Let p, q in mathbb{R} and (1-sqrt{3}i)^{200} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $(1-\sqrt{3}i)^{200} = 2^{199}(p + iq)$.Converting to polar form: $1-\sqrt{3}i = 2(\cos(\frac{-\pi}{3}) + i\sin(\frac{-\pi}{3}))$.So,$(1-\sq

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given $(1-\sqrt{3}i)^{200} = 2^{199}(p + iq)$.Converting to polar form: $1-\sqrt{3}i = 2(\cos(\frac{-\pi}{3}) + i\sin(\frac{-\pi}{3}))$.So,$(1-\sqrt{3}i)^{200} = 2^{200}(\cos(\frac{-200\pi}{3}) + i\sin(\frac{-200\pi}{3}))$.Since $\frac{-200\pi}{3} = -66\pi - \frac{2\pi}{3}$,we have $\cos(\frac{-200\pi}{3}) = \cos(\frac{-2\pi}{3}) = -\frac{1}{2}$ and $\sin(\frac{-200\pi}{3}) = \sin(\frac{-2\pi}{3}) = -\frac{\sqrt{3}}{2}$.Thus,$2^{200}(-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 2^{199}(p + iq)$.$2 	imes 2^{199}(-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 2^{199}(p + iq)$.$p + iq = -1 - i\sqrt{3}$,so $p = -1$ and $q = -\sqrt{3}$.Let $\alpha = p + q + q^2 = -1 - \sqrt{3} + 3 = 2 - \sqrt{3}$.Let $\beta = p - q + q^2 = -1 + \sqrt{3} + 3 = 2 + \sqrt{3}$.Sum of roots $\alpha + \beta = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$.Product of roots $\alpha \beta = (2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$.The quadratic equation is $x^2 - (\alpha + \beta)x + \alpha\beta = 0$,which is $x^2 - 4x + 1 = 0$.

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

Similar Questions

If $1, \omega, \omega^2$ are the cube roots of unity,then $(1 - 2\omega + \omega^2)^6$ is equal to

If $\alpha, \beta$ are the roots of $1+x+x^2=0$,then the value of $\alpha^4+\beta^4+\alpha^{-4}\beta^{-4}$ is

If $|x+iy|=\sqrt{x^2+y^2}$,then $|(1-\sqrt{3}i)^9+(\sqrt{3}+i)^9|=$

The sum of the products of the non-conjugate roots of $i^{1/4}$ taken two at a time is

If $1, \omega, \omega^2, \ldots, \omega^8$ are the roots of the equation $x^9-1=0$,then $\sum_{r=1}^8 \left(\omega^r\right)^{99} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module