Question

The real part of ${\left(\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right)}^6+{\left(\frac{1-i\sqrt{3}}{1+i\sqrt{3}}\right)}^6$

• A. $2$
• B. $-2$
• C. $1$
• D. $0$

Answer 1

The correct option is D ($0$)

Given $\frac{1+i\sqrt{3}}{1-i\sqrt{3}}=\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\times \frac{1+i\sqrt{3}}{1+i\sqrt{3}}$

$=\frac{1+3i^2+2i\sqrt{3}}{1-3i^2}$
[Since, $(a+b{)}^2=a^2+b^2+2ab$ and $(a+b)(a-b)=a^2-b^2$]

$=\frac{1-3+2i\sqrt{3}}{1+3}$ [Since, $i^2=-1$]

$=\frac{-2+2i\sqrt{3}}{4}$

$=\frac{1}{2}(i-1)$

$\therefore {\left(\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right)}^6=\frac{1}{2^6}{(i-1)}^6$
$=\frac{1}{2^6}{(i-1)}^{3\times 2}$
$=\frac{1}{2^6}[{(i-1)}^2{]}^3$

$=\frac{1}{64}{(i^2+1-2i)}^3$ [Since, $(a-b{)}^2=a^2+b^2-2ab$]

$=\frac{1}{64}{(-1+1-2i)}^3$

$=\frac{1}{64}{(-2i)}^3$
$=\frac{-8i}{64}$
$\therefore {\left(\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right)}^6=\frac{-i}{8}$

Similar way, consider$\frac{1-i\sqrt{3}}{1+i\sqrt{3}}=\frac{1-i\sqrt{3}}{1+i\sqrt{3}}\times \frac{1-i\sqrt{3}}{1-i\sqrt{3}}$

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

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

$=\frac{-2-2i\sqrt{3}}{4}$

$=\frac{-1}{2}(i+1)$

$\therefore {\left(\frac{1-i\sqrt{3}}{1+i\sqrt{3}}\right)}^6={\left(\frac{-1}{2}\right)}^6{(i+1)}^6$

$=\frac{1}{64}{(i^2+1+2i)}^3$

$=\frac{1}{64}{(-1+1+2i)}^3$

$=\frac{1}{64}{\left(2i\right)}^3$
$=\frac{-8i}{64}$
$\therefore {\left(\frac{1-i\sqrt{3}}{1+i\sqrt{3}}\right)}^6=\frac{-i}{8}$

$\therefore {\left(\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right)}^6+{\left(\frac{1-i\sqrt{3}}{1+i\sqrt{3}}\right)}^6=\frac{-i}{8}-\frac{i}{8}$
$=\frac{-2i}{8}$
$\therefore {\left(\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right)}^6+{\left(\frac{1-i\sqrt{3}}{1+i\sqrt{3}}\right)}^6=\frac{-i}{4}$