Constructing complex numbers

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 For any pair of real numbers, \(a, b\in\mathbb{R}\), we can define a complex number \(a + bi\) where \(i\) is the imaginary unit (\(i = \sqrt{-1}\)). However, the imaginary unit is not a real number. What do we exactly mean by the expression like \(a + bi\)? We answer this question by constructing a system of complex numbers from the set of pairs of real numbers.


Recall \(\mathbb{R}^2\), the set of ordered pairs of real numbers:

\[\mathbb{R}^2 = \{(a, b) \mid a, b\in\mathbb{R}\}.\]

Each element of \(\mathbb{R}^2\) is something like \((a,b)\) where \(a, b\in \mathbb{R}\).

Let us define addition in \(\mathbb{R}^2\) by

\[(a, b) + (c, d) = (a + c, b + d)\]

where \((a,b), (c, d) \in\mathbb{R}^2\). Note that \(+\) on the left-hand side is the addition in \(\mathbb{R}^2\) being defined, and \(+\) on the right-hand side is the addition of real numbers (\(\mathbb{R}\)), which is already defined. Thus defined addition is commutative and associative (verify this):

\[(a, b) + (c, d) = (c, d) + (a, b).\]

Next, the addition in \(\mathbb{R}^2\) is associative:

\[(a,b) +((c,d) + (e,f)) = ((a,b) + (c,d)) + (e, f).\]

Exercise. Prove this. □

Because of the associativity, we can simply write

\[(a, b) + (c,d) + (e,f)\]

to indicate the sum of three or more pairs of real numbers.

We can see that \((0,0)\) is the additive identity in the sense that

\[(a, b) + (0, 0) = (a, b) ~ \forall (a,b)\in\mathbb{R}^2.\]

For each \((a,b) \in \mathbb{R}^2\), its additive inverse is \((-a, -b)\) in the sense that

\[(a, b) + (-a, -b) = (0, 0).\]

Next, we define a ``multiplication'' in \(\mathbb{R}^2\) by

\[(a, b)\star (c, d) = (ac-bd, ad+bc).\]

This multiplication is commutative:

\[(c, d)\star(a,b) = (ac-bd, ad+bc) = (a,b) \star (c, d).\]

It is also associative:

\[\begin{eqnarray*} ((a,b)\star(c,d))\star(e,f) &=& (ac - bd, ad+bc)\star(e,f)\\ &=&((ac - bd)e - (ad+bc)f, (ac-bd)f + (ad+bc)e)\\ &=&(ace - bde - adf - bcf, acf-bdf + ade + bce)\\ &=&(a(ce - df) - b(cf + de), a(cf + de) + b(ce -df))\\ &=&(a, b)\star(ce - df, de + cf)\\ &=&(a, b)\star((c, d)\star (e, f)). \end{eqnarray*}\]

\((1, 0)\) is the multiplicative identity in the sense that

\[(a, b)\star(1, 0) = (a, b) ~ \forall(a,b) \in \mathbb{R}^2.\]

Furthermore, we have the multiplicative inverse of \((a,b) \neq (0,0)\) that is

\(\left(\frac{a}{a^2 + b^2}, -\frac{b}{a^2 + b^2}\right)\) because

\[(a, b)\star\left(\frac{a}{a^2 + b^2}, -\frac{b}{a^2 + b^2}\right) = \left(\frac{a^2+b^2}{a^2 + b^2}, \frac{-ab + ba}{a^2 + b^2}\right) = (1, 0).\]


Consider the map \(\theta: \mathbb{R} \to \mathbb{R}^2\) defined by \(\theta(r) = (r, 0)\). This map is an injection and embeds \(\mathbb{R}\) into \(\mathbb{R}^2\). We use boldface \(\mathbf{x}\) as shorthand for \((x,0)\). e.g., \(\mathbf{7} = (7, 0)\). Let \(\mathbf{R} = \{(x, 0)\mid x \in \mathbb{R}\}\). The elements of \(\mathbf{R}\) behave exactly like those of \(\mathbb{R}\) (i.e., real numbers). \(\mathbf{2} + \mathbf{3}\) \(= (2, 0) + (3, 0) = (5 , 0) = \mathbf{5}\), \(\mathbf{6}\cdot\mathbf{7} = (6, 0)\star(7,0)\) \(= (6\cdot 7 - 0\cdot 0, 6\cdot 0 + 7\cdot 0) = (42, 0) = \mathbf{42}\).

