Question

Given : $u=1+i\sqrt{3}$ and $v=\sqrt{3}+i$

Calculate $\frac{u^3}{v^4}$

• A. $\left(1/4\right)-i\sqrt{1/4}$
• B. $\left(3/4\right)-i\sqrt{3}/4$
• C. $\left(1/4\right)-i\sqrt{3}/4$
• D. none of these

Answer 1

The correct option is C $\left(1/4\right)-i\sqrt{3}/4$

$u=1+i\sqrt{3}⟹\sqrt{1+3}{e}^{i\left({\tan }^{-1}\sqrt{3}\right)}⟹2e^{i\pi /3}$

$v=i+\sqrt{3}⟹\sqrt{1+3}{e}^{i\left({\tan }^{-1}1/\sqrt{3}\right)}⟹2e^{i\pi /6}$

$\frac{u^3}{v^4}=\frac{2^3{e}^{i\pi }}{2^4{e}^{2i\pi /3}}=\left(1/2\right)e^{-i\pi /3}$

$⟹\frac{1}{2}\left[\cos \right(\pi /3)-i\sin (\pi /3\left)\right]⟹\frac{1}{2}(\frac{1}{2}-\frac{i\sqrt{3}}{2})$

$⟹(\frac{1}{4}-\frac{i\sqrt{3}}{4})$