If x = 1 + i, then what is the value of x^{6} + x^{4} + x² + 1?

Correct Answer is : −6i−3-6i - 3−6i−3

Correct Answer is : −6i−3-6i - 3−6i−3

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UPSC NDA 2 2019 Mathematics Solved Paper

Question : 21 of 120

Marks: +1, -0

x = 1 + i

x^{6} + x^{4} + x^{2} + 1

Solution:

Concept:

We use the property

i^2 = -1

and simplify higher powers of

x = 1 + i

by squaring repeatedly.

Explanation:

Step 1: Square

x = 1 + i

x^2 = (1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i

This gives

x^2 = 2i

Step 2: Square

x^2

to obtain

x^4

x^4 = (x^2)^2 = (2i)^2 = 4i^2 = 4(-1) = -4

Hence

x^4 = -4

Step 3: Write

x^6

(x^2)^3

and substitute

x^2 = 2i

x^6 = (2i)^3 = 8i^3 = 8(-i) = -8i

Note:

i^3 = i \cdot i^2 = i \cdot (-1) = -i

Step 4: Add all terms.

x^6 + x^4 + x^2 + 1 = (-8i) + (-4) + (2i) + 1

Combine real parts:

-4 + 1 = -3

Combine imaginary parts:

-8i + 2i = -6i

The sum simplifies to

-6i - 3

Answer:

The value is

-6i - 3

, which corresponds to option C.

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