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

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(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

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

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(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 $y = \frac{\sqrt{3}}{2}$.

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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