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MHTCET 10 May 2024 Shift 1 Paper

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Question : 134 of 150

Marks: +1, -0

If

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

\mathrm{i}=\sqrt{-1}

, then the value of

\left(3+w+3 w^{2}\right)^{4}

is

16
-16
16w
16w2

Solution:

Concept:

The given number

w

is a complex cube root of unity.

Explanation:

Let

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

.

For a complex cube root of unity, we know:

w^3 = 1

and

1 + w + w^2 = 0

.

From

1 + w + w^2 = 0

, we get

w^2 = -1 - w

.

Now substitute into the expression:

\left(3 + w + 3w^2\right)^4 = \left(3 + w + 3(-1 - w)\right)^4

= (3 + w - 3 - 3w)^4 = (-2w)^4

= 16w^4 = 16w^3 \cdot w = 16w

.

Thus the value of the expression is

16w

.

Answer:

Option C:

16w

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