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 deserves be called 'a number', i.e. whether we can associate with it some natural operations as addition or multiplication. Admitting for a moment that the positive answer is given for granted, we define

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

We then 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.

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.

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.