Math for Non-Geeks/Definition of complex numbers

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Here we will formally define the complex numbers $ℂ$ and prove that they form a field. First of all we will clarify what the addition and multiplication of complex numbers should look like.

Deriving for the formal definition of complex numbers

Derivation of the tuple notation

Complex numbers have the form $a+b\mathrm{i}$, where $a,b$ are real numbers and the imaginary unit $\mathrm{i}$ satisfies the equation ${\mathrm{i}}^2=-1$. However, we lack a mathematical definition for this new form of a number. So we will derive now a reasonable and precise definition.

A complex number $a+b\mathrm{i}$ is described by the two real numbers $a$ and $b$. Furthermore, complex numbers can be represented as points in a plane. $a$ is the $x$-coordinate of the point and the imaginary part $b$ is the $y$-coordinate:

Complex numbers are represented by dots on a plane

Now points within the plane can be described as tuples $(a,b)$ of the set $ℝ\times ℝ={ℝ}^2$. So we can assign to a tuple $(a,b)$ in $ℝ\times ℝ$ the complex number $a+b\mathrm{i}$. So we identify $(a,b)\widehat{=}a+b\mathrm{i}$, which provides a one-to-one identification of the complex set of numbers $ℂ$ with the set $ℝ\times ℝ$.

The tuple is a precisely defined mathematical concept, we can use it for the formal definition of the complex numbers. To this we say that complex numbers $a+b\mathrm{i}$ are in fact tuples $(a,b)$ within a special notation and that allow for a multiplication (to be defined later).

Deriving computational rules

Adding complex numbers works like adding vectors.

It would be nice to calculate with complex numbers like with real numbers, by adding and multiplying them. Let us first consider the addition of two complex numbers $a+b\mathrm{i}$ and $c+d\mathrm{i}$. The result should again be a complex number, i.e. of the form $x+y\mathrm{i}$. For this we add the two complex numbers, arrange the summands and factor out:

The result is again of the form $x+y\mathrm{i}$. The real and the imaginary parts are added up. For the formal definition of the addition we use the tuple notation $(a,b)$in $ℝ\times ℝ$, with identification $(a,b)\widehat{=}a+b\mathrm{i}$. With this we translate the above calculation into the tuple notation:

We see that summing nothing else than a component-wise addition in $ℝ\times ℝ$. This is exactly the vector addition in the plane $ℝ\times ℝ$. The multiplication of complex numbers is more complicated. We consider the product of two complex numbers $a+b\mathrm{i}$ and $c+d\mathrm{i}$ and multiply out the product:

Translated in the tuple-notation, we have:

Formal definition of complex numbers

Definition of complex numbers

The complex numbers are defined as tuples in $ℝ\times ℝ$ with the appropriate addition and multiplication.

Definition of real and imaginary part

A complex number $z=(a,b)$, can be described as a point in the plane. It is uniquely defined by its coordinates $a$ and $b$. These coordinates have special names. $a$ is called real part and $b$ imaginary part of the complex number.

The complex numbers form a field Template:Anchor

We can calculate with complex numbers as with real numbers. The addition corresponds to the vector addition in $ℝ\times ℝ$. Thus it inherits all properties of the addition in a vector space and fulfils for example the associative law $(z+w)+u=z+(w+u)$ and the commutative law $z+w=w+z$. The multiplication in the complex numbers has similar properties as the multiplication in the real numbers.

Like in the real numbers we can form fractions of the form $\frac{w}{z}$ in $ℂ$. This requires inverting a complex number $z=(a,b)$ to $z^{-1}=(c,d)$. The reciprocal number should satisfy the equation $z\cdot z^{-1}=1$. So we need to choose $(c,d)$ such that $(1,0)=(a,b)\cdot (c,d)=(ac-bd,ad+bc)$ holds. We will see that this system of equations is uniquely solvable for all $(a,b)\in ℂ\setminus \{0\}$.

Altogether the addition and the multiplication fulfil the so-called field axioms, as it does for the real numbers. Thus, calculations in $ℂ$ work with a similar structure compared to those within the real numbers.

$ℝ$ as a sub-field of $ℂ$

We identify the complex numbers $ℂ$ with the plane $ℝ\times ℝ$. Here the $x$ axis lying in the complex plane is the real number line. So it makes sense that the real numbers $ℝ$ are a subset of the complex numbers $ℂ$.

We also know that both $ℝ$ and $ℂ$ are fields. So $ℝ$ should be a sub-field of $ℂ$. In order to verify this, we have to show more than that $ℝ$ is a subset of $ℂ$. We must also prove that the addition and multiplication of real numbers in $ℂ$ again leads to real numbers. Mathematically, two statements have to be shown: $ℝ$ is a subset of $ℂ$ and the arithmetic operations preserve the real numbers in $ℂ$.

The first statement is easily confirmed. Strictly speaking, $ℝ$ is not subset of $ℂ$, since $ℂ=ℝ\times ℝ=\{(a,b)|a,b\in ℝ\}$ is a set of tuples and $ℝ$ just a set of single. So the elements of $ℝ$ and of $ℂ$ are different.

