4.4: Complex numbers

Up to this point, the chapter and solutions have largely avoided mentioning complex numbers. However, the present chapter would be incomplete were we not to interpret some of the earlier material in terms of complex numbers. Readers who have already met complex numbers will probably still find much in the next two sections that is new. Those for whom complex numbers are as yet unfamiliar should muddle through as best they can, and may then be motivated to learn more in due course.

We already know that the square x² of any real number x is ⩾ 0.

• If a = 0, then the equation x² = a has exactly one root, namely x = 0;
• if a > 0, then the equation x² = a has exactly two roots – namely ±$\sqrt{a}$, where $\sqrt{a}$ denotes the root that is positive;
• if a < 0, then the equation x² = a has no real roots.

And that is where the matter would have rested.

From a modern perspective, we can see that complex numbers are implicit in the formula for the roots of a quadratic equation: complex numbers become explicit as soon as the coefficients of a quadratic ax² + bx + c give rise to a negative discriminant b² − 4ac < 0.

But this may not have been quite how complex numbers were discovered. Contrary to oft-repeated myths, complex numbers may not have forced themselves on our attention by someone asking about “solutions to the quadratic equation x² = −1”. As long as we inhabit the domain of real numbers, we can be sure that no known number x could possibly have such a square, so we are unlikely to go in search of it.

New ideas in the history of mathematics tend to emerge when a fresh analysis of familiar entities forces us to consider the possible existence of some previously unsuspected universe. In the time from the ancient world up to the fifteenth century, the idea of “number”, and of calculation, was restricted to the world of real (usually positive) numbers. In such a world, quadratic equations with non-real solutions simply could not arise.

However, in the Brave New World of the Renaissance, where novelty, exploration, and discovery were part of the Zeitgeist, a general method for solving cubic equations was part of the as-yet-undiscovered “wild west” of mathematics, part of the mathematical New World which invited exploration. Notice that this was not a wildly speculative venture (like trying to solve the meaningless equation “x² = − 1”), since a cubic polynomial always has at least one real root. After three thousand years in which little progress had been made, the first half of the sixteenth century witnessed an astonishing burst of progress, resulting in the solution not only of cubic equations, but also of quartic equations. We postpone the details until Section 4.5. All we note here is that,

the general method for solving cubic equations published in 1545, was given as a procedure, illustrated by examples, that showed how to find genuinely real solutions to equations of the third degree having genuinely real (and positive) coefficients.

The procedure clearly worked. And it proceeded as follows:

Construct the real solution x as the sum x = u + v of two intermediate answers u and v - where the two summands u and v sometimes turned out to be what we would call “conjugate complex numbers”, whose imaginary parts cancelled out, leaving a real result for the required root x.

Those who devised the procedure had no desire to leave the real domain: they were focused on a problem in the real domain (a cubic equation with real coefficients, having a real root), and devised a general procedure to find that genuinely real root. But the procedure they discovered led the solver on a journey that sometimes “passed into the complex domain”, before returning to the real domain! (See Problem 129.)

Working with complex numbers depends on two skills - one very familiar, and one less so.

• The familiar skill is a willingness to work with a “number” in terms of its properties only, without wishing to evaluate it.

We are thoroughly familiar with this when we work with $\frac{2}{3}$ and other fractions: we know that $\frac{2}{3}=2\times \frac{1}{3}$ ; and all we know about $\frac{1}{3}$ is that “whenever we have 3 copies of $\frac{1}{3}$, we can simplify this to 1”. Much the same happens when we first learn to work with $\sqrt{2}$ , where we carry out such calculations as ${(1+\sqrt{2})}^2=3+2\sqrt{2}$ , based only on collecting up like terms and the fact that $\sqrt{2}\times \sqrt{2}$ can always be replaced by 2.

• The less familiar skill is easily overlooked. When, for whatever reason, we decide to allow solutions to the equation x² = − 1, three things need to be understood.

  − First, these new solutions come in pairs: if i is one solution of x² = −1 then −i is another (because (−1) x (−1) = 1 means that (−x)² = x² for all “numbers” x.

  − Second, the equation x² = − 1 has exactly two solutions - one the negative of the other (if x and y are both solutions, then x² = y², so x² − y² = (x − y)(x + y) = 0, so either x = y, or x = −y).

  − Third, we have no way of telling these two solutions apart: we know that each is the negative of the other, but there is no way of singling out one of them as “the main one” (as we could when defining the square root of a positive real such as 2). We can call them ±i, but they are each as good as the other. This important fact is often undermined by referring to one of these roots as $\sqrt{-1}$ (as if it were the dominant partner), and to the other as −$\sqrt{-1}$ (as if it were somehow just the “negative” of the main root).

