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How Cubic (and Not Quadratic) Led to Complex Numbers

J. B. Thoo Yuba College

2013 AMATYC Conference, Anaheim, California This presentation was produced usingLATEX with C. Campani’s BeamerLATEX class and saved as a PDF file: . See Norm Matloff’s web page for a quick tutorial. Disclaimer: Our slides here won’t show off what Beamer can do. Sorry. :-) Are you sitting in the right room?

This talk is about the solution of the general cubic , and how the solution of the general led mathematicians to study complex numbers seriously. This, in turn, encouraged mathematicians to forge ahead in to arrive at the modern theory of groups and rings. Hence, the solution of the general cubic equation marks a watershed in the .

We shall spend a fair amount of time on Cardano’s solution of the general cubic equation as it appears in his seminal work, (The Great Art). Outline of the talk

Motivation: square roots of negative numbers The : long known

The cubic formula: what is it? Brief history of the solution of the cubic equation Examples from Cardano’s Ars magna Bombelli’s famous example

Brief history of complex numbers The fundamental theorem of algebra References

Girolamo Cardano, The Rules of Algebra (Ars Magna), translated by T. Richard Witmer, Dover Publications, Inc. (1968).

Roger Cooke, The History of : A Brief Course, second edition, Wiley Interscience (2005).

Victor J. Katz, A : An Introduction, third edition, Addison-Wesley (2009).

Amy Shell-Gellasch and J. B. Thoo, Intersecting Mathematics and History: Topics in from an Historical Viewpoint, in preparation. Motivation: square roots of negative numbers Myth

Many of today’s popular algebra textbooks introduce imaginary numbers along these lines, saying1 The system allows us to solve equations such as x2 + 1 = 0 that have no solutions. The implication is that mathematicians became interested in complex numbers because they were unable to solve certain quadratic equations with only the real numbers.

1Elayn Martin-Gay, Beginning & Intermediate Algebra, 5th edition, p. 634. Myth buster

The truth is that, as late as the 16th century, when confronted with an equation like

x2 + 1 = 0 or x2 + 2x + 2 = 0,

one could simply say that the equation has no solution, and that would have been the end of it.

Square roots of negative numbers were considered to be “sophistic,” “impossible,” “imaginary,” and “useless” objects. Indeed, imaginary numbers were not understood as late as the 16th century beyond their being a convenient contrivance. The quadratic formula: long known Quadratic problems

Ancient Egypt Berlin Papyrus 6619: The of a square of 100 is equal to that of two smaller squares. The side of one is 2 4 the side of the other . Let me know the sides of 2 3 2 the two unknown squares[ x + ( 4 x) = 100]. Solved by the method of false position. In the solution, the of 1 2 16 is taken, yielding 1 4. Ancient Mesopotamia Cuneiform tablet AO 8862 (Louvre, Paris): I have multiplied the length and width so as to make the area. Then I added to the area the amount by which the length exceeds the width, obtaining 3,3. Then I added the length and width together, obtaining 27. What are the length, width, and area[ xy + x − y = 183; x + y = 27]?

Often in two unknown quantities; often solved by “completing the square.” The quadratic formula: rhetorical

Arabic: Abu Ja’far Muh.ammad ibn-Mus¯ ¯a al-Khw¯arizm¯ı (ca. 780–850) Al-kit¯ab al-muhtas.ar f¯ı h. is¯ab al-jabr wa’l muq¯abala (The Condensed Book on the Calculation of al-Jabr and al-Muqabala)

al-Khw¯arizm¯ı (a > 0, b > 0, c > 0)

Squares equal to roots ax 2 = bx Squares equal to numbers ax 2 = c Squares and roots equal to numbers ax 2 + bx = c Squares and numbers equal to roots ax 2 + c = bx Roots and numbers equal to squares bx + c = ax 2 x2 + bx = c: q q b b2 b2 b2 b2 b b → 2 → 4 → c + 4 → c + 4 → c + 4 − 2 = x

The solution is this: you halve the number of roots, which in the present instance yields five. This you multiply by itself; the product is twenty-five. Add this to thirty-nine; the sum is sixty-four. Now take the root of which, which is eight, and subtract from it half the number of roots, which is five; the remainder is three. This is the root of the square which you sought for; the square itself is nine.

One square, and ten roots of the same, amount to thirty-nine dirhems. . . . One square, and ten roots of the same, amount to thirty-nine dirhems. . . . The solution is this: you halve the number of roots, which in the present instance yields five. This you multiply by itself; the product is twenty-five. Add this to thirty-nine; the sum is sixty-four. Now take the root of which, which is eight, and subtract from it half the number of roots, which is five; the remainder is three. This is the root of the square which you sought for; the square itself is nine. x2 + bx = c: q q b b2 b2 b2 b2 b b → 2 → 4 → c + 4 → c + 4 → c + 4 − 2 = x S A

G Al-Khw¯arizm¯ı provided two geometric demonstrations of his solution. This one is most B widely known, and is often used to introduce the method 25 D of completing the square.

H b 2 x S A

x 2 + bx = c b 2 x 2 x x b x 2 + 2 · x = c 2 B b b2 b2 x 2 + 2 · x + = c + 2 4 4

2 2 b2 b b  b  b x x + = c + 4 2 2 2 4 H x2 + c = bx:

q q q b b2 b2 b2 b b2 b b2 b → 2 → 4 → 4 − c → 4 − c → 2 − 4 − c or 2 + c − 4 = x

[A] square and twenty-one in numbers are equal to ten roots of the same. . . . Solution: Halve the number of the roots; the half is five. Multiply this by itself; the product is twenty-five. Subtract from this the twenty-one which are connected with the square; the remainder is four. Extract its root; it is two. Subtract this from the half of the roots, which is five; the remainder is three. This is the root of the square which you required, and the square is nine. Or you may add the root to the half of the roots; the sum is seven; this is the root of the square which you sought for, and the square itself is forty-nine. [A] square and twenty-one in numbers are equal to ten roots of the same. . . . Solution: Halve the number of the roots; the half is five. Multiply this by itself; the product is twenty-five. Subtract from this the twenty-one which are connected with the square; the remainder is four. Extract its root; it is two. Subtract this from the half of the roots, which is five; the remainder is three. This is the root of the square which you required, and the square is nine. Or you may add the root to the half of the roots; the sum is seven; this is the root of the square which you sought for, and the square itself is forty-nine. x2 + c = bx:

q q q b b2 b2 b2 b b2 b b2 b → 2 → 4 → 4 − c → 4 − c → 2 − 4 − c or 2 + c − 4 = x (Handled negative numbers without fanfare.)

ax2 + bx = c √ p p 4ac + b2 − b c → 4ac → 4ac + b2 → 4ac + b2 → 4ac + b2 − b → = x 2a

India (598–670): tells how to solve quadratic equations by completing the square Take the absolute number from the side opposite to that from which the square and simple unkown are to be subtracted. To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the [value of the] middle term. √ p p 4ac + b2 − b c → 4ac → 4ac + b2 → 4ac + b2 → 4ac + b2 − b → = x 2a

India Brahmagupta (598–670): tells how to solve quadratic equations by completing the square Take the absolute number from the side opposite to that from which the square and simple unkown are to be subtracted. To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the [value of the] middle term.

