Complex number/Citable Version

The complex numbers $\mathbb {C}$ are numbers of the form $a+bi\scriptstyle$, obtained by adjoining the imaginary unit $i\scriptstyle$ to the real numbers (here $a\scriptstyle$ and $b\scriptstyle$ are reals). The number $i\scriptstyle$ can be thought of as a solution of the equation $x^{2}+1=0\scriptstyle$. In other words, its basic property is $i^{2}=-1\scriptstyle$. Of course, since the square of any real number is nonnegative, $i\scriptstyle$ cannot be a real number. A priori, it is not even clear whether such an object exists and that it can be called a number, i.e. whether we can associate with it some natural operations as addition or multiplication. Assuming, for a moment, that the answer is "yes", we may rewrite the first sentence above in symbols as

$\mathbb {C} =\{a+bi\;|\;a,b\in \mathbb {R} \}.$

Historical example

The need for complex numbers may have appeared for the first time during the sixteenth century, when Italian mathematicians like Scipione del Ferro, Niccolò Fontana Tartaglia, Gerolamo Cardano and Rafael Bombelli tried to solve cubic equations. Even for equations with three real solutions, the method they used sometimes required calculations with numbers whose squares are negative. Here is such an example (with modern notation). Let us consider the equation

$x^{3}=15x+4.\$

Cardano's method for solving it suggests looking for a solution by writing it as a sum $x=u+v$, where another condition on $u$ and $v$ is to be decided later. Recording this in the equation, we have, once the left member is expanded,

$u^{3}+3u^{2}v+3uv^{2}+v^{3}=15(u+v)+4,\$

which can be written as

$u^{3}+(3uv-15)(u+v)+v^{3}=4.\$

Now we choose the second condition on $\scriptstyle u$ and $\scriptstyle v$, namely $\scriptstyle 3uv-15=0$, or $\scriptstyle uv=5$. This implies that $\scriptstyle u^{3}$ and $\scriptstyle v^{3}$ are numbers which sum and product are given by

${\begin{cases}u^{3}v^{3}=125,\\u^{3}+v^{3}=4.\end{cases}}$

Now, it is a well-known fact that if a second degree polynomial $\scriptstyle x^{2}-sx+p$ has two roots, their sum is $\scriptstyle s$ and their product is $\scriptstyle p.$Hence we may find some values for $\scriptstyle u^{3}$ and $\scriptstyle v^{3}$ by solving the quadratic equation

$x^{2}-4x+125=0.\$

Its discriminant is $\scriptstyle \Delta =(-4)^{2}-4\cdot 125=-484=-22^{2}$, which is negative, so that the quadratic equation has no real solution: the usual formulae giving the solutions require taking the square root of the discriminant, which is undefined here.

Well, let us be bold and write $\scriptstyle \Delta =\left(22{\sqrt {-1}}\right)^{2}$. Here, the symbol $\scriptstyle {\sqrt {-1}}$ denotes an hypothetical number whose square would be $-1.$ At this stage, such a number has no meaning (squares of real numbers are always nonnegative), but we use it in a purely formal way. Using this symbol, we can write the "solutions" to the quadratic equation as

$u^{3}={\frac {4+22{\sqrt {-1}}}{2}}=2+11{\sqrt {-1}}$ and $v^{3}={\frac {4-22{\sqrt {-1}}}{2}}=2-11{\sqrt {-1}}.$

It remains to find cube roots of these "numbers". A straightforward calculation shows that $\scriptstyle u=2+{\sqrt {-1}}$ and $\scriptstyle v=2-{\sqrt {-1}}$ do the job. For instance, remembering the rule $\scriptstyle \left({\sqrt {-1}}\right)^{2}=-1$, we have

$\left(2+{\sqrt {-1}}\right)^{3}=2^{3}+3\cdot 2^{2}{\sqrt {-1}}+3\cdot 2\left({\sqrt {-1}}\right)^{2}+\left({\sqrt {-1}}\right)^{3}$

$=8+12{\sqrt {-1}}-6-{\sqrt {-1}}=2+11{\sqrt {-1}}.$

But now, going back to the original cubic equation, we get the real solution $\scriptstyle x=u+v=2+{\sqrt {-1}}+2-{\sqrt {-1}}=4$! One can verify it is indeed a solution, as $\scriptstyle 4^{3}=64=15\cdot 4+4$ (and once this solution is found, it is easy to find the two other solutions, which are also real).

