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

Answer: (D)

Let $z=\frac{3}{2}+i\frac{\sqrt{3}}{2}$
$r=\sqrt{\frac{9}{4}+\frac{3}{4}}=\sqrt{\frac{12}{4}}=\sqrt{3}$
$\theta ={\tan }^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}}\right)$
$={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$
$\therefore \frac{3}{2}+\frac{i\sqrt{3}}{2}=\sqrt{3}{e}^{\frac{i\pi }{6}}$
$\therefore {\left(\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)}^{50}={\left(\sqrt{3}{e}^{\frac{i\pi }{6}}\right)}^{50}$
$=(\sqrt{3}{)}^{50}{\left(e^{\frac{i\pi }{6}}\right)}^{50}=3^{25}{e}^{\frac{50\pi }{6}}$
$\Rightarrow {\left(\frac{3}{2}+\frac{i\sqrt{3}}{2}\right)}^{50}=3^{25}{e}^{\frac{i25\pi }{3}}$
$=3^{25}\left(\cos \frac{25\pi }{3}+i\sin \frac{25\pi }{3}\right)$
$=3^{25}(\cos 1500+i\sin 1500)$
$=3^{25}[\cos (360\times 4+60)+i\sin (360)\times 4+60)]$
$=3^{25}(\cos 60+i\sin 60)$
$\Rightarrow {\left(\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)}^{50}=3^{25}\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)$ ...(i)
According to question,
${\left(\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)}^{50}=3^{25}(x+iy)$
$3^{25}\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)=3^{25}(x+iy)$
which is true only when $x=\frac{1}{2},y=\frac{\sqrt{3}}{2}$