Question

${\left[\frac{-1+i\sqrt{3}}{2}\right]}^6+{\left[\frac{-1-i\sqrt{3}}{2}\right]}^6+{\left[\frac{-1+i\sqrt{3}}{2}\right]}^5+{\left[\frac{-1-i\sqrt{3}}{2}\right]}^5=$

• A. $-1$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (D)

($1$)

As we know that the cube roots of unity are $\frac{1\pm \sqrt{3}i}{2}$ and 1
Let $w=\frac{-1+i\sqrt{3}}{2}$ and $w^2=\frac{-1-i\sqrt{3}}{2}$

and also we have $w^3=1$ and $1+w+w^2=0$

Now consider, ${\left(\frac{-1+i\sqrt{3}}{2}\right)}^6+{\left(\frac{-1-i\sqrt{3}}{2}\right)}^6+{\left(\frac{-1+i\sqrt{3}}{2}\right)}^5+{\left(\frac{-1-i\sqrt{3}}{2}\right)}^5$

$=\left[\right(w){]}^6+[(w{)}^2{]}^6+[w^5]+[w^2{]}^5$
$=\left[\right(w^3){]}^2+[(w{)}^2{]}^{3\times 2}+[w^{3+2}]+[w^2{]}^{2+3}$
$=1^2+{\left(w^3\right)}^{(2\times 2)}+w^3\times w^2+w^4\times (w^3{)}^2$
$=1+1^4+1\left(w^2\right)+w^{3+1}\times 1^2$
$=1+1+w^2+w^3{w}^1$
$=1+1+w^2+w$
Since $1+w+w^2=0$
Thus, $1+1+w^2+w=1$
Therefore,
${\left(\frac{-1+i\sqrt{3}}{2}\right)}^6+{\left(\frac{-1-i\sqrt{3}}{2}\right)}^6+{\left(\frac{-1+i\sqrt{3}}{2}\right)}^5+{\left(\frac{-1-i\sqrt{3}}{2}\right)}^5=1$