The truth is that “$\sqrt{-1}$ ” is a serious abuse of notation, because there is no way to extend the definition of the function “$\sqrt{}$ ” in the way that this implies: when we try to “take square roots” of negative (or complex) numbers, the output is inescapably “two-valued”, so “$\sqrt{}$” is no longer a function. The two roots of x² = −1 are like Tweedledum and Tweedledee: we know there are two of them, and we know how they are related; but we have no way of distinguishing them, or of singling one of them out.

Once we accept this, we can write complex numbers in the form a + bi, where a and b are real numbers (just as we used to write numbers in the form $a+b\sqrt{2}$ where a and b are rational numbers). And we can proceed to add, subtract, multiply, and divide such expressions, and then collect up the “real” and “imaginary” parts to tidy up the answer.

Problem 128

(a) Write the inverse (a + bi)⁻¹ in the form c + di.

(b) Write down a quadratic equation with real coefficients, which has a + bi as one root (where a and b are real numbers).

Problem 129 Divide 10 into two parts, whose product is 40.

Problem 129 appears in Chapter XXXVII of Girolamo Cardano’s (1501–1576) book Ars Magna (1545). Having previously presented the general methods for solving quadratic, cubic, and quartic equations, he honestly confronts the phenomenon that his method for solving cubic equations (see Problem 135) produces the required real (and positive) solution x as a sum of complex conjugates u and v - involving not only negative numbers, but square roots of negative numbers. After presenting the formal solution of Problem 129, and having shown that the calculation works exactly as it should, he adds the bemused remark:

“So progresses arithmetic subtlety,
the end of which … is as refined as it is useless.”

Arithmetic with complex numbers in the form a + bi is done by carrying out the required operations, and then collecting up the “real” and “imaginary” parts as separate components - just as with adding vectors (a, b). We treat the two parts as Cartesian coordinates, and so identify the complex number a + bi with the point (a, b) in the complex plane.

The “Cartesian” representation a + bi is very convenient for addition. But the essential definition (and significance) of complex numbers is rooted in multiplication. And for multiplication it is often much better to work with complex numbers written in polar form. Suppose we mark the complex number w = a + bi in the complex plane.

The modulus |w| of w (often denoted by r) is the distance $r=\sqrt{a^2+b^2}$ of the complex number a + 6i from the origin in the complex plane.

The angle θ, measured anticlockwise from the positive real axis to the line joining the complex number w to the origin, is called the argument, $$Arg (w)=θ , of w.

It is then easy to check that a = r cos θ, b = r sin θ, and that

$$w=r(\cos \theta +i\sin \theta )$$

This is the polar form for w. Instead of focusing on the Cartesian coordinates a, b, the polar form pinpoints w in terms of

• its length, or modulus, r (which specifies the circle, with centre at the origin, on which the complex number w lies), and
• the argument 0 (which tells us where on this circle w is to be found).

Problem 130

(a) Given two complex numbers in polar form:

$$w=r(\cos \theta +i\sin \theta ),z=s(\cos \varphi +i\sin \varphi )$$

show that their product is precisely

$$wz=rs(\cos (\theta +\varphi )+i\sin (\theta +\varphi )).$$

(b) (de Moivre’s Theorem: Abraham de Moivre (1667–1754)) Prove that

$${(\cos \theta +i\sin \theta )}^n=\cos (n\theta )+i\sin (n\theta ).$$

(c) Prove that, if

$$z=r(\cos \theta +i\sin \theta )$$

satisfies zⁿ = 1 for some integer n, then r = 1.

The last three problems in this subsection look more closely at “roots of unity” − that is, roots of the polynomial equation xⁿ = 1. In the real domain, we know that:

(i) when n is odd, the equation xⁿ = 1 has exactly one root, namely x = 1; and

(ii) when n is even, the equation xⁿ = 1 has just two solutions, namely $x=\pm 1$.

In contrast, in the complex domain, there are n “n^th roots of unity”. Problem 130(c) shows that these “roots of unity” all lie on the unit circle, centered at the origin. And if we put $n\theta =2k\pi$ in Problem 130(b) we see that the n n^th roots of unity include the point “1 = cos0 + i sin0”, and are then equally spaced around that circle with $\theta =\frac{2k\pi }{n}(1⩽k⩽n-1)$ , and form the vertices of a regular n-gon.

Problem 131

(a) Find all the complex roots of unity of degree 3 (that is, the roots of x³ = 1) in surds form.

(b) Find all the complex roots of unity of degree 4 in surd form.

(c) Find all the complex roots of unity of degree 6 in surd form.

(d) Find all the complex roots of unity of degree 8 in surd form.

Problem 132 Use Problem 131(d) to factorise x⁴ + 1 as a product of four linear factors, and hence as a product of two quadratic polynomials with real coefficients.

Problem 133

(a) Find all the complex roots of unity of degree 5 in surd form.

(b) Factorise x⁵ − 1 as a product of one linear and two quadratic polynomials with real coefficients.