(Handled negative numbers without fanfare.) ax2 + bx = c √ p p 4ac + b2 − b → 4ac → 4ac + b2 → 4ac + b2 → 4ac + b2 − b → = x 2a

India Brahmagupta (598–670): tells how to solve quadratic equations by completing the square Take the absolute number from the side opposite to that from which the square and simple unkown are to be subtracted. To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the [value of the] middle term.

(Handled negative numbers without fanfare.) ax2 + bx = c

c √ p p 4ac + b2 − b → 4ac + b2 → 4ac + b2 → 4ac + b2 − b → = x 2a

India Brahmagupta (598–670): tells how to solve quadratic equations by completing the square Take the absolute number from the side opposite to that from which the square and simple unkown are to be subtracted. To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the [value of the] middle term.

(Handled negative numbers without fanfare.) ax2 + bx = c

c → 4ac √ p p 4ac + b2 − b → 4ac + b2 → 4ac + b2 − b → = x 2a

India Brahmagupta (598–670): tells how to solve quadratic equations by completing the square Take the absolute number from the side opposite to that from which the square and simple unkown are to be subtracted. To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the [value of the] middle term.

(Handled negative numbers without fanfare.) ax2 + bx = c

c → 4ac → 4ac + b2 √ p 4ac + b2 − b → 4ac + b2 − b → = x 2a

India Brahmagupta (598–670): tells how to solve quadratic equations by completing the square Take the absolute number from the side opposite to that from which the square and simple unkown are to be subtracted. To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the [value of the] middle term.

(Handled negative numbers without fanfare.) ax2 + bx = c

p c → 4ac → 4ac + b2 → 4ac + b2 √ 4ac + b2 − b → = x 2a

India Brahmagupta (598–670): tells how to solve quadratic equations by completing the square Take the absolute number from the side opposite to that from which the square and simple unkown are to be subtracted. To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the [value of the] middle term.

(Handled negative numbers without fanfare.) ax2 + bx = c

p p c → 4ac → 4ac + b2 → 4ac + b2 → 4ac + b2 − b India Brahmagupta (598–670): tells how to solve quadratic equations by completing the square Take the absolute number from the side opposite to that from which the square and simple unkown are to be subtracted. To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the [value of the] middle term.

(Handled negative numbers without fanfare.) ax2 + bx = c √ p p 4ac + b2 − b c → 4ac → 4ac + b2 → 4ac + b2 → 4ac + b2 − b → = x 2a 4a2x2 + 4abx = 4ac 4a2x2 + 4abx + b2 = 4ac + b2 (2ax + b)2 = 4ac + b2 p 2ax + b = 4ac + b2 √ 4ac + b2 − b x = 2a

Bhaskara II (1114–1185): Vija-ganita (or Bijaganita)

131. Śríd’hara’s rule on this point: Multiply both sides of the equation by a number equal to four times the [coefficient] of the square, and add to them a number equal to the square of the original [coefficient] of the unknown quantity. (Then extract the root.)

ax2 + bx = c 4a2x2 + 4abx + b2 = 4ac + b2 (2ax + b)2 = 4ac + b2 p 2ax + b = 4ac + b2 √ 4ac + b2 − b x = 2a

Bhaskara II (1114–1185): Vija-ganita (or Bijaganita)

131. Śríd’hara’s rule on this point: Multiply both sides of the equation by a number equal to four times the [coefficient] of the square, and add to them a number equal to the square of the original [coefficient] of the unknown quantity. (Then extract the root.)

ax2 + bx = c 4a2x2 + 4abx = 4ac (2ax + b)2 = 4ac + b2 p 2ax + b = 4ac + b2 √ 4ac + b2 − b x = 2a

Bhaskara II (1114–1185): Vija-ganita (or Bijaganita)

131. Śríd’hara’s rule on this point: Multiply both sides of the equation by a number equal to four times the [coefficient] of the square, and add to them a number equal to the square of the original [coefficient] of the unknown quantity. (Then extract the root.)

ax2 + bx = c 4a2x2 + 4abx = 4ac 4a2x2 + 4abx + b2 = 4ac + b2 p 2ax + b = 4ac + b2 √ 4ac + b2 − b x = 2a

Bhaskara II (1114–1185): Vija-ganita (or Bijaganita)

131. Śríd’hara’s rule on this point: Multiply both sides of the equation by a number equal to four times the [coefficient] of the square, and add to them a number equal to the square of the original [coefficient] of the unknown quantity. (Then extract the root.)

ax2 + bx = c 4a2x2 + 4abx = 4ac 4a2x2 + 4abx + b2 = 4ac + b2 (2ax + b)2 = 4ac + b2 Bhaskara II (1114–1185): Vija-ganita (or Bijaganita)

131. Śríd’hara’s rule on this point: Multiply both sides of the equation by a number equal to four times the [coefficient] of the square, and add to them a number equal to the square of the original [coefficient] of the unknown quantity. (Then extract the root.)

ax2 + bx = c 4a2x2 + 4abx = 4ac 4a2x2 + 4abx + b2 = 4ac + b2 (2ax + b)2 = 4ac + b2 p 2ax + b = 4ac + b2 √ 4ac + b2 − b x = 2a q 1 8 16 2 2 2 S + 9 S + 2 = S → x + 9 x + 2 = 2x 2x2 − 9x = 18

ax2 + bx = c

The square-root of half the number of a swarm of bees is gone to a shrub of jasmine; and so are eight-ninths of the whole swarm; a female is buzzing to one remaining male, that is humming within a lotus, in which he is confined, having been allured to it by its fragrance at night. Say, lovely woman, the number of bees. 2x2 − 9x = 18

ax2 + bx = c

The square-root of half the number of a swarm of bees is gone to a shrub of jasmine; and so are eight-ninths of the whole swarm; a female is buzzing to one remaining male, that is humming within a lotus, in which he is confined, having been allured to it by its fragrance at night. Say, lovely woman, the number of bees.

q 1 8 16 2 2 2 S + 9 S + 2 = S → x + 9 x + 2 = 2x ax2 + bx = c

The square-root of half the number of a swarm of bees is gone to a shrub of jasmine; and so are eight-ninths of the whole swarm; a female is buzzing to one remaining male, that is humming within a lotus, in which he is confined, having been allured to it by its fragrance at night. Say, lovely woman, the number of bees.

q 1 8 16 2 2 2 S + 9 S + 2 = S → x + 9 x + 2 = 2x 2x2 − 9x = 18 The square-root of half the number of a swarm of bees is gone to a shrub of jasmine; and so are eight-ninths of the whole swarm; a female is buzzing to one remaining male, that is humming within a lotus, in which he is confined, having been allured to it by its fragrance at night. Say, lovely woman, the number of bees.

q 1 8 16 2 2 2 S + 9 S + 2 = S → x + 9 x + 2 = 2x 2x2 − 9x = 18

ax2 + bx = c The quadratic formula: modern

x2 + bx + c = 0

√ b b2 − 4c x = − ± 2 2

D = b2 − 4c : two real roots† if D > 0, one if D = 0, no real root if D < 0

†We will always mean distinct real roots. The cubic formula: what is it? Cubic problems

Ancient Mesopotamia

Cuneiform tablets found containing the sum of the square and the of an integer for many different integers. This suggests they may have solved problems that reduced to x3 + x2 = a, and perhaps even problems that reduced to ay 3 + by 2 + cy = d.