Let \(\mathbf{i} = (0, 1)\) (Note \(\mathbf{i} \not\in \mathbf{R}\)). Then \(\mathbf{i}^2 = \mathbf{i}\cdot\mathbf{i}\)\(= (0, 1)\star(0, 1) = (0\cdot 0 - 1\cdot 1, 0\cdot 1 + 1\cdot 0)\)\(= (-1, 0) = \mathbf{-1}\).

Note that every element of \(\mathbb{R}^2\) can be expressed in terms of bold symbols and \(\mathbf{i}\)
\[(a,b) = (a, 0) + (0, b) = (a,0) + (0, 1)\star(b,0) = \mathbf{a} + \mathbf{i}\mathbf{b}\]
where \(\mathbf{a},\mathbf{b}\in\mathbf{R}\)  (we often omit ``\(\cdot\)'' in multiplication).

Definition (Complex numbers)

The set \(\mathbb{C}\) of complex numbers is the set \(\mathbb{R}^2\) equipped with the addition (\(+\)) and multiplication (\(\star\)) as above.

Thus, complex numbers are something like \(\mathbf{x} + \mathbf{iy}\) where \(\mathbf{x},\mathbf{y}\in \mathbf{R}\). We can express the addition and multiplication of complex numbers using only the expressions like this rather than explicitly resorting to expressions like \((x,y)\) in \(\mathbb{R}^2\). For example,
\[\begin{eqnarray*} (\mathbf{3} -\mathbf{2}\mathbf{i}) + (\mathbf{3/2} + \mathbf{i}) &=& (\mathbf{3} + \mathbf{3/2}) + (\mathbf{-2} + \mathbf{1})\mathbf{i}\\ &=& \mathbf{9/2} - \mathbf{i},\\ (\mathbf{3} -\mathbf{2}\mathbf{i}) \cdot (\mathbf{3/2} + \mathbf{i}) &=& \mathbf{3}\cdot\mathbf{3/2} +\mathbf{3}\cdot\mathbf{i} -\mathbf{2}\mathbf{i}\cdot\mathbf{3/2} -\mathbf{2i}\cdot\mathbf{i}\\ &=&\mathbf{9/2} + (\mathbf{3} - \mathbf{3})\mathbf{i} - \mathbf{2}\mathbf{i}^2\\ &=&\mathbf{9/2} + \mathbf{0}\mathbf{i} - \mathbf{2}\cdot(\mathbf{-1})\\ &=&\mathbf{13/2}. \end{eqnarray*}\]
Thus, we can treat \(\mathbf{x} + \mathbf{i}\mathbf{y}\) as a polynomial of \(\mathbf{i}\), and whenever we have \(\mathbf{i}^2\) we replace it with \(\mathbf{-1}\).

We don't need \(\mathbb{R}\) anymore. The new \(\mathbf{R}\) not only behaves exactly like \(\mathbb{R}\), but also is a subset of \(\mathbb{C}\). (Note that \(\mathbb{R}\) is not a subset of \(\mathbb{R}^2\) (\(\mathbb{R}\not\subset \mathbb{R}^2\)); hence \(\mathbb{R}\) is not a subset of \(\mathbb{C}\) either.) So, we can just discard \(\mathbb{R}\) and work with \(\mathbf{R}\). But then, we don't need to use boldface numbers like \(\mathbf{3}\) anymore. From now on, we use the ordinary fonts like \(3\) or \(a + ib\) to mean \(\mathbf{3}\) or \(\mathbf{a} + \mathbf{ib}\), respectively, and write \(\mathbb{R}\) to mean \(\mathbf{R}\). With this new convention, we now have \(\mathbb{R} \subset \mathbb{C}\). We can write a complex number as \(a + ib\) to mean \((a,b)\) with specific algebraic rules.

Definition (Real and imaginary parts)

If \(a, b \in \mathbb{R}\), then the real and imaginary parts of \(z = a + ib\) are \(a\) and \(b\), respectively. We write \(\Re(z) = a\) and \(\Im(z) = b\).
  We say that the complex number \(z\) is real if \(\Im(z) = 0\), and that \(z\)is purely imaginary if \(\Re(z) = 0\).


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