Question

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

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

Answer 1

The correct option is D $(\frac{1}{2},\frac{\sqrt{3}}{2})$

Let $z=\frac{3}{2}+i\frac{\sqrt{3}}{2}$

$r=\sqrt{\frac{9}{4}+\frac{3}{4}}=\sqrt{\frac{12}{4}}=\sqrt{3}$

$\theta ={\tan }^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}}\right)={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$

$\therefore \frac{3}{2}+\frac{i\sqrt{3}}{2}={\sqrt{3e}}^{\frac{i\pi }{6}}$

$\therefore {(\frac{3}{2}+i\frac{\sqrt{3}}{2})}^{50}={\left(\sqrt{3}{e}^{\frac{i\pi }{6}}\right)}^{50}$

$=(\sqrt{3}{)}^{50}{\left(e^{\frac{i\pi }{6}}\right)}^{50}=3^{25}{e}^{\frac{i50\pi }{6}}$

$\Rightarrow {(\frac{3}{2}+\frac{i\sqrt{3}}{2})}^{50}=3^{25}{e}^{\frac{i25\pi }{3}}$

$=3^{25}(\cos \frac{25\pi }{3}+i\sin \frac{25\pi }{3})$

$=3^{25}(\cos 1500+i\sin 1500)$

$=3^{25}\left[\cos \right(360\times 4+60)+i\sin (360\times 4+60)$

$=3^{25}(\cos 60+i\sin 60)$

$\Rightarrow {(\frac{3}{2}+i\frac{\sqrt{3}}{2})}^{50}=3^{25}(\frac{1}{2}+i\frac{\sqrt{3}}{2})$

According to question,

${(\frac{3}{2}+i\frac{\sqrt{3}}{2})}^{50}=3^{25}(x+iy)$

$3^{25}(\frac{1}{2}+i\frac{\sqrt{3}}{2})=3^{25}(x+iy)$

which is true only when $x=\frac{1}{2},y=\frac{\sqrt{3}}{2}$