Bombelli father of complex arithmetic?

May 26th, 2005 ~ Posted in: Mathematics

I have often wondered what caused complex arithmetic to be defined the way it is. The definition of the addition seems obvious, but why is it that
(a, b)(c,d) = (ac-bd, bc+ad)? The answer according to Tristan Needham in his book Visual Complex Analysis:

Apparently, the mathematician Bombelli realized that the solution to the general cubic equation x^3 = 3px+2q given in Cardano’s Ars Magna:

\displaystyle x = \sqrt[3]{q + \sqrt{q^2 - p^3}} + \sqrt[3]{q - \sqrt{q^2 - p^3}}

gives complex solutions if p^3 > q^2. For the example x^3 = 15x + 4, the solution is x = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}. In general, when x is complex, that implies the equation x^3 = 3px + 2q has no solution, but by graphing the line and the cubic, you can see that there is a solution in this case: x=4.

From here, Bombelli had his “wild idea”: what if he worked backwards to determine what x = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i} was? First, he needed to determine what \sqrt[3]{2 \pm 11i} is; he did so by assuming \sqrt[3]{2 \pm 11i} = 2 \pm ni (Why? I don’t know; it doesn’t seem obvious— maybe he tried different numbers besides 2, and this was one that worked). From here, he assumed the standard laws of complex arithmetic, to solve the equation (2 \pm ni)^3 = 2 \pm 11i, yielding x=4 as expected.

This was the first example of complex arithmetic at work. More specifically, before Bombelli considered this instance in which complex arithmetic was needed to solve a cubic equation (consider the graph of x^3 and a line; there is always an intersection, so there is always a real solution), complex numbers were encountered only in the solution to quadratic equations. There, they could be ignored, since the presence of a complex number indicates that the given line and the parabola x^2 do not intersect.

Mathematics ~

5 Responses to “Bombelli father of complex arithmetic?”

  • 1. Mikael Johansson
    May 26th, 2005 at 4:21 pm

    This may just be me being algebraic, arrogant and ignoring the historical part of it all - but if you start out supposing i^2=-1, then (a+bi)(c+di)=ac+i^2bd+i(ac+bd)=ac-bd+i(ac+bd) as witnessed.

  • 2. Mikael Johansson
    May 26th, 2005 at 4:22 pm

    Eeeek. Here’s to comment previews over the world. I forgot to slash my [tex] up there. =/

  • 3. Alex
    May 26th, 2005 at 4:55 pm

    Note that you’re defining (a,b)(c,d) = (ac - bd, ad + bc); you can’t say that is the result of any other axiom. The question is why was multiplication defined that way? As opposed to say (a,b)(c,d) = (ac, 0)? The only thing you ‘have’ to keep is that if you multiply two real numbers together you get another, which the alternative satisfies. So why didn’t we choose that?

  • 4. Mikael Johansson
    May 27th, 2005 at 11:28 am

    I’d doubt that the origin of complex arithmetic would have been to start multiplying pairs of numbers - which can be made in a number of ways admittedly - but on the other hand I haven’t really delved deep into mathematical history. My comment was more towards that if you instead of just taking R\times R and give it an algebra structure start out by saying “Well, we seem to get negative square roots. Suppose that isn’t just a weird impossible idea and let’s see what happens.”

    Given the general outlook during 15-18th century on mathematics and numbers, this is a slightly more plausible mindset for originators - and in that mindset defining i by i^2=-1 and then calculating (a+bi)(c+di) is a more natural way of thinking than putting algebra structures on ordered pairs.

  • 5. Alex
    May 27th, 2005 at 2:16 pm

    Excellent point— that is certainly the natural way of thinking.

    And after this discussion, I realize I did a poor job expressing why Bombelli could in a sense be considered the father of complex arithmetic; I added that onto the post.

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