Birth of Complex Numbers in Solving Cubic Equations

CardanBombeliEvaluated.nb 1

Math 421 è Fall 2006

Birth of complex numbers in solving cubic equations

Revised 9/6/06

The problem

To solve the cubic equation x3 + Ax2 + Bx + K = 0.

Strategy: è change-of-variable Ø new cubic with no quadratic term; è solve new cubic; è use the solutions of that to solve the original.

Reduction of cubic to depressed cubic (Anonymous, end of 14th century)

Temporarily replace x by u and rename the constant term K:

In[1]:= cubic = u3 + Au2 + Bu + K

Out[1]= K + Bu+ Au2 + u3

Make linear substitution

1 u = x - ÅÅÅÅ3 A

1 In[2]:= depressed = cubic .u→ x − A 3 A A 2 A 3 Out[2]= K + B − + x + A − + x + − + x 3 ê3 3

Collect coefficients of the powers of x: I M I M I M CardanBombeliEvaluated.nb 2

In[3]:= Collect depressed, x

2A3 AB A2 Out[3]= − + K + − + B x + x3 27 3 3 @ D That is the depressed cubic: no x2 term, so of form J N x3 + bx + c where

A3 2 A3 AB b =-ÅÅÅÅÅÅÅ3 + B, c = ÅÅÅÅÅÅÅÅÅÅÅ27 - ÅÅÅÅÅÅÅÅÅ3 + K.

In[4]:= depressedCubic = x3 + bx + c

Out[4]= c + bx+ x3

1 Exercise. The linear substitution just used was u = x - ÅÅÅÅ3 A. Among all possible linear substitutions u = x - cst, 1 why use cst = ÅÅÅÅ3 A?

del Ferro & Tartaglia solution of depressed cubic (Scipione del Ferro, 1515, and Niccolò Fontana aka "Tartaglia")

Scipione del Ferro and Niccolò Tartaglia discovered formula for a root of a depressed cubic x3 + bx + c. In Mathematica:

c c2 b3 c c2 b3 In[5]:= delFerroTartagliaRoot b_, c_ := 3 − + + + 3 − − + 2 4 27 2 4 27

In[6]:= delFerroTartagliaRoot@b, c D &''''''''''''''''''''''''''''''''$%%%%%%%%%%%%%%%%''''''''%%%%%%%''''' &''''''''''''''''''''''''''''''''$%%%%%%%%%%%%%%%%''''''''%%%%%%%''''' 1 3 1 3 c b3 c2 c b3 c2 Out[6]= − − + + − + + 2 27 4 @ 2 D 27 4 ê ê i y i y j z j z j $%%%%%%%%%%%%%%%%%%%%% z j $%%%%%%%%%%%%%%%%%%%%% z The del Ferro-Tartagliaj solutionz will appearj simpler if in thez depressed cubic you take b = 3 p and c = 2 q to obtain the form: k { k {

x3 + 3 px + 2 q

In[7]:= delFerroTartagliaRoot b, c . b −> 3 p, c → 2 q

1 3 1 3 Out[7]= −q − p3 + q2 + −q + p3 + q2 @ Dê 8 < ê ê Nicer formula from…è!!!!!!!!!!!!!!! è!!!!!!!!!!!!!!! I M I M CardanBombeliEvaluated.nb 3

In[8]:= niceDepressedCubic = depressedCubic . b → 3 p, c → 2 q

Out[8]= 2q+ 3px+ x3 ê 8 < …by new Mathematica function:

In[9]:= niceDelFerroTartagliaRoot p_, q_ := delFerroTartagliaRoot b, c . b −> 3 p, c → 2 q

So del Ferro-Tartaglia formula for root of a depressed cubic of form x3 + 3 px + 2 q is: @ D @ Dê 8 < In[10]:= niceDelFerroTartagliaRoot p, q

1 3 1 3 Out[10]= −q − p3 + q2 + −q + p3 + q2 @ D ê ê è!!!!!!!!!!!!!!! è!!!!!!!!!!!!!!! I M I M Cardan's general solution of the cubic (Girolamo Cardano, Ars magna, 1545)

Cardan uses the linear substitution that reduces a general cubic to a depressed cubic along with the del Ferro-Tartaglia formula for one solution of the depressed cubic to obtain a general formula for solving any cubic. Mathematica can do it, too:

