Complex number/Citable Version

The complex numbers $\mathbb {C}$ are numbers of the form a+bi, obtained by adjoining the imaginary unit i to the real numbers (here a and b are reals). The number i can be thought of as a solution of the equation $x^{2}+1=0$. In other words, its basic property is $i^{2}=-1$. Of course, since the square root of any real number is positive, $i\notin \mathbb {R}$. 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 write

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

Of course, we do not formally define complex numbers this way but, rather, define them as ordered pairs of real numbers. The above notation is, however, traditional.

Aside on notation: 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 as pairs, i.e., use $(a,b)$ rather than $a+bi$? Thee 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

$4+5\equiv 2(mod7)$

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 $(x+1)(x+2)=x^{2}+3x+2$, but $x^{2}+3x+2=1\cdot (x^{2}+1)+(3x+1)$, so when we divide by $x^{2}+1$, the remainder is $3x+1$. 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.

Working with Complex Numbers

Basic Operations

We define addition and multiplication in the obvious way, using $i^{2}=-1$ to rewrite results in the form $a+bi$:

$(a+bi)+(c+di)=(a+c)+(b+d)i\$

$(a+bi)(c+di)=(ac-bd)+(bc+ad)i\$

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

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

and, in particular,

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

It turns out that with addition and multiplication defined this way, $\mathbb {C}$ satisfies the axioms for a field, and is called the field of complex numbers. If $c=a+bi$ is a complex number, we call $a$ the real part of $c$ and write $a=Re(c)$. Similarly, $b$ is called the imaginary part of $c$ and we write $b=Im(c)$. If the imaginary part of a complex number is $0$, the number is said to be real, and we write $a$ instead of $a+0i$. We thus identify $\mathbb {R}$ with a subset (and, in fact, a subfield) of $\mathbb {C}$.

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 ${\bar {z}}=x-iy$. Using the identity $(a+b)(a-b)=a^{2}-b^{2}$ 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 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

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

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

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

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

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

(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 expomential 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 ${\bar {e^{i\theta }}}=e^{-i\theta }$

so,

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

Geometric Interpretation

Since a complex number $z=x+iy$ 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., $\mathbb {R} ^{2})$. 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, but also 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 $\mathbb {R} ^{2}$. Similarly, if $\alpha \in \mathbb {R}$ is real, multiplication by </math>\alpha</math> is just scalar multiplication. In $\mathbb {C}$ 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, $\mathbb {C}$ 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.

Algebraic Closure

An important property of $\mathbb {C}$ is that it is algebraically closed. This means that any non-constant real polynomial must have a root in $\mathbb {C}$. 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. To illustrate, we consider a proof based on Liouville's theorem: If $p(z)$ is a polynomial function of a complex variable then both $p(z)$ and $1/p(z)$ will be holomorphic in any domain where $p(z)\not =0$. But, by the triangle inequality, we know that outside a neighborhood of the origin $|p(z)|>|p(0)|$, so if there is no $z_{0}$ such that $p(z_{0})=0$, we know that $1/p(z)$ is a bounded entire (i.e., holomorphic in all of $\mathbb {C}$) function. By Liouville's theorem, it must be constant, so $p(z)$ must also be constant.

There are also proofs that do not depend on complex analysis, but they require more algebraic or topological machinery. The starting point here is that $\mathbb {R}$ 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 $\mathbb {C} =\mathbb {R} [i]$ is the splitting field of $x^{2}+1$, so if we can show that $\mathbb {C}$ has no finite extensions. We are done. Suppose $K/\mathbb {C}$ is a finite normal extension with Galois group G. A Sylow 2-subgroup H must correspond to an intermeiate field L, such that L is an extension of $\mathbb {R}$ of odd degree, but we know no such extensions exist. This contradiction establishes the theorem.

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).

Notational Variants

This article follows the usual convention in mathematics (and physics) of using $i$ as the imaginary unit. Complex numbers are frequently used in electrical engineering, but in that discipline it is usual to use $j$ instead, reserving $i$ for electrical current. This usage is found in some programming languages, notably Python.

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. 
• Jacobson, Nathan (1974). Basic Algebra I. W.H. Freeman and Company. ISBN 0-7167-0453-6.