Question

If ${(\frac{3}{2}+i\frac{\sqrt{3}}{2})}^{100}=3^{50}(x-iy)$ where x,y are real,$i=\sqrt{-1},$ then the value of $\frac{y}{x}$ equals

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

Answer 1

The correct option is C $\sqrt{3}$

${(\frac{3}{2}+i\frac{\sqrt{3}}{2})}^{100}=3^{50}(x-iy)$

$(\sqrt{3}{)}^{100}{(\frac{\sqrt{3}}{2}+i\frac{1}{2})}^{100}=3^{50}(x-iy)$

$(3{)}^{50}{(\frac{\sqrt{3}}{2}+i\frac{1}{2})}^{100}=3^{50}(x-iy)$

${(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})}^{100}=(x-iy)$

$\cos \frac{100\pi }{6}+i\sin \frac{100\pi }{6}=(x-iy)$ [ By De Moivre's theorem ]

$\cos (16\pi +\frac{2\pi }{3})+i\sin (16\pi +\frac{2\pi }{3})=(x-iy)$

$\cos \frac{2\pi }{3}+isin\frac{2\pi }{3}=(x-iy)$

$\therefore x=-\frac{1}{2},y=-\frac{\sqrt{3}}{2}$

$⟹\frac{y}{x}=\sqrt{3}$