The fact that the formal calculations managed to give a real solution suggests that the "number" $\scriptstyle {\sqrt {-1}}$ may have some sense. But to really give it a legitimate status, one has to construct a new set of numbers, containing the real numbers, but also other numbers whose squares may be negative real numbers. This will be the set of complex numbers. A rigorous construction of this set was given much later by Carl Friedrich Gauss in 1831.

Formal definition

Complex numbers were defined by Hamilton as ordered pairs of real numbers, i.e.

$\mathbb {C} =\{(a,b)\mid a,b\in \mathbb {R} \}.$

To call them 'numbers' we need to introduce some operations on such pairs. So we define

• addition (a, b) + (c, d) = (a + c, b + d)
• multiplication (a, b)(c, d) = (ac - bd, bc + ad)

While this definition can look arbitrary and artificial at the first sight, it turns out to be very natural one. In particular, the basic properties of the usual operations are preserved and we can employ many formulas from the elementary algebra we are accustomed to. More specifically, the sum (or the product) of two numbers does not depend on the order of terms; the sum (product) of three or more elements does not depend on order of operations ('we can suppress the parentheses'); the product of a complex number with a sum of two other numbers expands in the usual way. In mathematical language this means that with addition and multiplication defined this way, $\scriptstyle \mathbb {C}$ satisfies the axioms for a field, and is called the field of complex numbers.

Now we are ready to understand the 'real' meaning of $\scriptstyle {\sqrt {-1}}$ and its usage in the above historical example. Observe that the pairs of type (a,0) are identical to the set of reals, so we write (a, 0)=a. Observe also that by definition (0,1)(0,1) = (-1,0)=-1. It follows that $\scriptstyle {\sqrt {-1}}$, the hypothetical number whose square root gives -1, is well defined as (0,1). It is so characteristic that we usually denote it by i and call it the imaginary unit. Now, the historical example computation is fully justified: since the basic operations on complex numbers behave like the usual addition and multiplication, we can find the roots of a second degree polynomial with the well-known formulas.

Beyond the notation

Next step is to observe that any complex number (a,b) can be expressed as a+bi and, conversely, any sum of this type represents a complex number. The notation using the imaginary unit i is called the algebraic form or rectangular form or yet Cartesian form of a complex number. It is both traditional and commonly used to perform computations. The importance of such formally trivial rewriting is difficult to overestimate and we discuss it in more detail.

There is a well established tradition in mathematics of adopting notation that is suggestive, even if it is, in some ways, unnatural or awkward. For example, if complex numbers are ordered pairs of real numbers, why not represent them simply as pairs, i.e., use $(a,b)$ rather than $a+bi$? There are several ways of answering this question. One is that our notation tends to guide our thinking, and writing $x=x+0i$ emphasizes the idea that the real number x is a complex number, whereas writing $(x,0)$ for the same number suggests that, as a complex number, x is something fundamentally different (perhaps it is). A second, and rather different, reason for using the notation $a+bi$ is that it suggests a parallel with another part of mathematics. In elementary number theory, we learn to perform arithmetic modulo a number base. For example, we may write

$4+5\equiv 2{\pmod {7}}$

to indicate that when we add 4 and 5 and then divide the result by 7, the remainder is 2. We can do something similar with polynomials in a single variable x. We know that $\scriptstyle (x+1)(x+2)=x^{2}+3x+2$, but $\scriptstyle x^{2}+3x+2=1\cdot (x^{2}+1)+(3x+1)$, so when we divide by $\scriptstyle x^{2}+1$, the remainder is $\scriptstyle 3x+1$. And by the same token,

$(1+i)(2+i)=2+3i+i^{2}=1+3i\$

so, when we add or multiply complex numbers, we are just doing modular arithmetic! Of course, there are also times when we wish to focus on the geometric or analytic aspects of complex numbers rather than the algebraic ones, but there is a tendency to want to retain the same notation where possible, and there is no question but that mathematical notation also tends to be dictated by tradition and historical accident.

