1.1 • THE ORIGIN OF COMPLEX NUMBERS 3general cubic equation. Surprisingly, this development played a significant role
in helping to establish the legitimacy of imaginary numbers. Roots of negative
numbers, of course, had come up earlier in the simplest of quadratic equations
such as x^2 + 1 = 0. The solutions we know today as x = ±.;=I, however,
were easy for mathematicians to ignore. In Cardano's time, negative numbers
were still being treated with some suspicion, as it was difficult to conceive of any
physical reality corresponding to them. Taking square roots of such quantities
was surely all the more ludicrous. Nevertheless, Ca.rdano made some genuine
attempts to deal with .;=I. Unfortunately, his geometric thinking made it
hard to make much headway. At one point he commented that the process of
arithmetic that deals with quantities such as .;=I "involves mental tortures and
is truly sophisticated." At another point he concluded that the process is "as
refined as it is useless." Many mathematicians held this view, but finally there
was a breakthrough.
In his 1572 treatise L'Algebro, Rafael Bombelli showed that roots of negativenumbers have great utility indeed. Consider the depressed cubic x^3 -15x -4 = 0.
Using Formula (1-3), we compute x = {h + v'-ffi + {/2 - J - 121 or, in a
somewhat different form, x = {/2 + 11.J=I + {/2 - nFI.
Simplifying this expression would have been very difficult if Bombelli had
not come up with what he called a "wild thought." He suspected that if the
original depressed cubic had real solutions, then the two parts of x in the pre-
ceding equation could be written as u + vFI and u - vR for some real
numbers u and v. That is, Bombelli believed u + v.;=I = {/2 + 11 .J=I and
u - vR = V2- 11FI, which would mean (u+ vV-1)3
= 2 + llV-1,
and (u -vV-1)
3
= 2 - 11.J=T. Then, using the well-known algebraic identity
(a+ b)^3 = a^3 + 3a^2 b + 3ab^2 + b3, and assuming that roots of negative numbers
obey the rules of algebra, he obtained(u + vH)^3 = u^3 + 3(u^2 )vH + 3(u)(vH)^2 + (vv'=I)^3
= u^3 + 3(u)(vH)^2 + 3(u^2 )vFI + (v!=l)^3
= (u^3 - 3uv^2 ) + (3u^2 v -v^3 )H
= u(u^2 - 3v^2 } + v(3u^2 - v^2 )H
= 2 +11v=I.(1-4}
(1-5)By equating like parts of Equations (1-4} and (1-5) Bombelli reasoned that
u(u^2 - 3v^2 } = 2 and v(3u^2 - v^2 } = 11. Perhaps thinking even more wildly,
Bombelli then supposed that u and v were integers. The only integer factors of
2 are 2 and 1, so the equation u(u^2 - 3v^2 ) = 2 led Bombelli to conclude thatu = 2 and u^2 - 3v^2 = 1. From this conclusion it follows that v^2 = 1, or v = ±1.
Amazingly, u = 2 and v = 1 solve the second equation v(3u^2 - v^2 ) = 11, so
Bombelli declared the values for u and v to be u = 2 and v = 1, respectively.