If (sqrt{3}+i)^{100}=2^{99}(p+iq),then p and | vedclass

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(A) Given $(\sqrt{3}+i)^{100}=2^{99}(p+iq)$.Express $\sqrt{3}+i$ in polar form: $2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$.Using De Moivre's theor

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$x^{2}-(\sqrt{3}-1)x-\sqrt{3}=0$

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(A) Given $(\sqrt{3}+i)^{100}=2^{99}(p+iq)$.Express $\sqrt{3}+i$ in polar form: $2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$.Using De Moivre's theorem: $(2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}))^{100} = 2^{100}(\cos \frac{100\pi}{6} + i \sin \frac{100\pi}{6}) = 2^{100}(\cos \frac{50\pi}{3} + i \sin \frac{50\pi}{3})$.Since $\frac{50\pi}{3} = 16\pi + \frac{2\pi}{3}$,we have $\cos \frac{50\pi}{3} = \cos \frac{2\pi}{3} = -\frac{1}{2}$ and $\sin \frac{50\pi}{3} = \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}$.Thus,$2^{100}(-\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 2^{99}(p+iq)$.Dividing by $2^{99}$,we get $2(-\frac{1}{2} + i \frac{\sqrt{3}}{2}) = p+iq$,so $p = -1$ and $q = \sqrt{3}$.The quadratic equation with roots $p$ and $q$ is $x^2 - (p+q)x + pq = 0$.$p+q = \sqrt{3}-1$ and $pq = -\sqrt{3}$.Therefore,the equation is $x^2 - (\sqrt{3}-1)x - \sqrt{3} = 0$.

If $(\sqrt{3}+i)^{100}=2^{99}(p+iq)$,then $p$ and $q$ are roots of the equation :

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