Question 15 of 28intermediate🔧 ApplyNumerical2 marks

Evaluate: [i18+(1i)25]3\left[i^{18} + \left(\frac{1}{i}\right)^{25}\right]^3.

Correct Answer

2 - 2i

Exercise: Miscellaneous Exercise on Chapter 4 | Q: 1 | (Page 10)
For More Understanding

Explanation

This question tests the understanding of powers of i. The key concept from the textbook is that for any integer k, i^(4k) = 1, i^(4k+1) = i, i^(4k+2) = -1, and i^(4k+3) = -i. Also, 1/i = -i. The solution requires evaluating each power separately, then combining and cubing the result.

Solution Steps

  1. Step 1: Evaluate i18i^{18}. Since 18 = 4(4) + 2, we have i18i^{18} = i^(4k+2) = -1.

  2. Step 2: Evaluate (1/i)^25. From the context, 1/i = -i. So (1/i)^25 = (-i)^25 = (-1)^25 × i25i^{25} = -1 × i^(4×64 \times 6+1) = -1 × i = -i.

  3. Step 3: Find the sum inside the bracket. i18i^{18} + (1/i)^25 = -1 + (-i) = -1 - i.

  4. Step 4: Evaluate (-1 - i)^3. Using (a+b)^3 = a3a^{3} + 3a²b + 3ab² + b³: (-1)³ + 3(-1)²(-i) + 3(-1)(-i)² + (-i)³ = -1 - 3i + 3(-1)(-1) + i = -1 - 3i + 3 + i = 2 - 2i.

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