In[11]:= Solve x3 + Ax2 + Bx + K 0, x TraditionalForm

Out[11]//TraditionalForm= A 1 x Ø-ÅÅÅÅÅÅ@+ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ -2 A3 + 9 BA- 27 K + 3Dêê3 4 KA3 - B2 A2 - 18 BKA+ 4 B3 + 27 K2 ^ 1 3 - 3 3 3 2 3 2 3 B - A2 3 -2 A3 + 9 BA- 27 K + 3 3 4 KA3 - B2 A2 - 18 BKA+ 4 B3 + 27 K2 ^ 1 3 , è!!!! è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!! !!!!!!!! !!!!! 99 A è!!!!1 II M H ê LM x Ø-ÅÅÅÅÅÅ - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 1 -Â 3 -2 A3 + 9 BA- 27 K + 3 3 4 KA3 - B2 A2 - 18 BKA+ 4 B3 + 27 K2 ^ 1 3 + è!!!!3 6 3 2 è!!!! è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! I H LM ë I I M H ê LM= 1 +Â 3 3 B - A2 322 3 -2 A3 + 9 BA- 27 K + 3 3 4 KA3 - B2 A2 - 18 BKA+ 4 B3 + 27 K2 ^ 1 3 , è!!!! è!!!! è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 9 II MI M H ê LM A è!!!!1 3 x Ø-ÅÅÅÅÅÅ - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 1 +Â 3 -2ê A + 9 BA- 27 K + 3 3 4 KA3 - B2 A2 - 18 BKA+ 4 B3 + 27 K2 ^ 1 3 + 3 è!!!!6 3 2 è!!!! è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! II MH LM ë I I M H ê LM= 1 -Â 3 3 B - A2 322 3 -2 A3 + 9 BA- 27 K + 3 3 4 KA3 - B2 A2 - 18 BKA+ 4 B3 + 27 K2 ^ 1 3 è!!!! è!!!! è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!! !!!!!!!!!!!!! 9 è!!!! II MI M H ê LM ê è!!!! è!!!! è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!! !!!!!!!! !!!!! II MH LM ë I I M H ê LM== CardanBombeliEvaluated.nb 4

Bombeli: use square-roots of negative numbers to obtain real roots of cubics (Rafael Bombeli, L'algebra, 1572)

ü Paradox

è A depressed cubic always has a real solution: x3 + 3 px + 2 q = 0 is equivalent to x3 =-3 px - 2 q, and the cube function y = x3 always intersects the line y =-3 px - 2 q in at least one point, no matter what p and q are!

In[12]:= Plot x3, x, −5, 5 , PlotStyle → Red ;

@ 8 < 2 D

-4 -2 2 4

-1

-2

1 3 1 3 è Yet del Ferro-Tartaglia formula -q - p3 + q2 + -q + p3 + q2

involves square-roots of negative numbers when q2 ê<-p3. ê è!!!!!!!!!!!!!!! è!!!!!!!!!!!!!!! I M I M

ü Example

In the depressed cubic equation x3 + 3 px + 2 q, take p = 5 and q =-2:

x3 - 15 x - 4 = 0 CardanBombeliEvaluated.nb 5

In[13]:= eqn = x3 − 15 x − 4 0

Out[13]= −4 − 15 x + x3 0

There must be a solution—the cubic y = x3 and the line y = 15 x + 4 must intersect:

In[14]:= Plot x3,15 x + 4 , x, −7, 7 , PlotStyle → Red, Blue ;

75 @8 < 8 < 8

-6 -4 -2 2 4 6 -25

-50

-75

In fact, x = 4 is a solution:

In[15]:= eqn .x→ 4

Out[15]= True ê

ü Bombeli's "wild thought"

In[16]:= niceDelFerroTartagliaRoot p, q

1 3 1 3 Out[16]= −q − p3 + q2 + −q + p3 + q2 @ D ê ê The depressed cubicè!!!!!!! has!!!!!!!! p =-5, q =-è!!!!!!!2. So…!!!!!!!! I M I M In[17]:= p3 + q2 . p →−5, q →−2

Out[17]= −121 ê 8 < … then del Ferro-Tartaglia solution in this example is:

