Since <em>i</em> = √(-1), it follows that <em>i</em> ² = -1, <em>i</em> ³ = -<em>i</em>, and <em>i</em> ⁴ = 1.
q1. We have
<em>i </em>³¹ = <em>i</em> ²⁸ × <em>i</em> ³ = (<em>i</em> ⁴)⁷ × <em>i</em> ³ = 1⁷ × (-<em>i</em> ) = -<em>i</em>
q2. Approach each square root individually:
√(-20) = √(-1 × 2² × 5) = √(-1) × √(2²) × √5 = 2<em>i</em> √5
√(-12) = √(-1 × 2² × 3) = √(-1) × √(2²) × √3 = 2<em>i</em> √3
Then
√(-20) × √(-12) = (2<em>i</em> √5) × (2<em>i</em> √3) = (2<em>i</em> )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
√<em>a</em> × √<em>b</em> ≠ √(<em>a</em> × <em>b</em>)
for complex numbers <em>a</em> and <em>b</em>. Otherwise, we would have nonsensical claims like 1 = -1 :
√(-1) × √(-1) = <em>i</em> ² = -1
whereas
√(-1) × √(-1) ≠ √((-1)²) = √1 = 1
q3. Nothing tricky about this one:
3<em>i</em> × 4<em>i</em> = 12<em>i</em> ² = -12