China: Xugu Suanjing (Continuation of Ancient Mathematics; 7th century) compute the length of a leg of a right triangle given that the product of the 1 other leg and the hypotenuse is 1337 20 and the difference between the 1 3 1 2 1 64 hypotenuse and the leg is 1 10 [x + 4 x + 50 x − 8938513 125 = 0]. Generally used numerical methods (“Horner’s method”). b x = y − 3 Reduced cubic: y 3 + py + q = 0,

where b2 2b3 bc p = c − 3 and q = 27 − 3 + d.

The cubic formula: modern

General cubic: x3 + bx2 + cx + d = 0 Reduced cubic: y 3 + py + q = 0,

where b2 2b3 bc p = c − 3 and q = 27 − 3 + d.

The cubic formula: modern

General cubic: x3 + bx2 + cx + d = 0

b x = y − 3 The cubic formula: modern

General cubic: x3 + bx2 + cx + d = 0

b x = y − 3 Reduced cubic: y 3 + py + q = 0,

where b2 2b3 bc p = c − 3 and q = 27 − 3 + d. r q r q 3 q q2 p3 3 q q2 p3 r = − 2 + 4 + 27 , s = − 2 − 4 + 27 ,

q2 p3 ∆ = 4 + 27 : three real roots if ∆ < 0 (irreducible case), two real roots if ∆ = 0, one if ∆ > 0

Let p and q be any real numbers. Then the reduced or depressed cubic equation x3 + px + q = 0 has solutions x1, x2, and x3 given by

x1 = r + s, x2 = rω2 + sω3, x3 = rω3 + sω2, where

√ √ 1 3 1 3 ω2 = − 2 + 2 i, ω3 = − 2 − 2 i. q2 p3 ∆ = 4 + 27 : three real roots if ∆ < 0 (irreducible case), two real roots if ∆ = 0, one if ∆ > 0

Let p and q be any real numbers. Then the reduced or depressed cubic equation x3 + px + q = 0 has solutions x1, x2, and x3 given by

x1 = r + s, x2 = rω2 + sω3, x3 = rω3 + sω2, where r q r q 3 q q2 p3 3 q q2 p3 r = − 2 + 4 + 27 , s = − 2 − 4 + 27 , √ √ 1 3 1 3 ω2 = − 2 + 2 i, ω3 = − 2 − 2 i. Let p and q be any real numbers. Then the reduced or depressed cubic equation x3 + px + q = 0 has solutions x1, x2, and x3 given by

x1 = r + s, x2 = rω2 + sω3, x3 = rω3 + sω2, where r q r q 3 q q2 p3 3 q q2 p3 r = − 2 + 4 + 27 , s = − 2 − 4 + 27 , √ √ 1 3 1 3 ω2 = − 2 + 2 i, ω3 = − 2 − 2 i.

q2 p3 ∆ = 4 + 27 : three real roots if ∆ < 0 (irreducible case), two real roots if ∆ = 0, one if ∆ > 0 Brief history of the solution of the cubic equation

The principal players were the Italian mathematicians

Luca Pacioli (1445–1514) Scipione del Ferro (1465–1526) Niccolò Tartaglia (1500–1557) Girolamo Cardano (1501–1576) – Ludovico Ferrari (1522–1565) Raphael Bombelli (1526–1572) Luca Pacioli (1445–1514)

Summa de arithmetica geometria proportioni et proportionalita (1494) In the Summa, we find the abbreviations co.(cosa) for x (cosa means “thing”), ce.(censo) for x2, and cu.(cubo) for x3 Declared at the end that solving the cubic equation would be as impossible as solving the ancient problem of squaring the circle Scipione del Ferro (1465–1526)

Within twenty years of the Summa, he found a method for solving the “cube and first power equal to the number,”

x3 + px = q,

where p and q are any positive real numbers. Revealed solution to his pupils Annibale della Nave (1500–1558) and Antonio Maria Fiore (first half of the sixteenth century) at his death Niccolò Tartaglia (1500–1557)

Born Niccolò Fontana In 1535 burst onto cubic scene with announcement he discovered solution of “cube and square equal to the number,”

x3 + px2 = q Tartaglia’s announcement prompted del Ferro’s pupil Fiore, who believed Tartaglia to be bluffing, to challenge Tartaglia to a public problem-solving contest Each contestant proposed 30 problems to the other and had 50 days to solve as many of the problems as he could All of the 30 problems that Fiore had proposed to Tartaglia were of the form x3 + px = q, which, after having worked doggedly, Tartaglia finally discovered how to solve just before the end of the 50 days Fiore could not solve any of the problems proposed by Tartaglia, most of which led to equations of the form x3 + px2 = q Fiore was humiliated; Tartaglia’s renown grew Girolamo Cardano (1501–1576)

Also commonly called Cardan Artis magnae, sive de regulis algebraicis (The Great Art, or the Rules of Algebra; 1545), commonly referred to as the Ars magna Ars magna gives solution of the cubic and the quartic, the latter by his pupil Ludovico Ferrari (1522–1565) In Ars magna, Cardano gives credit to del Ferro and Tartaglia for the solution of “cube and first power equal to a constant,” and to Ferrari for the solution of the quartic Tartaglia-Cardano “cubic dispute”

After Tartaglia beat Fiore in a public problem-solving contest, Cardano immediately set out to obtain the secret of the cubic solution from Tartaglia Cardano offered to publish Tartaglia’s solution in a work, giving full credit to Tartaglia; however, Tartaglia refused to divulge his secret, saying that he would publish it in his own work at the right time Tartaglia had a change of heart, perhaps in the hope of securing a favor from Cardano, and provided Cardano with his solution of the cubic; but this was only after Cardano “swore a most solemn oath, by the Sacred Gospels and his word as a gentleman, never to publish the method, and he pledged by his Christian faith to put it down in cipher, so that it would be unintelligible to anyone after his death” It seems that after pledging to Tartaglia that he would never divulge the secret of Tartaglia’s method, Cardano and his young pupil Ferrari, who had been a servant in Cardano’s household, discovered that del Ferro had originally solved the “cube and first power equal to the number” After having confirmed this by inspecting the papers of the late del Ferro, Cardano apparently no longer felt bound to his oath of secrecy to Tartaglia because del Ferro, and not Tartaglia, was the first discoverer Nevertheless, Cardano gives credit to Tartaglia in the Ars magna; and, nevertheless, Tartaglia was furious that Cardano had broken his oath. This was followed by bitter exchanges between Tartaglia and Cardano, with Ferrari being Cardano’s champion Examples from Cardano’s Ars magna

