Constructing complex numbers

 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

$$(a,b)+(0,0)=(a,b)\mathrm{\forall }(a,b)\in {\mathbb{R}}^2.$$

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:

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

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

$$(a,b)\star (1,0)=(a,b)\mathrm{\forall }(a,b)\in {\mathbb{R}}^2.$$

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

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

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

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 $\mathbf{i}\notin \mathbf{R}$). 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{-}\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,

$$\begin{array}{rcl}(\mathbf{3}-\mathbf{2}\mathbf{i})+(\mathbf{3}\mathbf{/}\mathbf{2}+\mathbf{i})& =& (\mathbf{3}+\mathbf{3}\mathbf{/}\mathbf{2})+(\mathbf{-}\mathbf{2}+\mathbf{1})\mathbf{i}\\ & =& \mathbf{9}\mathbf{/}\mathbf{2}-\mathbf{i},\\ (\mathbf{3}-\mathbf{2}\mathbf{i})\cdot (\mathbf{3}\mathbf{/}\mathbf{2}+\mathbf{i})& =& \mathbf{3}\cdot \mathbf{3}\mathbf{/}\mathbf{2}+\mathbf{3}\cdot \mathbf{i}-\mathbf{2}\mathbf{i}\cdot \mathbf{3}\mathbf{/}\mathbf{2}-\mathbf{2}\mathbf{i}\cdot \mathbf{i}\\ & =& \mathbf{9}\mathbf{/}\mathbf{2}+(\mathbf{3}-\mathbf{3})\mathbf{i}-\mathbf{2}{\mathbf{i}}^2\\ & =& \mathbf{9}\mathbf{/}\mathbf{2}+\mathbf{0}\mathbf{i}-\mathbf{2}\cdot (\mathbf{-}\mathbf{1})\\ & =& \mathbf{13}\mathbf{/}\mathbf{2}.\end{array}$$

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 $\mathbf{-}\mathbf{1}$.

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 $\mathrm{\Re }(z)=a$ and $\mathrm{\Im }(z)=b$.

  We say that the complex number $z$ is real if $\mathrm{\Im }(z)=0$, and that $z$is purely imaginary if $\mathrm{\Re }(z)=0$.