How Solving Cubic Equations Led to the Discovery of Imaginary Numbers

In math, a complex number has a real and an imaginary component combined in the form of a + bi.

a and b are real numbers, which means that they can include whole numbers, repeating decimals, terminating decimals and those that go on forever without patterns– the transcendental numbers. Examples of transcendental numbers include e, π, eπ, probably Catalan’s constant ( 1 – 1/9 + 1/25 – 1/49 +1/64- …) and ii. i without its exponent is an imaginary number, the square root of -1.

Above, to the left, Gerolamo Cardano (1501 – 1576) was a Renaissance mathematician and inventor of the universal joint, the cardan. He published Tartaglia’s and Ferrari’s solution for 3rd(cubic) and 4th degree equations. The first European to use negative numbers, he also realized, while investigating the cubic, that imaginary numbers had to be considered. To the right, Raffaele Bombelli(1526 – 15) first treated the square root of -1 as a variable. The Italian mathematician also worked as a hydraulic engineer.

Many math students have their 1st encounter with i typically when solving second degree differential equations. They learn that a seemingly nonsensical entity becomes very useful in coming up with real solutions to real scientific and engineering problems. But it’s when trying to solve somewhat simpler equations that a 16th century Italian mathematicians first stumbled upon i.

Consider x3 + 3x2 – 2x – 2 = 0. This is a 3rd degree equation, dubbed cubic because its highest power is three. It can’t be solved by isolating the variable. Factoring won’t work either. Well, sort of. In this particular case, by fluke, you can use trial and error and quickly discover that x = 1 is one of the solutions. Then, as we will show later, you can subsequently use long division and find the other two solutions. But mathematicians wanted a more systematic technique because most numbers are not whole, so trial and error will, for cubic equations, fail miserably most of the time.

It turns out that the above can be reduced to something simpler if we could somehow eliminate the power of 2 term. How? Let’s say we replaced each x with a y+k term, where k is some constant that we will find.

x = y + k

Now consider a general case where we have ax3 + bx2 + …, If we replace each x with y + k then

a(y + k)3 + b(y + k)2 +….. = 0. Expanding this we will get
y3 + 3y2k + 3yk2 + k3 + by2 + 2byk + bk2 +…..

Since k will just be a number, we don’t care if k itself has a power of 2. But we do want to eliminate the square of any variable—here, terms with y2. Essentially we want 3y2k +by2 = 0, and that will happen if, in our substitution, we choose k=-b/3. Why? Well, it’s the value of k if we solve the above equation:

3y2k +by2 = 0. The y2 cancels; then 3k + b = 0, so k=-b/3.

Let’s apply it to x3 + 3x2 – 2x – 2 = 0. We let x = y + k, where k =-b/3= -3/3 =-1.

Hence, x = y -1 and we substitute it into the equation:

(y -1)3 + 3(y -1)2 – 2(y -1) – 2 = 0

Simplifying we get a 3rd degree equation free of a squared term, what’s called a depressed cubic:

y3 – 5y =- 2. But how do we solve that?

(Instead of y3+py=q, above is another way of representing a depressed cubic. )

We begin with the observation that if we expand the cube of difference of any two numbers, we obtain the following:

(s-t)3 = s3 – 3s2t +3st2 – t3. If we factor the st terms and leave the individual cubes by themselves on the right hand side of the equation, we obtain:

(s -t)3+3st(s -t) = s3 – t3. Now compare this to our equation:

y3 -5y = -2.

We notice that y = s – t; 3st=-5 and that s3 – t3 = -2.

We eliminate one variable, s. Since s =-5/(3t), we substitute it into s3 – t3 = -2 and after eliminating the variable from the denominator, we obtain

t6 – 2t3 + 125/27 = 0. If we let u = t3, we will get a simple quadratic equation!

u2– 2u + 125/27 = 0. With the quadratic formula we get u= 1± (7/9)√(-6).

Cardano knew that an expression such as u= 1± (7/9)√(-6) was meaningful and that it was somehow connected to the real solution, but he didn’t know what to do. It’s thanks to another Italian mathematician, Raffaele Bombelli who realized that we can do the following: let the square root of -6 be expressed as √(6)*√(-1), where√(-1) was defined by Bombelli to be the imaginary number i.

So u= 1± (7/9)i√(6). Now it seems that the solution to our problem does not have a real answer, but it’s merely an illusion. If we work with i, we can work our way backwards to the value of x, and we will discover that it gives us the trial-and-error real solution of x=1. How?

Recall that u = t3. Thus t = u(1/3), which we discovered is t= [ 1± (7/9)i√(6) ]1/3

That’s a real monstrosity, so there must be a simpler way to express the complex number, t.

Let t = a +bi =[ 1± (7/9)i√(6) ]1/3

That means that (a +bi)3 = 1 ± (7/9)i√(6). Now expand the left side and we get

a3 + 3a2bi -3ab2 – b2i = 1 ± (7/9)i√(6). Let’s separate the real and imaginary parts from the left hand side of the equation and factor:

a(a2-3b2) +bi (3a2-b) = 1 ± (7/9)i√(6).

If we focus on the real part of each equation we notice that a(a2-3b2)= 1. Could we be so lucky as to have a = 1 and a2-3b2= 1, given that 1(1) =1? Well, no. If we replace a with 1 and solve for b, we would get a b value of 0, which is impossible since we already know that t is a complex number. Maybe a= -1 and a2-3b2= -1, given that -1(-1) =1. If that’s the case, solving for b yields b=√(2/3) =(1/3)√(6). To check if these values make sense, see if they also satisfy the imaginary part: bi (3a2-b2) =(7/9)i√(6). Cancel the i‘s. Substitute in a = -1 and b=(1/3)√(6):

(1/3)√(6)[3(-1)2-(1/3)2√(6)2]=√(6)-(1/27)(6)√(6)=√(6)-(1/9)(2)√(6)= (7/9)√(6), which checks out as being the coefficient for the i in our expression!

Given, then, that t = a +bi = -1 +(1/3)√(6)i

Working backwards we had s =-5/(3t) =-5/(3[-1 +(1/3)√(6)i]) = 1 +(1/3)√(6)i

y = s -t = 1 +(1/3)√(6)i – [-1 +(1/3)√(6)i] = 2

The imaginary terms have cancelled out and led us to a real answer!

Finally we had x = y -1, so x = 2-1 =1.

We’re almost done.

If we consider the roots ( x intercepts) of the graph of y = x3 + 3x2 – 2x – 2, we need to set the x terms to zero and thus we’re back to our problem. y = x3 + 3x2 – 2x – 2 = 0. A third degree equation can have up to 3 x intercepts. This one does indeed have 3.

To obtain the remaining two, use the fact that if our first root is x=1, then x-1 is a factor of x3 + 3x2 – 2x – 2. By using long division, we can get that x3 + 3x2 – 2x – 2 =(x-1)(x2 + 4x + 2) = 0. Use the quadratic formula to solve x2 + 4x + 2 and we find that x = -2 +√(2)~ = -0.586 and x =-2 -√(2)~ = -3. 414, as confirmed by the graph.

Source for leading image: Matheepan Panchalingam, via Flickr. Biographical pictures are form Wikipedia.

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