Question

If ${(\frac{3}{2}+i\frac{\sqrt{3}}{2})}^{48}=3^{24}(x+iy)$, then the ordered pair $(x,y)$ is

• A. $(0,2)$
• B. $(0,1)$
• C. $(1,0)$
• D. $(1,1)$

Answer 1

Answer: (C)

$(1,0)$

${(\frac{3}{2}+\frac{i\sqrt{3}}{3})}^{48}={\left\{\sqrt{3}(\frac{\sqrt{3}}{2}+\frac{i}{2})\right\}}^{48}$

$=3^{24}{(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})}^{48}$

$=3^{24}{\left(e^{i\pi /6}\right)}^{48}=3^{24}\left(e^{i8\pi }\right)$

$=3^{24}(\cos \pi +i\sin 8\pi )=3^{24}(1+i.0)$

$=3^{24}(x+iy)$

$\therefore (x,y)=(1,0)$