In Ars magna (The Great Art) we find the solution of the general cubic equation and the general published for the first time. “For I had been deceived by the words of Luca Paccioli, who denied that any more general rule could be discovered than his own. Notwithstanding the many things which I had already discovered, as is well known, I had despaired and had not attempted to look any further. Then, however, having received Tartaglia’s solution and seeking for the proof of it, I came to understand that there were a great many other things that could also be had.” p, q, a, b, c > 0

XI Cube and first power equal to the number x3 + px = q XII Cube equal to the first power and number x3 = px + q XIII Cube and number equal to the first power x3 + q = px XIV Cube equal to the square and number x3 = bx2 + d XV Cube and square equal to the number x3 + bx2 = d XVI Cube and number equal to the square x3 + d = bx2 XVII Cube, square, and first power equal to the number x3 + bx2 + cx = d XVIII Cube and first power equal to the square and number x3 + cx = bx2 + d XIX Cube and square equal to the first power and number x3 + bx2 = cx + d XX Cube equal to the square, first power, and number x3 = bx2 + cx + d XXI Cube and number equal to the square and first power x3 + d = bx2 + cx XXII Cube, first power, and number equal to the square x3 + cx + d = bx2 XXIII Cube, square, and number equal to the first power x2 + bx2 + d = cx In treating the different cases of the cubic in Chapters XI to XXIII of the Ars magna, with the exception of Chapter XVI, Cardano first provides a proof of the method (“demonstratio”), then he states the rule (“regula”) for solving the particular case, and follows the rule with one or more examples (“exemplum”); in Chapter XVI, Cardano states the rule before providing a proof of the method. His proofs are in the Greek geometrical tradition. To ease the reading in the following examples, we write, for instance, x3 + 6x = 20 where Cardano writes “a cube and 6 things equal 20” (“cubus & 6 positiones, æquantur 20”), and

q3 √ q3 √ 108 + 10 − 108 − 10 instead of Cardano’s

“R/ v : cub : R/ 108 p : 10 m : R/ v : cubica R/ 108 m : 10”. Ch XI, “On the Cube and First Power Equal to the Number”

Rule

Cube one-third the coefficient of x; add to it the square of one-half the constant of the equation; and take the square root of the whole. You will duplicate this, and to one of the two you add one-half the number you have already squared and from the other you subtract one-half the same. You will then have a binomium and its apotome. Then, subtracting the of the apotome from the cube root of the binomium the remainder [or] that which is left is the value of x. Direct substitution shows 2 is a solution (the only real solution).

For example,

cubus & 6 positiones, æquantur 20 x 3 + 6x = 20.

Cube 2, one-third of 6, making 8; square 10, one-half the constant; 100 results. Add 100 and 8, making√ 108, the square root of which is 108. This you will duplicate: to one add 10, one-half the constant, and from the other subtract the √same. Thus you will obtain the√ binomium 108 + 10 and its apotome 108 − 10. Take the cube roots of these. Subtract [the cube root of the] apotome from that of the binomium and you will have the value of x:

q3 √ q3 √ 108 + 10 − 108 − 10 R/ v : cub : R/ 108 p : 10 m : R/ v : cubica R/ 108 m : 10 For example,

cubus & 6 positiones, æquantur 20 x 3 + 6x = 20.

Cube 2, one-third of 6, making 8; square 10, one-half the constant; 100 results. Add 100 and 8, making√ 108, the square root of which is 108. This you will duplicate: to one add 10, one-half the constant, and from the other subtract the √same. Thus you will obtain the√ binomium 108 + 10 and its apotome 108 − 10. Take the cube roots of these. Subtract [the cube root of the] apotome from that of the binomium and you will have the value of x:

q3 √ q3 √ 108 + 10 − 108 − 10 R/ v : cub : R/ 108 p : 10 m : R/ v : cubica R/ 108 m : 10

Direct substitution shows 2 is a solution (the only real solution). x3 + 6x = 20 =⇒ x3 + 6x − 20 = 0

p = 6, q = −20

q2 p3 ∆ = 4 + 27 = 108 > 0 =⇒ one real root

q3 √ q3 √ x1 = 108 + 10 − 108 − 10 = 2,

x2 = −1 + 3i, x3 = −1 − 3i = −1 + 3i

= −1 − 3i

The roots Cardano missed:

q √ q √  1 3 3 x2 = − 2 10 + 108 + 10 − 108

√ q √ q √  3 3 3 + i 2 10 + 108 − 10 − 108

q √ q √  1 3 3 x3 = − 2 10 + 108 + 10 − 108

√ q √  q √  3 3 3 − i 2 10 + 108 − 10 − 108 The roots Cardano missed:

q √ q √  1 3 3 x2 = − 2 10 + 108 + 10 − 108

√ q √ q √  3 3 3 + i 2 10 + 108 − 10 − 108 = −1 + 3i

q √ q √  1 3 3 x3 = − 2 10 + 108 + 10 − 108

√ q √  q √  3 3 3 − i 2 10 + 108 − 10 − 108 = −1 − 3i For example, x 3 = 6x + 40. Raise 2, one-third the coefficient of x, to the cube, which makes 8; subtract this from 400, the square of 20, one-half the constant, making 392; the square √ root of this added to 20 makes 20 + 392, and subtracted from 20 makes √ p3 √ p3 √ 20 − 392; the sum of the cube roots of these, 20 + 392 + 20 − 392, is the value of x.

p3 √ p3 √ Exercise: Show that 20 + 392 + 20 − 392 = 4.

Ch XII, “On the Cube Equal to the First Power and Number”

The rule, therefore, is: When the cube of one-third the coefficient of x is not greater than the square of one-half the constant of the equation, subtract the former from the latter and add the square root of the remainder to one-half the constant of the equation and, again, subtract it from the same half, and you will have, as was said, a binomium and its apotome, the sum of the cube roots of which constitutes the value of x. p3 √ p3 √ Exercise: Show that 20 + 392 + 20 − 392 = 4.

