Question:
If $(\sqrt{3}+\mathrm{i})^{100}=2^{99}(\mathrm{p}+\mathrm{i} \mathrm{q})$, then $\mathrm{p}$ and $\mathrm{q}$ are roots of the equation:
Correct Option: 1
Solution:
$\left(2 e^{i \pi / 6}\right)^{100}=2^{99}(p+i q)$
$2^{100}\left(\cos \frac{50 \pi}{3}+\mathrm{i} \sin \frac{50 \pi}{3}\right)=2^{99}(\mathrm{p}+\mathrm{i} \mathrm{q})$
$p+i q=2\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)$
$p=-1, q=\sqrt{3}$
$x^{2}-(\sqrt{3}-1) x-\sqrt{3}=0$
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