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MHTCET 18 Apr 2026 Shift 1 Paper

Question : 101 of 150

Marks: +1, -0

The value of

\left(\frac{-1+i \sqrt{3}}{2}\right)^{18}+\left(\frac{-1-i \sqrt{3}}{2}\right)^{18}

Solution:

Concept:

These two complex numbers are the non-real cube roots of unity, and any integer power of

\omega

reduces using

\omega^3 = 1

Explanation:

Let

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

and

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

These are the standard complex cube roots of unity, so

\omega^3 = 1

and

(\omega^2)^3 = \omega^6 = 1

Now simplify each term of the given expression.

First term:

\left(\frac{-1+i\sqrt{3}}{2}\right)^{18} = \omega^{18} = (\omega^3)^6 = 1^6 = 1

Second term:

\left(\frac{-1-i\sqrt{3}}{2}\right)^{18} = (\omega^2)^{18} = \omega^{36} = (\omega^3)^{12} = 1^{12} = 1

Therefore the required sum is

1 + 1 = 2

Answer:

2

, which matches Option A.

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