Ch XII, “On the Cube Equal to the First Power and Number”

The rule, therefore, is: When the cube of one-third the coefficient of x is not greater than the square of one-half the constant of the equation, subtract the former from the latter and add the square root of the remainder to one-half the constant of the equation and, again, subtract it from the same half, and you will have, as was said, a binomium and its apotome, the sum of the cube roots of which constitutes the value of x. For example, x 3 = 6x + 40. Raise 2, one-third the coefficient of x, to the cube, which makes 8; subtract this from 400, the square of 20, one-half the constant, making 392; the square √ root of this added to 20 makes 20 + 392, and subtracted from 20 makes √ p3 √ p3 √ 20 − 392; the sum of the cube roots of these, 20 + 392 + 20 − 392, is the value of x. Ch XII, “On the Cube Equal to the First Power and Number”

The rule, therefore, is: When the cube of one-third the coefficient of x is not greater than the square of one-half the constant of the equation, subtract the former from the latter and add the square root of the remainder to one-half the constant of the equation and, again, subtract it from the same half, and you will have, as was said, a binomium and its apotome, the sum of the cube roots of which constitutes the value of x. For example, x 3 = 6x + 40. Raise 2, one-third the coefficient of x, to the cube, which makes 8; subtract this from 400, the square of 20, one-half the constant, making 392; the square √ root of this added to 20 makes 20 + 392, and subtracted from 20 makes √ p3 √ p3 √ 20 − 392; the sum of the cube roots of these, 20 + 392 + 20 − 392, is the value of x.

p3 √ p3 √ Exercise: Show that 20 + 392 + 20 − 392 = 4. x3 = 6x + 40 =⇒ x3 − 6x − 40 = 0

p = −6, q = −40

q2 p3 ∆ = 4 + 27 = 392 > 0 =⇒ one real root

q3 √ q3 √ x1 = 20 + 392 + 20 − 392 = 4,

√ √ x2 = −2 + i 6, x3 = −2 − i 6 The roots Cardano missed:

q √ q √  1 3 3 x2 = − 2 20 + 392 + 20 − 392

√ q √ q √  √ 3 3 3 + i 2 20 + 392 − 20 − 392 = −2 + i 6

q √ q √  1 3 3 x3 = − 2 20 + 392 + 20 − 392

√ q √ q √  √ 3 3 3 − i 2 20 + 392 − 20 − 392 = −2 − i 6 Example: x 3 + 3 = 8x. Solving y 3 = 8y + 3

1 according to the preceding rule, I obtain 3. The square of one-half of this is 2 4 , 3 which multiplied by 3 is 6 4 . Subtracting this from 8, the coefficient of x, leaves 1 1 1 4 , the square root of which added to or subtracted from 1 2 , which is one-half the solution for the cube equal to the first power and constant, gives both 1 q 1 1 q 1 solutions which were being sought. One is 1 2 + 1 4 , the other 1 2 − 1 4 .

Ch XIII, “On the Cube and Number Equal to the First Power”

The rule, therefore, is: When the cube and the constant are equal to the first power, find the solution for the cube equal to the same number of y’s and the same constant; take three times the square of one-half of this and subtract it from the coefficient of the first power; and the square root of the remainder added to or subtracted from one-half the solution for the cube equal to y plus the constant gives the solution for the cube and constant equal to x. Ch XIII, “On the Cube and Number Equal to the First Power”

The rule, therefore, is: When the cube and the constant are equal to the first power, find the solution for the cube equal to the same number of y’s and the same constant; take three times the square of one-half of this and subtract it from the coefficient of the first power; and the square root of the remainder added to or subtracted from one-half the solution for the cube equal to y plus the constant gives the solution for the cube and constant equal to x. Example: x 3 + 3 = 8x. Solving y 3 = 8y + 3

1 according to the preceding rule, I obtain 3. The square of one-half of this is 2 4 , 3 which multiplied by 3 is 6 4 . Subtracting this from 8, the coefficient of x, leaves 1 1 1 4 , the square root of which added to or subtracted from 1 2 , which is one-half the solution for the cube equal to the first power and constant, gives both 1 q 1 1 q 1 solutions which were being sought. One is 1 2 + 1 4 , the other 1 2 − 1 4 . x3 + 3 = 8x =⇒ x3 − 8x + 3 = 0

p = −8, q = 3

q2 p3 1805 ∆ = 4 + 27 = − 108 < 0 =⇒ three real roots

q 1 1 x1 = 1 2 + 1 4 ,

1 q 1 x2 = −3, x3 = 1 2 − 1 4 The root Cardano missed:

r q r q  1 3 3 1805 3 3 1805 x2 = − 2 − 2 + i 108 + − 2 − i 108

√ r q r q  3 3 3 1805 3 3 1805 + i 2 − 2 + i 108 − − 2 − i 108 = −3 Cardano’s rules in modern notation

Ch XI, “On the Cube and First Power Equal to the Number”

rq rq 3 3 q 2 p 3 q 3 q 2 p 3 q x + px = q : x = 2 + 3 + 2 − 2 + 3 − 2

Ch XII, “On the Cube Equal to the First Power and Number”

r q r q 3 3 q q 2 p 3 3 q q 2 p 3 x = px + q : x = 2 + 2 + 3 − 2 + 2 − 3

Ch XIII, “On the Cube and Number Equal to the First Power” q 3 3 y y 2 x + q = px → y = py + q : x = 2 ± p − 3 2 Compare to the cubic formula: For any real numbers p and q,

x3 + px + q = 0 : has solutions x1, x2, and x3 given by

x1 = r + s, x2 = rω2 + sω3, x3 = rω3 + sω2, where r q r q 3 q q2 p3 3 q q2 p3 r = − 2 + 4 + 27 , s = − 2 − 4 + 27 ,

√ √ 1 3 1 3 ω2 = − 2 + 2 i, ω3 = − 2 − 2 i. Cubics not in reduced form

Cardano tells how to transform general a cubic equation,

x3 = bx2 + d, x3 + bx2 + cx = d, &c., into a reduced cubic equation,

x3 + px = q, x3 = px + q, or x3 + q = px,

b essentially by making the change of x = y − 3 . For example, x 3 + 6x 2 = 100. The cube of 2 is 8 which, being multiplied by 2, is 16. Subtract this from 100 and you will have y 3 = 84 + 12y. [. . . ]

Ch XV, “On the Cube and Square Equal to the Number”

Cube one-third the coefficient of x 2, multiply the result by 2, and take the difference between this and the constant of the equation. [The result is the constant of a new equation.] Then multiply the square of one-third the coefficient of x 2 by 3 and you will have [the number of] y’s which are equal to the cube and the constant, if twice the cube is greater than the constant of the equation, or [the number of] y’s which equal the cube, if the difference between these two numbers is zero. Having derived the solution [for this equation], subtract one-third the coefficient of x 2 from it and the remainder is the value of x. Ch XV, “On the Cube and Square Equal to the Number”

