sqrt i^4 Solution

If you haven’t seen the √(i4) puzzle, I recommend you go check it out. It’s good stuff. Read on for the answer.

To recap,1: √(i) = √(i5)
2: √(i) = √(i4) * √(i)
3: √(i) = √(i2) * √(i2) * √(i)
4: √(i) = i * i * √(i)
5: √(i) = -1 * √(i)
6: 1 = -1

The error is in the third step from the last with this assumption:
√(i2) * √(i2) = √(i4)
If you spell that out, it becomes immediately apparent:√(i2) * √(i2) ?= √(i4)
√(-1) * √(-1) ?= √(i2 * i2)
i * i ?= √(-1 * -1) = √(1)
-1 != 1

When taking square roots, it’s never certain whether you’re specifying the positive or the negative root. For example, √(4) is either 2 or -2. The only requirement is that (√(4))2 = 4. The assumption that √(x2) * √(x2) = √(x4) is valid iff all of the roots are positive. For example, -2 * 2 != 4. When dealing with real numbers, we always make the assumption that we’re talking about positive roots. However, that same assumption cannot be safely applied to complex numbers. When dealing with any root of i, we must assume that the root is the “negative” root. So, we must apply the rule that √(i2) * √(i2) = -√(i4). Apply that to rule the third step above, and we’re left with this:
1: √(i) = √(i5)
2: √(i) = √(i4) * √(i)
3: √(i) = -√(i2) * √(i2) * √(i)
4: √(i) = -i * i * √(i)
5: √(i) = 1 * √(i)
6: 1 = 1
Not as exciting, but a lot more correct. (In that it is correct, instead of being not correct.)

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