Question

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

• A. $(-3, 0)$
• B. $(0, 3)$
• C. $(0, -3)$
• D. $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$

Answer 1

The Correct Option is D

Let $z=\frac{3}{2}+i \frac{\sqrt{3}}{2}$
$r=\sqrt{\frac{9}{4}+\frac{3}{4}}=\sqrt{\frac{12}{4}}=\sqrt{3}$
$\theta =\tan ^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}}\right)$
$=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}$
$\therefore \frac{3}{2}+\frac{i \sqrt{3}}{2}=\sqrt{3} e^{\frac{i \pi}{6}}$
$\therefore \left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{50}=\left(\sqrt{3} e^{\frac{i \pi}{6}}\right)^{50}$
$=(\sqrt{3})^{50}\left(e^{\frac{i \pi}{6}}\right)^{50}=3^{25} e^{\frac{50 \pi}{6}}$
$\Rightarrow\left(\frac{3}{2}+\frac{i \sqrt{3}}{2}\right)^{50}=3^{25} e^{\frac{i 25 \pi}{3}}$
$=3^{25}\left(\cos \frac{25 \pi}{3}+i \sin \frac{25 \pi}{3}\right)$
$=3^{25}(\cos 1500+i \sin 1500)$
$\left.=3^{25}[\cos (360 \times 4+60)+i \sin (360) \times 4+60)\right]$
$=3^{25}(\cos 60+i \sin 60)$
$\Rightarrow\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{50}=3^{25}\left(\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)$ ...(i)
According to question,
$\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{50}=3^{25}(x +i y)$
$3^{25}\left(\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)=3^{25}(x +i y)$
which is true only when $x=\frac{1}{2}, y=\frac{\sqrt{3}}{2}$