Cube one-third the coefficient of x 2, multiply the result by 2, and take the difference between this and the constant of the equation. [The result is the constant of a new equation.] Then multiply the square of one-third the coefficient of x 2 by 3 and you will have [the number of] y’s which are equal to the cube and the constant, if twice the cube is greater than the constant of the equation, or [the number of] y’s which equal the cube, if the difference between these two numbers is zero. Having derived the solution [for this equation], subtract one-third the coefficient of x 2 from it and the remainder is the value of x. For example, x 3 + 6x 2 = 100. The cube of 2 is 8 which, being multiplied by 2, is 16. Subtract this from 100 and you will have y 3 = 84 + 12y. [. . . ] Example . . . 1 2 1 3 x + 3 x + 9 x = 19. Turn these into whole numbers and you will have

x 3 + 3x 2 + 9x = 171,

for all are multiplied by 9.. . . Therefore

y 3 + 6y = 178

[. . . ]

Ch XVII, “On the Cube, Square, and First Power Equal to the Number”

Cube one-third the coefficient of the square (which we show by this sign, Tpqd˜ [Tertia pars numeri quadratorum]) and add it to the constant. Then multiply the coefficient of the square by one-third itself, and the difference between this product and the coefficient of x is the number of y’s to be added to the cube if the product is less than the given coefficient of x or to be added to the constant if the product is greater than the given coefficient of x. [. . . ] Ch XVII, “On the Cube, Square, and First Power Equal to the Number”

Cube one-third the coefficient of the square (which we show by this sign, Tpqd˜ [Tertia pars numeri quadratorum]) and add it to the constant. Then multiply the coefficient of the square by one-third itself, and the difference between this product and the coefficient of x is the number of y’s to be added to the cube if the product is less than the given coefficient of x or to be added to the constant if the product is greater than the given coefficient of x. [. . . ] Example . . . 1 2 1 3 x + 3 x + 9 x = 19. Turn these into whole numbers and you will have

x 3 + 3x 2 + 9x = 171,

for all are multiplied by 9.. . . Therefore

y 3 + 6y = 178

[. . . ] Problem III An oracle ordered a prince to build a sacred building whose space should be 400 cubits, the length being six cubits more than the width, and the width three cubits more than the height. These quantities are to be found. Let the altitude be x; the width be x + 3; and the length be x + 9. Multiplied in turn, you will have x 3 + 12x 2 + 27x = 400. Add the cube of 4, one-third the coefficient of x 2, which is 64, to 400 and 464 results. Multiply 12, the coefficient of x 2, by one-third itself, making 48, the difference between which and 27 is 21, the number of y’s which equal the cube and constant. Therefore multiply 21 by 4, one-third the coefficient of x 2, and 84 results; take the difference between this and 464, which is 380; add this to y, since the first sum of numbers was greater than the second number produced; and you will have

y 3 = 21y + 380.

[. . . ] The quartic formula Ch XXXIX, “On the Rule by Which We Find an Unknown Quantity in Several Stages”

Rule II 2. There is another rule, more noble than the preceding. It is ’s, who gave it to me on my request. Through it we have all the solutions for equations of the fourth power, square, first power, and number, or of the fourth power, cube, square, and number.. . . [. . . ] Reducing this to a rule: It is always sufficient to have b3 plus one and a quarter times the coefficient of x 2 [for the coefficient of b2] plus such a number of b’s as was the original constant of the equation. Thus, if we had

x 4 + 12x 2 + 36 = 6x 2 + 60x,

we will have b3 + 15b2 + 36b = 450, .... that has solutions

1 √ √ √ 1 √ √ √ y1 = 2 ( z1 + z2 + z3), y3 = − 2 ( z1 − z2 + z3), 1 √ √ √ 1 √ √ √ y2 = 2 ( z1 − z2 − z3), y4 = − 2 ( z1 + z2 − z3),

where z1, z2, and z3 satisfy the cubic resolvant

z3 + 2pz2 + (p2 − 4r)z − q2 = 0

2 and z1z2z3 = q > 0.

Given the general quartic equation

x4 + bx3 + cx2 + dx + e = 0, make the linear substitution x = y − b/4 to obtain the reduced quartic equation y 4 + py 2 + qy + r = 0 that has solutions

1 √ √ √ 1 √ √ √ y1 = 2 ( z1 + z2 + z3), y3 = − 2 ( z1 − z2 + z3), 1 √ √ √ 1 √ √ √ y2 = 2 ( z1 − z2 − z3), y4 = − 2 ( z1 + z2 − z3),

where z1, z2, and z3 satisfy the cubic resolvant

2 and z1z2z3 = q > 0.

Given the general quartic equation

x4 + bx3 + cx2 + dx + e = 0, make the linear substitution x = y − b/4 to obtain the reduced quartic equation y 4 + py 2 + qy + r = 0

z3 + 2pz2 + (p2 − 4r)z − q2 = 0 Given the general quartic equation

x4 + bx3 + cx2 + dx + e = 0, make the linear substitution x = y − b/4 to obtain the reduced quartic equation y 4 + py 2 + qy + r = 0 that has solutions

1 √ √ √ 1 √ √ √ y1 = 2 ( z1 + z2 + z3), y3 = − 2 ( z1 − z2 + z3), 1 √ √ √ 1 √ √ √ y2 = 2 ( z1 − z2 − z3), y4 = − 2 ( z1 + z2 − z3), where z1, z2, and z3 satisfy the cubic resolvant

z3 + 2pz2 + (p2 − 4r)z − q2 = 0

2 and z1z2z3 = q > 0. The quintic formula That is, there is not one for the general quintic, but particular quintics may be solved using only radicals. E.g., Wolfram2 gives the example x5 + 20x + 32 = 0 that has for one solution

There is not one using only radicals.

2 There is not one using only radicals.

That is, there is not one for the general quintic, but particular quintics may be solved using only radicals. E.g., Wolfram2 gives the example x5 + 20x + 32 = 0 that has for one solution

2 (r 1 5 √ q √ q √ − 2500 5 + 250 50 − 10 5 − 750 50 + 10 5 5 r 5 √ q √ q √ − 2500 5 − 750 50 − 10 5 − 250 50 + 10 5 r 5 √ q √ q √ + 2500 5 + 750 50 − 10 5 + 250 50 + 10 5 r ) 5 √ q √ q √ + 2500 5 − 250 50 − 10 5 + 750 50 + 10 5 . First proposed proof by Italian Paolo Ruffini (1765–1822) in 1798; no one could understand it Complete proof by Niels Henrik Abel (1802–1829) in mid-1820s Abel went on to attempt (i) to find all equations of any given degree that can be solved by radicals; (ii) to decide whether a given equation can be solved by radicals. Not successful before his death Picked up by Evariste Galois (1811–1832). This led to what is now called that describes exactly when a equation is solvable by radicals Complex numbers Ch XXXVII, “On the Rule (Triple) for Postulating a Negative”

