21
$\begingroup$

I ran into a problem dividing by imaginary numbers recently. I was trying to simplify:

$2 \over i$

I came up with two methods, which produced different results:

Method 1: ${2 \over i} = {2i \over i^2} = {2i \over -1} = -2i$

Method 2: ${2 \over i} = {2 \over \sqrt{-1}} = {\sqrt{4} \over \sqrt{-1}} = \sqrt{4 \over -1} = \sqrt{-4} = 2i$

I know from using the formula from this Wikipedia article that method 1 produces the correct result. My question is: why does method 2 give the incorrect result? What is the invalid step?

$\endgroup$
5
  • 1
    $\begingroup$ See also: math.stackexchange.com/questions/438/… $\endgroup$
    – Isaac
    Jul 29, 2010 at 22:18
  • 5
    $\begingroup$ For the right interpretation of the square root, only the last step is wrong; remember, -2i is also a square root of -4. $\endgroup$ Jul 29, 2010 at 22:19
  • 1
    $\begingroup$ @Qiaochu Yuan: "a square root" and the result of the radial operator are not the same. The radical operator means the principal root. $\endgroup$
    – Isaac
    Jul 29, 2010 at 22:23
  • 1
    $\begingroup$ @Isaac: there is a more important point to be made here. The "right interpretation" I'm talking about is the one where you take the multiplicative property of the square root seriously. If you do that, you sometimes end up with different answers for the same computation. This is because you are trying to analytically continue the same function along different paths and this doesn't always lead to a consistent answer, and now you are in a perfect position to learn about the fundamental group, covering spaces, monodromy... $\endgroup$ Jul 29, 2010 at 22:37
  • 1
    $\begingroup$ @Qiaochu Yuan: I don't disagree that it could have nice properties as you interpret it, but to interpret the radical-indicated square root on a real number to be anything but the nonnegative real root for nonnegative reals and i*sqrt(|x|) on negative reals is inconsistent with every single text for calculus or earlier courses that I've ever seen. Further, if you are constructing the complex numbers from the reals by taking square-roots of negative real numbers, the construction is independent of the choice of root for -1, but then requires that all other square roots are chosen accordingly. $\endgroup$
    – Isaac
    Jul 29, 2010 at 23:01

3 Answers 3

28
$\begingroup$

The incorrect step is saying:

$\sqrt{4}/\sqrt{-1} = \sqrt{4/-1}$

The identity:

$\sqrt{a}/\sqrt{b} = \sqrt{a/b}$

is only justified when $a$ and $b$ are positive.

$\endgroup$
5
  • 8
    $\begingroup$ The problem comes from thinking of the square root as a function (on R or C). The square root is not really a function; it can be tortured into pretending to be one on the non-negative real line for our needs, but its natural habitat is on its Riemann surface. $\endgroup$ Jul 29, 2010 at 22:18
  • 3
    $\begingroup$ Technically, the identity is also justified if a=0, but that's not really a useful technicality. $\endgroup$
    – Isaac
    Jul 29, 2010 at 22:19
  • 4
    $\begingroup$ @Qiaochu Yuan: It's perfectly reasonable to think of the square root as a function on C (though you need to pick a definition of principal root). You cannot, however, expect a general function f to satisfy f(a)/f(b)=f(a/b). $\endgroup$
    – Isaac
    Jul 29, 2010 at 22:21
  • 7
    $\begingroup$ @Isaac: that was a rhetorical flourish. My point is that one cannot pick a canonical branch, and if like me you care about canonical choices, you ignore such things. This isn't just stubbornness; it is a really important realization in Galois theory, for example, that you can't distinguish i and -i without making a noncanonical choice somewhere. (More precisely, there is no canonical isomorphism between the algebraic closure of R and C.) $\endgroup$ Jul 29, 2010 at 22:33
  • $\begingroup$ @Qiaochu Your last comment: Yes. Yes yes yes yes yes. $\endgroup$
    – BBischof
    Jul 30, 2010 at 4:15
5
$\begingroup$

The only soundproof way to be sure to find the right result while dividing two complex numbers

$$\frac{a+bi}{c+di}$$

is reducing it to a multiplication. The answer is of the form $x+yi$; therefore

$$(c+di)(x+yi) = a+bi$$

and you will end up with two linear equations, one for the real coefficient and another for the imaginary one. As Simon and Casebash already wrote, taking a square root leads to problems, since you cannot be sure which value must be chosen.

$\endgroup$
1
$\begingroup$

This is exactly the same issue as in this question. Each non-zero complex number has two numbers that square to give it, with the same magnitude, but with opposite sign. When we define the square root function, we have to decide which of the roots we want. For positive numbers, it is obvious to choose the positive root. For negative number, we choose to have the positive imaginary values, although because of symmetry the choice doesn't mean much anyway.

So, to see if the standard multiplication and division laws apply, then we have to consider domain the numbers are in. We already know they apply for non-negative real numbers. It is easy enough to verify that for negative numbers $$\sqrt{a}*\sqrt{b}=-\sqrt{ab}$$ and $$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$$ We also see that, if $a$ is positive and $b$ negative, then $$\sqrt{ab}=\sqrt{a}\sqrt{b}$$ $$\sqrt{\frac{a}{b}}=-\frac{\sqrt{a}}{\sqrt{b}}$$ and $$\sqrt{\frac{b}{a}}=\frac{\sqrt{b}}{\sqrt{a}}$$

$\endgroup$
2
  • $\begingroup$ Even two negative numbers is an incorrect usage of the identity, per my answer on that question (it happens to work with division, but fails with multiplication). $\endgroup$
    – Isaac
    Jul 29, 2010 at 22:26
  • $\begingroup$ @Isaac: You are right $\endgroup$
    – Casebash
    Jul 29, 2010 at 22:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .