somewhere near the beginning.

Bombelli father of complex arithmetic?

Filed under: Mathematics — Alex @ 2:28 pm 5/26/2005

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.

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