Imaginary numbers were not understood during Cardano’s day beyond their being a convenient contrivance. When confronted with an equation like

x2 + 1 = 0 or x2 + 2x + 2 = 0,

one could simply say that the equation has no solution. Nevertheless, Cardano did address the appearance of the square root of negative numbers in Chapter XXXVII of the Ars magna. This was the first time that the square root of a was written down explicitly. In “Rule II” of that chapter, Cardano considers a problem that is equivalent to solving the equation x(10 − x) = 30 or x(10 − x) = 40: The second species of negative assumption involves the square root of a negative. I will give an example: If it should be said, Divide 10 into two parts the product of which is 30 or 40, it is clear that this case is impossible. Nevertheless, we will work thus: We divide 10 into two equal parts, making each 5. These we square, making 25. Subtract 40, if you will, from the 25 thus produced . . . leaving a remainder of −15, the square root of which added to or subtracted from√5 gives parts the√ product of which is 40. These will be 5 + −15 and 5 − −15. y

(5, 25)

We can see from the graph of

x(10 − x) = y

to the left why the solutions of

x(10 − x) = 40

are complex: √ √ 5 + −15 and 5 − −15.

x So progresses subtlety the end of which, as is said, is as refined as it is useless.

Cardano goes on to give a geometrical demonstration of the solution, in which he says: √ . . . you will have to imagine√ −15 . . . and√ you will have that√ which you seek,√ namely 5 + 25 − 40 and 5 − 25 − 40, or 5 + −15 and √5 − −15. Putting√ aside the mental tortures involved, multiply 5 + −15 by 5 − −15, making 25 − (−15) which is +15. Hence this product is 40.. . . Cardano goes on to give a geometrical demonstration of the solution, in which he says: √ . . . you will have to imagine√ −15 . . . and√ you will have that√ which you seek,√ namely 5 + 25 − 40 and 5 − 25 − 40, or 5 + −15 and √5 − −15. Putting√ aside the mental tortures involved, multiply 5 + −15 by 5 − −15, making 25 − (−15) which is +15. Hence this product is 40.. . . So progresses arithmetic subtlety the end of which, as is said, is as refined as it is useless. Exit Ars magna Enter Bombelli’s famous example Raphael Bombelli (1526–1572)

L’Algebra, first printed in 1569 First serious study of complex numbers is motivated by solutions of the cubic equation, not the quadratic “I have found another kind of cubic root of a polynomial which is very different from the others. This [cubic root] arises in the chapter dealing with the equation of the kind x3 = px + q, when p3/27 > q2/4, as we will show in that chapter.” The Cardano’s formula produced the solution q √ q √ 3 2 + −121 + 3 2 − −121

Direct substitution shows that 4 is a solution Bombelli showed that, in fact, q √ q √ 3 2 + −121 + 3 2 − −121 = 4

Thus, somehow real numbers can be expressed using square roots of negative numbers in a way that is not obvious.

It was the combination of Cardano’s cubic formula and Bombelli’s famous equation, x3 = 15x + 4, that began a serious study of imaginary and complex numbers. Direct substitution shows that 4 is a solution Bombelli showed that, in fact, q √ q √ 3 2 + −121 + 3 2 − −121 = 4

Thus, somehow real numbers can be expressed using square roots of negative numbers in a way that is not obvious.

It was the combination of Cardano’s cubic formula and Bombelli’s famous equation, x3 = 15x + 4, that began a serious study of imaginary and complex numbers.

The Cardano’s formula produced the solution q √ q √ 3 2 + −121 + 3 2 − −121 It was the combination of Cardano’s cubic formula and Bombelli’s famous equation, x3 = 15x + 4, that began a serious study of imaginary and complex numbers.

The Cardano’s formula produced the solution q √ q √ 3 2 + −121 + 3 2 − −121

Direct substitution shows that 4 is a solution Bombelli showed that, in fact, q √ q √ 3 2 + −121 + 3 2 − −121 = 4

Thus, somehow real numbers can be expressed using square roots of negative numbers in a way that is not obvious. Recall Cardano’s cubic formula

Let p and q be any real numbers. Then the reduced or depressed cubic equation x3 + px + q = 0

has solutions x1, x2, and x3 given by

x1 = r + s, x2 = rω2 + sω3, x3 = rω3 + sω2, where r q r q 3 q q2 p3 3 q q2 p3 r = − 2 + 4 + 27 , s = − 2 − 4 + 27 , √ √ 1 3 1 3 ω2 = − 2 + 2 i, ω3 = − 2 − 2 i.

q2 p3 ∆ = 4 + 27 : three real roots if ∆ < 0 (irreducible case), two real roots if ∆ = 0, one if ∆ > 0 “This kind of square root has in its calculation [algorismo] different operations than the others and has a different name. Since when p3/27 > q2/4, the square root of their difference can be neither positive nor negative, therefore I will call it ‘more than minus’ [‘plus of minus’] when it should be added and ‘less than minus’ [‘minus of minus’] when it should be subtracted.. . . I will first treat multiplication, giving the law of plus and minus”

Plus times more than minus makes more than minus (+)(+i) = +i Minus times more than minus makes less than minus (−)(+i) = −i Plus times less than minus makes les than minus (+)(−i) = −i Minus times less than minus makes more than minus (−)(−i) = +i More than minus times more than minus makes minus (+i)(+i) = − More than minus times less than minus makes plus (+i)(−i) = + Less than minus times more than minus makes plus (−i)(+i) = + Less than minus times less than minus makes minus (−i)(−i) = − Bombelli’s famous equation: x 3 = 15x + 4

L’Algebra: Ars magna:

x 3 = 6x + 40. Raise 2, one-third the coefficient of x, to the cube, which makes 8; subtract this from 400, the square of 20, one-half the constant, making 392; the square root of this added to √ 20 makes 20 + 392, and subtracted from 20 √ makes 20 − 392; the sum of the cube roots p3 √ p3 √ of these, 20 + 392 + 20 − 392, is the value of x.

q3 √ q3 √ x = 2 + −121 + 2 − −121 x3 = 15x + 4 cubic formula produces the solution q √ q √ x = 3 2 + −121 + 3 2 − −121 √ √ = 3 2 + 11i + 3 2 − 11i direct substitution shows 4 is a solution could it be that √ √ 3 2 + 11i + 3 2 − 11i = 4 ? Bombelli’s genious was to presume that the two cube roots are, themselves, complex conjugates, that is to say, √ √ 3 2 + 11i = a + bi and 3 2 − 11i = a − bi.

(There was no reason at that time to think that the cube root of a complex number may be a complex number.) Then, by cubing the relations and setting equal real and imaginary parts, he proceeded to show that, in fact, √ √ 3 2 + 11i = 2 + i and 3 2 − 11i = 2 − i, so that √ √ 3 2 + 11i + 3 2 − 11i = (2 + i) + (2 − i) = 4. And now complex numbers could no longer be swept under the rug, for the cubic formula expresses some real number solutions of cubic equations using square roots of negative numbers: a real puzzler! Thus, these “sophistic,” “impossible,” “imaginary,” and “useless” objects no longer could be ignored.