Another reason, this time of practical nature, is elaborated in the next section.

Working with complex numbers

It is not very compelling to compute using the ordered pair definition, especially when it concerns the product of two complex numbers. It turns out that the imaginary unit comes in handy with its property $i^{2}=-1$. Algebraic operations become as natural as for the reals.

Basic operations

Addition is straightforward, $\scriptstyle (a+bi)+(c+di)=(a+c)+(b+d)i\ .$ More notably, we can rewrite multiplication using $i^{2}=-1$ to obtain results in the form $a+bi$:

$(a+bi)(c+di)=ac+adi+bci+bdi^{2}=(ac-bd)+(bc+ad)i.\$

To handle division, we simply note that $(c+di)(c-di)=c^{2}+d^{2}$, so

${\frac {1}{c+di}}={\frac {c-di}{c^{2}+d^{2}}},$

from which it follows that

${\frac {a+bi}{c+di}}={\frac {(ac+bd)+(bc-ad)i}{c^{2}+d^{2}}}.$

If $z=a+bi$ is a complex number, we call $a$ the real part of $z$ and write $\scriptstyle a=Re(z)$. Similarly, $b$ is called the imaginary part of $z$ and we write $\scriptstyle b=Im(z)$. If the imaginary part of a complex number is $0$, the number is said to be real. As mentioned earlier, we write $a$ instead of $a+0i$ and thus we identify $\scriptstyle \mathbb {R}$ with a subset of $\scriptstyle \mathbb {C}$.

Going a bit further, we can introduce the important operation of complex conjugation. Given an arbitrary complex number $z=x+iy$, we define its complex conjugate to be $\scriptstyle {\bar {z}}=x-iy$. Using the identity $\scriptstyle (a+b)(a-b)=a^{2}-b^{2}$ we derive the important formula

$z{\bar {z}}=x^{2}+y^{2}$

and we define the modulus of a complex number z to be

$|z|={\sqrt {z{\bar {z}}}}$

Note that the modulus of a complex number is always a nonnegative real number. The modulus (also called absolute value) satisfies three important properties that are completely analogous to the properties of the absolute value of real numbers

• $|z|\geq 0$ and $|z|=0$ if and only if $z=0$
• $|z_{1}z_{2}|=|z_{1}||z_{2}|\$
• $|z_{1}+z_{2}|\leq |z_{1}|+|z_{2}|$

The last inequality is known as the triangle inequality.

The complex exponential

Recall that in real analysis, the ordinary exponential function may be defined as

$\exp x=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\cdots$

The same series may be used to define the complex exponential function

$\exp z=1+z+{\frac {z^{2}}{2!}}+{\frac {z^{3}}{3!}}+\cdots$

(where, of course, convergence is defined in terms of the complex modulus, instead of the real absolute value).

Notation: The expressions $\exp \ z$ and $e^{z}\$ mean the same thing, and may be used interchangeably.

The complex exponential has the same multiplicative property that holds for real numbers,namely

$e^{z_{1}z_{2}}=e^{z_{1}}e^{z_{2}}\$

The complex exponential function has the important property that

$e^{i\theta }=\cos \theta +i\sin \theta \$

as may be seen immediately by substituting $z=i\theta$ and comparing terms with the usual power series expansions of $\sin \theta$ and $\cos \theta$.

The familiar trigonometric identity

$\sin ^{2}\theta +\cos ^{2}\theta =1\$

immediately implies the important formula

$|e^{i\theta }|=1$, for any $\theta \in \mathbb {R} .$

Of course, there is no reason to assume this identity. We only need note that ${\overline {e^{i\theta }}}=e^{-i\theta }$, so

$|e^{i\theta }|^{2}=e^{i\theta }e^{-i\theta }=e^{0}=1.\$

Geometric interpretation

Graphical representation of a complex number

Since a complex number $\scriptstyle z=x+iy$ corresponds (essentially by definition) to an ordered pair of real numbers $(x,y)$, it can be interpreted as a point in the plane (i.e. $\scriptstyle \mathbb {R} ^{2})$. When complex numbers are represented as points in the plane, the resulting diagrams are known as Argand diagrams, after Robert Argand. The geometric representation of complex numbers turns out to be very useful, both as an aid to understanding the properties of complex numbers and as a tool in applying complex numbers to geometrical and physical problems.