3 3 3 3 2 + -121 + 2 - -121 = 2 + 11 -1 + 2 - 11 -1 The known solution x = 4 could be recovered from the del Ferro-Tartaglia if the two terms on the right have the forms "###############è!!!!!!!#### ####!!!!!### "###############è!!!!!!!#### ####!!!!!### "###############è######## !!!!!!!#### "###############è######## !!!!!!!####

3 3 2 + 11 -1 = m + n -1, 2 - 11 -1 = m - n -1. (*)

for suitable"############### m and n.è####### # !!!!!!!#### è!!!!!!! "###############è######## !!!!!!!#### è!!!!!!! How? Well, the sum of these two solutions has the form… CardanBombeliEvaluated.nb 6

In[18]:= m + n −1 + m − n −1

Out[18]= 2m è!!!!!!! è!!!!!!! I M I M …so here we must have: 2 m = 4 (the known solution).

Thus to obtain the solution x = 4, Bombeli wants to find n for which (*) holds with m = 2, in other words, 3 3 2 + n -1 = 2 + 11 -1, 2 - n -1 = 2 - 11 -1

Pretend usual rules of algebra hold for expressions involving è!!!!!!! è!!!!!!! è!!!!!!! è!!!!!!! I M I M Â= -1, and assume the special rule è!!!!!!! 2 Â2 = -1 =-1. 3 Then multiply out to calcluate 2 + n -1 : è!!!!!!! I M 3 In[19]:= theCube Expand 2 n 1 = +è!!!!!!! − I M Out[19]= 8 + 12 n − 6n2 −n3 è!!!!!!! AI M E To compare result with 2 + 11 -1 , separate the "real part" from the part involving Â:

In[20]:= ComplexExpand theCube è!!!!!!! Out[20]= 8 − 6n2 + 12 n − n3 @ D That is supposed to equal 2 + 11 -1: H L In[21]:= Solve 8 − 6 n2,12 n − n3 2, 11 ,n è!!!!!!! Out[21]= n → 1 @8 < 8 < D

Exercise.88 The "usual<< rules of algebra" include such identities as: x + y + u + v = x + u + y + v , x + y u + v = xu + yv + xv + yu, k x + y = kx + ky, kH xyL = Hkx y L= x Hky, L H L Hx + yLH+ z =L x + y + z . These identitiesH Lhold for real numbers x, y, u, v, k, z. Assume suchH LidentitiesH L hold forH "complexL numbers" as well—for numbers of the form a + b Â where a and b are real. And stillH assumeL that Â2 =ÂÂ=-H 1.L Then put each of the following into the form y=u +Âv with u and v real: a + b Â + c + d Â , a + b Â c + d Â

H L H L H LH L CardanBombeliEvaluated.nb 7

ü The moral

Square-roots of negative numbers are useful (essential?) in obtaining real roots of certain cubic equations.

(But what are such "complex" numbers? That's what's next in this course!)

Appendices

ü Appendix 1: Obtaining real imaginary parts of complex numbers involving symbolic quantities

How in Mathematica can you solve for n the equation 8 - 6 n2 +Â12 n - n3 = 2 + 11 -1 without copying and pasting, or reading off and retyping, the real and imaginary parts (as was done above)? è!!!!!!! H L Use Re and Im.

In[22]:= ?Re

Re z gives the real part of the complex number z. More…

In[23]:= ?Im @ D Im z gives the imaginary part of the complex number z. More…

In[24]:= theCube @ D Out[24]= 8 + 12 n − 6n2 −n3

In[25]:= Re theCube

Out[25]= 8 − 12 Im n + Im n3 − 6Re n2 @ D Try some more: @ D @ D @ D In[26]:= Simplify Re theCube

Out[26]= 8 − 12 Im n + Im n3 − 6Re n2 @ @ DD Stymied! Why? Because Mathematica doesn't know n should be real. Tell it so: @ D @ D @ D In[27]:= ComplexExpand Re theCube

Out[27]= 8 − 6n2 @ @ DD CardanBombeliEvaluated.nb 8

In[28]:= realPart, imaginaryPart = ComplexExpand Re theCube , ComplexExpand Im theCube

Out[28]= 8 − 6n2,12n− n3 8 < 8 @ @ DD @ @ DD< Now once again you could solve 8 - 6 n2 +Â12 n - n3 = 2 + 11 -1: 8 < In[29]:= Solve realPart, imaginaryPart == 2, 11 ,n è!!!!!!! H L Out[29]= n → 1 @8 < 8 < D

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