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

Answer: (A)

$ (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 $x ^2-4 x +1=0$