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2016ers Chit-Chat Thread (2 Viewers)

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leehuan

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Actually, what is the proof that √a = √b = √(ab) breaks down over the imsginary numbers?
 
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leehuan said:
Actually, what is the proof that √a = √b = √(ab) breaks down over the imsginary numbers?
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We need to ask ourselves what the symbol √(z) actually means for a complex number z. In general, it is not true that √(zw) = √(z)√(w) for z, w complex. Recall that sqrt(·) is a multivalued function on the complex plane. E.g. sqrt(1) = +1 or -1. Now, we assign a principal value to this in the reals by assigning the positive root to the symbol √1. But it turns out our choice of principal values cannot be made continuous over the whole of the complex plane (further reading: http://mathworld.wolfram.com/BranchCut.html ).

I believe the convention for principal square root for complex numbers is as follows: If z is a negative real, then √z is defined as the square root with the positive imaginary part, e.g. √(-4) = 2i, not -2i. If z is a non-negative real, you already know the convention. If z is a non-real complex number, we define √z to be the square root with positive real part (such a square root will always exist and be unique, since the two square roots are negatives of each other and exist as can be seen using de Moivre's Law to verify things for example).

Now, let's return to the proposed equation √(zw) = √(z) * √(w). It is easy to check that these are both indeed square roots of the complex number zw, by squaring each side. So the only question is whether they are the same square root (either they are, or they are negatives of each other).

Now, recall that the LHS has positive real part by definition of √ (assuming zw isn't a non-positive number, which we'll deal with afterwards). Hence the RHS will equal the LHS if and only if it also has positive real part. Suppose √(z) = a + bi and √(w) = c + di, where a, b > 0 and c, d are real. Then the RHS has positive real part iff ac – bd > 0. So this is the condition for which the equation of interest will hold when zw isn't a non-negative real. It is clear to see that if b = 0 = d (which happens iff √(z) and √(w) are positive reals, which happens iff z and w are positive reals), then this condition is met. This explains why the equation is always true for positive real numbers z, w. But it is also easy to see that the equation won't hold in general, since we don't always have ac – bd > 0 (easy to cook up counterexamples). Here, the example √(1) = √(-1) √(-1) (z = -1 = w) fails, because here, √z = i = 0+1i = √w. Hence here, ac – bd = 0*0 – 1*1 = -1, which is not positive.

Now, let's consider the case when zw is a non-positive real. If zw = 0, then one of z or w is 0, and since √0 = 0, the equation clearly holds. Now, let's consider the case where zw is a strictly negative real. Then √(zw) = bi, for some strictly positive real number b. So √(zw) = √(z)√(w) will hold iff the RHS has positive imaginary part. Suppose z = A + Bi and w = C + Di. Then zw = (AC – BD) + (AD + BC)i. Since we assumed zw is a negative real, the conditions on z and w are that AC – BD < 0 and AD + BC = 0. If z is a positive real and w is a negative real, then √(zw) = √(z)√(w) can be seen to hold. In general it won't hold though. If √(z) = a + bi and √(w) = c + di, then √(z)√(w) has imaginary part ad + bc, so we need this to be positive for the equation to hold. This doesn't hold in general though.

To see this, let zw = -A, where A > 0. We are asking whether √(zw) = √(z)√(w). Now, since z = -A/ w = (-A)•u, where u = 1/w, we are asking whether √(-A•u) = √(-A)√u (if it is, we can equivalently rearrange to get what we want, using the fact that √(1/w) = 1/√w. This is indeed a true fact for w not a negative real, because the LHS here has non-negative real part, and the RHS is the reciprocal of a number with non-negative real part, which also must have non-negative real part. In the case where z or w is a negative real number, since zw = -A, one of z or w must also be a positive real number, and it is easy to show that √(zw) = √(z)√(w) will hold in this case.).

Now, consider if u = cis(3pi/4), then √(u) = cis(3pi/8). Also, √(-A) = √A * i. Hence √(-A) √u = √(A) * i * cis(3pi/8) = √(A) * cis(3pi/8 + pi/2) = √(A) * cis(7pi/8), which has negative real part, so cannot equal √(-A*u).
 
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Paradoxica

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State the fallacy in the following argument:

If a person is wearing a hat, they have a head.

I am not wearing a hat.

Therefore, I do not have a head.
 
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