There are no real surprises when we look at addition and subtraction in isolation: addition of complex numbers is not essentially different from addition of vectors in $\scriptstyle \mathbb {R} ^{2}$. Similarly, if $\scriptstyle \alpha \in \mathbb {R}$ is real, multiplication by $\scriptstyle \alpha$ is just scalar multiplication. In $\scriptstyle \mathbb {C}$ we have

$z_{1}+z_{2}=(x_{1}+iy_{1})+(x_{2}+iy_{2})=(x_{1}+x_{2})+i(y_{1}+y_{2})\$

and

$\alpha z=\alpha (x+iy)=\alpha x+i\alpha y.\,$

To put it succintly, $\scriptstyle \mathbb {C}$ is a 2-dimensional real vector space with respect to the usual operations of addition of complex numbers and multiplication by a real number. There doesn't seem to be much more to say. But there is more to say, and that is that the multiplication of complex numbers has geometric significance. This is most easily seen if we take advantage of the complex exponential, and write complex numbers in polar form

$z=re^{i\theta }.$

Here, r is simply the modulus $\scriptstyle {\sqrt {x^{2}+y^{2}}}$ or vector length. The number $\theta$ is just the angle formed with the x-axis, and is called the argument. Now, when complex numbers are written in polar form, multiplication is very interesting

$z_{1}z_{2}=(r_{1}e^{i\theta _{1}})(r_{2}e^{i\theta _{2}})=r_{1}r_{2}e^{i(\theta _{1}+\theta _{2})}.$

In other words, multiplication by a complex number z has the effect of effect of simultaneously scaling by the numbers' modulus and rotating by its argument. This is really astounding. Translation corresponds to complex addition, scaling to multiplication by a real number, and rotation to multiplication by a complex number of unit modulus. The one type of coordinate transformation that is missing from this list is reflection. On the other hand, there is an arithmetic operation we have not considered, and that is division. Recall that

${\frac {1}{z}}={\frac {\bar {z}}{|z|^{2}}}.$

In other words, up to a scaling factor, division by z is just complex conjugation. Returning to the representation of complex numbers in rectangular form, we note that complex conjugation is just th transformation (or map) $\scriptstyle x+iy\;\mapsto \;x-iy$ or, in vector notation, $\scriptstyle (x,y)\;\mapsto \;(x,-y)$. This is nothing other than reflection in the x-axis, and any other reflection may be obtained by combining that transformation with rotations and translations.

Historically, this observation was very important and led to the search for higher dimensional algebras that could "arithmetize" Euclidean geometry. It turns out that there are such generalizations in dimensions 4 and 8, known as the quaternions and octonions (also known as Cayley numbers). At that point, the process stops, but the ideas developed in this process have played an important role in the development of modern differential geometry and mathematical physics).

Algebraic closure

An important property of $\scriptstyle \mathbb {C}$ is that it is algebraically closed. This means that any non-constant polynomial with complex coefficients has a root in $\scriptstyle \mathbb {C}$. This result is known as the fundamental theorem of algebra. There are many proofs of this theorem. Many of the simplest depend crucially on complex analysis. But it is by no means necessary to rely on complex analysis here. We offer two separate proofs, one that relies on complex analysis and one that does not. As an aside, it is interesting to note that avoiding the methods of one branch of mathematics (complex analysis), requires the use of more advanced methods from another branch of mathematics (in this case, field theory).

