Question

${\left(\frac{-1+\sqrt{-3}}{2}\right)}^{100}+{\left(\frac{-1-\sqrt{-3})}{2}\right)}^{100}$is equal to

• A. $2$
• B. $0$
• C. $-1$
• D. $1$

Answer 1

The correct option is C

$-1$

Explanation for the correct answer:

Step 1. Find the value of given expression:

${\left(\frac{-1+\sqrt{-3}}{2}\right)}^{100}+{\left(\frac{-1-\sqrt{-3}}{2}\right)}^{100}$

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

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

where, $\omega$ is cube root of unity

Step 2. Put the values of ${\left(\frac{-1+\sqrt{-3}}{2}\right)}^{100}$and ${\left(\frac{-1-\sqrt{3}}{2}\right)}^{100}$in given equation:

${\left(\frac{-1+\sqrt{-3}}{2}\right)}^{100}+{\left(\frac{-1-\sqrt{-3}}{2}\right)}^{100}$

$={\left(\frac{-1+\sqrt{3}i}{2}\right)}^{100}+{\left(\frac{-1-\sqrt{3}i}{2}\right)}^{100}$ $\left[\mathbf{\because }\mathit{i}\mathbf{=}\sqrt{\mathbf{-}\mathbf{1}}\right]$

$={\omega }^{100}+{\omega }^{200}$

$=\omega +{\omega }^2$ $\left[\mathbf{\because }{\mathit{\omega }}^{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{1}\mathbf{+}\mathit{\omega }\mathbf{+}{\mathit{\omega }}^{\mathbf{2}}\mathbf{=}\mathbf{0}\right]$

$=\mathbf{-}\mathbf{1}$

Hence, Option ‘C’ is Correct.