About all this, Bombelli says interestingly:

It was a wild thought in the judgement of many; and I too for a long time was of the same opinion. The whole matter seemed to rest on sophistry rather than on truth. Yet I sought so long, until I actually proved this to be case. About all this, Bombelli says interestingly:

It was a wild thought in the judgement of many; and I too for a long time was of the same opinion. The whole matter seemed to rest on sophistry rather than on truth. Yet I sought so long, until I actually proved this to be case.

And now complex numbers could no longer be swept under the rug, for the cubic formula expresses some real number solutions of cubic equations using square roots of negative numbers: a real puzzler! Thus, these “sophistic,” “impossible,” “imaginary,” and “useless” objects no longer could be ignored. Brief history of complex numbers Complex numbers: 17th century

Even as late as the seventeenth century, prominent mathematicians like René Descartes (1596–1650) referred to positive roots of an equation as “true” roots and negative roots as “false” roots; Descartes called roots that contained a square root of a negative number “imaginary.” By this time, finally, negative numbers were beginning to gain full acceptance in Europe, even though they had long ago been accepted in India and China, but it would be a while longer (not until the nineteenth century) before imaginary or complex numbers would enjoy an equivalent standing. Complex numbers: 18th century

Leonhard Euler√ (1707–1783) introduced using i for −1, a notation that he adopted late in his life. Even so, he did not consider complex numbers to be “true” numbers. In his Elements of Algebra (1770), Euler wrote: 143. And, since all numbers which it is possible to conceive, are either greater or less than 0, or are 0 itself, it is evident that we cannot rank the square root of a negative number amongst possible numbers, and we must therefore say that is an impossible quantity. In this manner we are led to the idea of numbers, which from their nature are impossible; and therefore they are usually called imaginary quantities, because they exist merely in the imagination. √ √ √ √ 144. All such expressions, as −1, −2, −3, −4, &c. are consequently impossible, or imaginary numbers, since they represent roots of negative quantities; and of such numbers we may truly assert that they are neither nothing, nor greater than nothing, nor less than nothing; which necessarily constitutes them imaginary, or impossible. 145. But notwithstanding this, these numbers present themselves to the mind; they exists in our imagination, and we√ still have a sufficient idea of them; since we know that by −4 is meant a number which, multiplied by itself, produces −4; for this reason also, nothing prevents us from making use of these imaginary numbers and employing them in calculation. Complex numbers: 19th century

The idea to represent complex numbers in a plane occurred to three persons from three different parts of Europe all about the turn of the nineteenth century: Caspar Wessel (1745–1818), a Norwegian surveyor and cartographer; Jean Robert Argand (1768–1822), a French-Swiss bookkeeper; and Carl Friedrich Gauss (1777–1855), arguably the greatest of all German mathematicians. Both Wessel and Argand proposed using line segments or “vectors” to represent complex numbers. Wessel presented his idea first to the Royal Academy of Sciences of Denmark in 1797, and then in a paper published in the Philosophical Transactions of the Academy in 1799. Argand published his idea in a booklet titled Essai sur une manière de représenter les quantités imaginarie dans les constructions géométriques that was privately printed in 1806. However, perhaps because neither Wessel nor Argand was well known in mathematics, their ideas went largely overlooked. It was not until Gauss proposed representing a complex number as a point in the plane that the plane geometric interpretation took hold. Apparently, Gauss had used his idea implicitly in his 1799 doctoral dissertation on the fundamental theorem of algebra; it was not until 1831 before Gauss’s idea was described publicly in a commentary on his paper Theoria Residuorum Biquadraticorum. The fundamental theorem of algebra

Today, the fundamental theorem of algebra may be stated thus:

Let a0, a1,..., an−1 be complex numbers. If n ≥ 1, then the equation

n n−1 2 x + an−1x + ··· + a2x + a1x + a0 = 0

has at least one complex number solution. Albert Girard (1595–1632)

The first explicit statement of the theorem, without a proof, was by Albert Girard in his work Invention nouvelle en l’algèbre (A New Discovery in Algebra, 1629): Theorem. Every . . . admits of as many solutions as the denomination of the highest quantity indicates.. . . By the “denomination of the highest quantity” is meant the degree of the equation, so Girard asserts that a polynomial of degree n has n solutions. Katz* tells us that Girard realized that some solutions may be repeated (multiplicity greater than one) and some may be imaginary (“impossible”). According to Katz, “In answer to the anticipated question of the value of these impossible solutions, Girard answered that ‘they are good for three things: for the certainty of the general rule, for being sure that there are no other solutions, and for its utility.’ ” Katz indicates that Girard provides√ the example √ x4 + 3 = 4x that has solutions 1, 1, −1 + i 2, and −1 − i 2.

*Victor J. Katz, A History of Mathematics: An Introduction, 3rd ed., Addision-Wesley (2009). René Descartes (1596–1650)

Some eight years later, Descartes, in his Third Book of La géométrie, would enunciate essentially Girard’s theorem, writing: Every equation can have as many distinct roots (values of the unknown quantity) as the number of dimensions of the unknown quantity in the equation. Katz remarks that Descartes uses the phrase “can have” instead of Girards “admits of” regarding the number of roots or solutions because, unlike Girard, Descartes considers only distinct roots and, at least earlier on, does not consider imaginary roots. Carl Friedrich Gauss (1777–1855)

Gauss was the first to prove the fundamental theorem of algebra; he did so in his doctoral thesis of 1799. Katz informs us that “Gauss was so intrigued with the fundamental theorem—that every polynomial p(x) with real coefficients has a real or complex root—that he published four different proofs of it, in 1799, 1815, 1816, and 1848. Each proof used in some form or other the geometric intepretation of complex numbers, although in the first three proofs Gauss hid this notion by considering the real and imaginary parts of the numbers separately.. . . It was only in his final proof in 1848 that Gauss believed mathematicians would be comfortable enough with the geometric interpretation of complex numbers so that he could use it explicitly. In fact, in that proof, similar to his first one, he even permitted the coefficients of the polynomial to be complex.” Ironically, Gauss himself did not consider a geometric interpretation to be a sufficient justification for complex numbers. References

Girolamo Cardano, The Rules of Algebra (Ars Magna), translated by T. Richard Witmer, Dover Publications, Inc. (1968).

Roger Cooke, The History of Mathematics: A Brief Course, second edition, Wiley Interscience (2005).

Victor J. Katz, A History of Mathematics: An Introduction, third edition, Addison-Wesley (2009).

Amy Shell-Gellasch and J. B. Thoo, Intersecting Mathematics and History: Topics in Elementary Mathematics from an Historical Viewpoint, in preparation.