Using complex analysis

A startlingly simple proof is based on Liouville's theorem: If $\scriptstyle p(z)$ is a polynomial function of a complex variable then both $\scriptstyle p(z)$ and $\scriptstyle 1/p(z)$ will be holomorphic in any domain where $\scriptstyle p(z)\not =0$. But, by the triangle inequality, we know that outside a neighborhood of the origin $\scriptstyle |p(z)|>|p(0)|$, so if there is no $\scriptstyle z_{0}$ such that $\scriptstyle p(z_{0})=0$, we know that $\scriptstyle 1/p(z)$ is a bounded entire (i.e., holomorphic in all of $\scriptstyle \mathbb {C}$) function. By Liouville's theorem, it must be constant, so $p(z)$ must also be constant.

Using algebra (and a bit of real analysis)

There are also proofs that do not depend on complex analysis, but they require more algebraic or topological machinery.

Formal argument

The starting point here is that $\scriptstyle \mathbb {R}$ is a real closed field (i.e., an ordered field containing positive square roots and in which odd degree polynomials always do posess a root). The starting point is to note that $\scriptstyle \mathbb {C} =\mathbb {R} [i]$ is the splitting field of $x^{2}+1$, so if we can show that $\scriptstyle \mathbb {C}$ has no finite extensions, then we are done. Suppose $\scriptstyle K/\mathbb {C}$ is a finite normal extension with Galois group G. A Sylow 2-subgroup H must correspond to an intermediate field L, such that L is an extension of $\scriptstyle \mathbb {R}$ of odd degree, but we know no such extensions exist. This contradiction establishes the theorem.

Intuition

We need to show that any polynomial $\scriptstyle p(x)\in \mathbb {R} [x]$ has a root in $\scriptstyle \mathbb {C}$. If the degree of $p$ is odd, we can appeal to the intermediate value theorem, because as $x$ approaches $\scriptstyle +\infty$ and $\scriptstyle -\infty$, $p(x)$ must become large and positive at one limit and large and negative at the other. So, what is needed is a method of reducing the case of arbitrary $n$, to this case. (Well, not quite, because in general, the roots of $p$ will lie in $\scriptstyle \mathbb {C}$, not $\scriptstyle \mathbb {R}$). It turns out that if we write $n=2^{k}n_{0}$ where $n_{0}$ is odd (i.e., if we factor $n$ into its even and odd pieces) we can dispense with the odd piece as above, and then rely on a purely algebraic argument based on Galois theory (after its discoverer Évariste Galois) to handle the even part.

Complex numbers in physics

Complex numbers appear everywhere in mathematical physics, but one area where the role of complex numbers is especially difficult to ignore is in quantum mechanics. There are a number of ways of formulating the basic laws of quantum mechanics, but here we consider just one: the Schrödinger equation, discovered by Erwin Schrödinger in 1926. In rectangular coordinates, it may be written

$i\hbar {\frac {\partial \psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}\Delta \psi +V(x,y,z)\psi ,$

where

$\Delta ={\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}$

is known as the Laplacian operator and $V(x,y,z)$ is the potential function. (As a practical example, the potential function might represent the attractive force per unit mass between the nucleus of a hydrogen atom and an electron). Now, there is some subtlety in the interpretation of $\psi$ because a system can be affected by observation, and the functions $\psi$ we "see" must be eigenstates of the operator defined by the Schrödinger equation, but when we do measure, say, the position of a particle, the probability of finding it in a small region R is just

$\int _{R}\nolimits |\psi |^{2}.$

Further reading

• Ahlfors, Lars V. (1979). Complex Analysis, 3rd edition. McGraw-Hill, Inc.. ISBN 0-07-000657-1. 
• Apostol, Tom M. (1974). Mathematical Analysis, 2nd edition. Addison-Wesley. ISBN 0-201-00-288-4. 
• Conway, John H.; Derek A. Smith (2003). On Quaternions and Octonions: Their Geometry, Arithmetic and Symmetry. A K Peters, Ltd.. ISBN 1-56881-134-9. 
• Jacobson, Nathan (1974). Basic Algebra I. W.H. Freeman and Company. ISBN 0-7167-0453-6. 
• Williams, Floyd (2003). Topics in Quantum Mechanics. Birkhäuser. ISBN 0-8176-4311-7.