Question

The value of ${\left(\frac{1+\sqrt{3}i}{1-\sqrt{3}i}\right)}^{64}+{\left(\frac{1-\sqrt{3}i}{1+\sqrt{3}i}\right)}^{64}$ is

• A. $0$
• B. $-1$
• C. $1$
• D. $i$

Answer 1

The correct option is B $-1$

We know that,

$\omega =\frac{-1+\sqrt{3}i}{2}\Rightarrow 1-\sqrt{3}i=-2\omega$
and

${\omega }^2=\frac{-1-\sqrt{3}i}{2}\Rightarrow 1+\sqrt{3}i=-2{\omega }^2$

Now,

${\left(\frac{1+\sqrt{3}i}{1-\sqrt{3}i}\right)}^{64}+{\left(\frac{1-\sqrt{3}i}{1+\sqrt{3}i}\right)}^{64}$

$={\left(\frac{-2{\omega }^2}{-2\omega }\right)}^{64}+{\left(\frac{-2\omega }{-2{\omega }^2}\right)}^{64}$

$={\omega }^{64}+\frac{1}{{\omega }^{64}}=\omega +{\omega }^2$