If left(frac{3}{2}+i frac{sqrt{3}}{2}right)^{ | vedclass

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Question

(D) Let $z = \frac{3}{2} + i \frac{\sqrt{3}}{2} = \sqrt{3} \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = \sqrt{3} \left( \cos \frac{\pi}{6} + i

Vedclass · Accepted Answer

$\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$

Vedclass · Answer

(D) Let $z = \frac{3}{2} + i \frac{\sqrt{3}}{2} = \sqrt{3} \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = \sqrt{3} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) = \sqrt{3} e^{i \frac{\pi}{6}}$.Then, $\left( \frac{3}{2} + i \frac{\sqrt{3}}{2} \right)^{50} = (\sqrt{3})^{50} e^{i \frac{50\pi}{6}} = 3^{25} e^{i \frac{25\pi}{3}}$.Since $\frac{25\pi}{3} = 8\pi + \frac{\pi}{3}$, we have $e^{i \frac{25\pi}{3}} = e^{i \frac{\pi}{3}} = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$.Thus, $3^{25} (x + iy) = 3^{25} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right)$.Comparing real and imaginary parts, we get $x = \frac{1}{2}$ and $y = \frac{\sqrt{3}}{2}$.

List-$I$	List-$II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$	$1.$ True
$Q.$ There exists a $k \in \{1, 2, \ldots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\|1-z_1\|\|1-z_2\| \ldots \|1-z_9\|}{10}$ equals	$3.$ $1$
$S.$ $1 - \sum_{k=1}^9 \cos \left(\frac{2k\pi}{10}\right)$ equals	$4.$ $2$

If $\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{50}=3^{25}(x+i y)$ where $x$ and $y$ are real, then the ordered pair $(x, y)$ is

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