: \[ 1 - \sqrt{3}i = 2 \left( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right). \] Raise it to the power of 200: \[ (1 - \sqrt{3}i)^{200} = 2^{200} \left( \cos \frac{200 \cdot 5\pi}{3} + i \sin \frac{200 \cdot 5\pi}{3} \right). \] Simplify the angle modulo \( 2\pi \): \[ \frac{200 \cdot 5\pi}{3} = 1000\pi/3 = 333\cdot 2\pi + \pi/3 = \pi/3. \] Thus: \[ (1 - \sqrt{3}i)^{200} = 2^{200} \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right). \] Comparing with \( 2^{199}(p + iq) \), we find: \[ p = 1, \quad q = \sqrt{3}. \] Now, calculate: \[ p + q + q^2 = 1 + \sqrt{3} + 3 = 4 + \sqrt{3}, \] \[ p - q + q^2 = 1 - \sqrt{3} + 3 = 4 - \sqrt{3}. \] The quadratic equation with roots \( 4 + \sqrt{3} \) and \( 4 - \sqrt{3} \) is: \[ x^2 - (p + q) x + (p + q + q^2) = x^2 - 4x + 1 = 0. \] Thus, the equation is: \[ \boxed{x^2 - 4x + 1 = 0}. \]