However, we can identify the real numbers with a subset of the complex numbers, which behaves similar to $ℝ$. To find this subset, we use the visualization of the complex numbers in the plane. The subset we are looking for is the real axis in the complex plane. A complex number $z=(a,b)\in ℂ$ lies on this axis exactly when its imaginary part is zero, i.e. $b=0$. Thus the real line is identified with the set $\{(a,0)|a\in ℝ\}=ℝ\times \{0\}\subseteq ℂ$.

Now, a quick mathematical investigation of the identification of $ℝ\times \{0\}$ with the real numbers follows. Intuitively, there is nothing to do: the real line looks like the real axis within the complex plane. Mathematically, there is still some work to be done: we need a one-to-one relationship (bijective mapping) of $ℝ$ to $ℝ\times \{0\}$. Or we define an injective mapping $\iota :ℝ\to ℂ$ (called embedding) with $\iota (ℝ)=ℝ\times \{0\}$. Then $\iota$ bijectively maps the real numbers to $ℝ\times \{0\}$.

And we need that $ℝ\times \{0\}$ has the same structure as the real numbers. Our embedding map $\iota$ should preserve the structure of $ℝ$ in the image. This means, sums in $ℝ$ should be mapped from $\iota$ to sums in $ℂ$ and the same with products. And the neutral elements $0$ and $1$ shall be mapped from the real numbers to the corresponding neutral elements in the complex numbers. (A mapping with such properties is also called field homomorphism).

How should we choose $\iota$? Let us look again at our visualization of the complex plane. We want to map the real number line $ℝ$ to the real axis $ℝ\times \{0\}$. The easiest way to do this is to just embed the number line into the two-dimensional plane. In other words, map a real number $a\in ℝ$ to $(a,0)\in ℝ\times \{0\}$:

It remains to be shown that our picture fulfils the characteristics of an injective field homomorphism. Such an injective body homohorphism is also called field monomorphism:

Due to the properties of a field monomorphism, the structure of a field is preserved in the image of the embedding. Simply put, the image of the field monomorphism fulfils the field axioms and thus defines a field again. Since the image of the embedding $\iota$ is a subset of the field of complex numbers, we can regard the image $\iota (ℝ)$ as a sub-field of $ℂ$. Furthermore, the image $\iota :ℝ\to \mathrm{Imag}(\iota )$ gives a field isomorphism, i.e. a bijective field homomorphism between $ℝ$ and $\mathrm{Imag}(\iota )$. This justifies the notion $ℝ\subseteq ℂ$ and we view from now on all real numbers $a\in ℝ$ as equal to the complex number $a:=\iota (a)$.

Definition of the notation $a+b\mathrm{i}$

We would like to write a complex number as $a+b\mathrm{i}$. According to our definition with $ℂ=ℝ\times ℝ$, this number is the tuple $(a,b)$. To simplify calculations, we would like to introduce the notation $a+b\mathrm{i}$ without the tuple. For this we have to define $\mathrm{i}$ mathematically. Since $\mathrm{i}$ lies in the complex plane on the $y$ axis at coordinate $1$, we choose $\mathrm{i}:=(0,1)$:

In the beginning we looked for the solution of the equation $x^2=-1$ and with $\mathrm{i}$ we found one of these solutions. We can verify that ${\mathrm{i}}^2=-1$ for $\mathrm{i}=(0,1)$ by explicit computation:

Here, we use the embedding $\iota$ of the real numbers in $ℂ$ and the notation $a=\iota (a)$ for $a\in ℝ$. So there indeed is ${\mathrm{i}}^2=-1$. Now we show that the notation $a+b\mathrm{i}$ for $(a,b)$ indeed makes sense. Using $c:=\iota (c)=(c,0)$ for $c\in ℝ$ we show $(a,b)=a+b\mathrm{i}$. Thanks to this proof, we can then calculate with the complex number $a+b\mathrm{i}$ as if it was a sum:

$ℂ$ is not an ordered field Template:Anchor

It would be nice to have an ordering of complex numbers, that means a larger/smaller relation for complex numbers. Let us consider the numbers $1$ and $\mathrm{i}$. We notice that they lie on the unit circle. This is the set of all points which have the distance $1$ to zero:

Unit circle in the complex plane with 1 and i

Is now $1<\mathrm{i}$, $1=\mathrm{i}$ or $1>\mathrm{i}$? At first this case seems to be ambiguous, because both numbers have the same absolute value. What about $1$ and $-2\mathrm{i}$? The number $-2\mathrm{i}$ is further away from zero than the number $1$. Is $-2\mathrm{i}>1$ then also valid? Can the product of a negative number with the imaginary unit really be greater than a positive number?

From these small examples we can already see that establishing an ordering of complex numbers is difficult. In fact this is not possible. The following theorem proves this:

$ℂ$ is algebraically closed

With $ℂ$ we constructed a field in which the equation $x^2=-1$ is solvable, so the polynomial $x^2+1$ has a zero. In the complex numbers $ℂ$ we even have that every polynomial (with coefficients in $ℂ$) of degree greater or equal to $1$ has at least one zero. This excludes only constant polynomials, which of course (except the zero polynomial) have no zeros. This property is not valid in the real numbers: for instance, $x^2+1$ has no real zeros.

This property of complex numbers is called algebraic closure and is treated in algebra. The algebraic closure of $ℂ$ is proved in a theorem with the majestic name Fundamental Theorem of Algebra.

Exercises

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