Complex Number Question 1
Question 1 - 24 January - Shift 1
Let $p, q \in \mathbb{R}$ and $(1-\sqrt{3} i)^{200}=2^{199}(p+i q)$,
$i=\sqrt{-1}$ Then $p+q+q^{2}$ and $p-q+q^{2}$ are roots of the equation.
(1) $x^{2}+4 x-1=0$
(2) $x^{2}-4 x+1=0$
(3) $x^{2}+4 x+1=0$
(4) $x^{2}-4 x-1=0$
Show Answer
Answer: (2)
Solution:
Formula: Demoivre’s theorem
$(1-\sqrt{3} i)^{200}=2^{199}(p+i q)$
$2^{200}(\cos \frac{\pi}{3}-i \sin \frac{\pi}{3})^{200}=2^{199}(p+i q)$
$2(-\frac{1}{2}-i \frac{\sqrt{3}}{2})=p+iq$
$p=-1, q=-\sqrt{3}$
$\alpha=p+q+q^{2}=2-\sqrt{3}$
$\beta=p-q+q^{2}=2+\sqrt{3}$
$\alpha+\beta=4$
$\alpha \beta=1$
equation is given by $x^2-(\alpha + \beta)x + \alpha \beta$
$\Rightarrow x^{2}-4 x+1=0$
