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Sagot :
Since i = √(-1), it follows that i ² = -1, i ³ = -i, and i ⁴ = 1.
q1. We have
i ³¹ = i ²⁸ × i ³ = (i ⁴)⁷ × i ³ = 1⁷ × (-i ) = -i
q2. Approach each square root individually:
√(-20) = √(-1 × 2² × 5) = √(-1) × √(2²) × √5 = 2i √5
√(-12) = √(-1 × 2² × 3) = √(-1) × √(2²) × √3 = 2i √3
Then
√(-20) × √(-12) = (2i √5) × (2i √3) = (2i )² √(5 × 3) = -4√15
You may have been tempted to combine the square roots immediately, but that would have given the wrong answer.
√(-20) × √(-12) ≠ √((-20) × (-12)) = √240 = 4√15
More generally, we have
√a × √b ≠ √(a × b)
for complex numbers a and b. Otherwise, we would have nonsensical claims like 1 = -1 :
√(-1) × √(-1) = i ² = -1
whereas
√(-1) × √(-1) ≠ √((-1)²) = √1 = 1
q3. Nothing tricky about this one:
3i × 4i = 12i ² = -12
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