Complex Algebra 1: Complex Numbers

 A complex number is a number of the form $z=x+iy$, where $x,y\in\mathbb{R}$ and $i$ is the number whose square is $-1$. $i$ is called the pure imaginary number. $i$ is symbolically written as $i=\sqrt{-1}$. The real numbers $x$ and $y$ are called, respectively, the real part and imaginary part of $z$ and we also write them as $x=\mathrm{Re}\ z$ and $y=\mathrm{Im}\ z$. Since there is no real number $i$ satisfying $i^2=-1$, it is called an imaginary number. But the number $i$ is as real as any real number. It just does not live in the one-dimensional world (the number line) but actually in two-dimensional world! The set of all complex numbers is denoted by $\mathbb{C}$ and is called the complex plane. It turns out the complex plane can be identified with the real plane $\mathbb{R}^2$ via the map
$$x+iy\in\mathbb{C}\longmapsto (x,y)\in\mathbb{R}^2$$
Notice that under this identification, $i=(0,1)$. This identification is topological, i.e. $\mathbb{C}$ is homeomorphic to $\mathbb{R}^2$. From an algebraic point of view, however, $\mathbb{C}$ and $\mathbb{R}^2$ are different. As we will see, $\mathbb{C}$ with operations $+$ and $\cdot$ is an algebraic structure called a field but $\mathbb{R}^2$ is not.

Complex Plane versus Real Plane

Let $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ be two complex numbers. They also can be represented as ordered pairs as $z_1=(x_1,y_1)$ and $z_2=(x_2,y_2)$. Two complex numbers $z_1$ and $z_2$ are equal if and only if $x_1=x_2$ and $y_1=y_2$. The sum of two complex numbers $z_1$ and $z_2$ is defined by
\begin{equation}
\label{eq:sum}
z_1+z_2=x_1+x_2+i(y_1+y_2)
\end{equation}
The multiplication $z_1\cdot z_2$ is defined by
\begin{equation}
\label{eq:product}
\begin{aligned}
z_1\cdot z_2&=(x_1+iy_1)(x_2+iy_2)\\
&=x_1x_2-y_1y_2+i(x_1y_2+x_2y_1)
\end{aligned}
\end{equation}
The addition $+$ is associative and commutative. $0$ is the additive identity and for each $z\in\mathbb{C}$, $-z$ is an  additive inverse. This means that $(\mathbb{C},+)$ is an abelian group. The multiplication $\cdot$ is associative and commutative. $1$ is the multiplicative identity and for each nonzero complex number $z=x+iy$, there is a multiplicative inverse $\frac{1}{z}$.
\begin{equation}
\label{eq:inverse}
\begin{aligned}
\frac{1}{z}&=\frac{1}{x+iy}\\
&=\frac{x-iy}{(x+iy)(x-iy)}\\
&=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}
\end{aligned}
\end{equation}
This means that $(\mathbb{C}\setminus\{0\},\cdot)$ is an abelian group. There are also distributive laws hold in usual sense. In this case, we say $(\mathbb{C},+,\cdot)$ is a field.

For $z=x+iy$, $\bar z=x-iy$ is called the conjugate of $z$. As seen above,
\begin{equation}
\begin{aligned}
z\bar z&=x^2+y^2\\
&=(\mathrm{Re}\ z)^2+(\mathrm{Im}\ z)^2
\end{aligned}
\end{equation}
and clearly,
\begin{equation}
\bar{\bar z}=z
\end{equation}
The norm, length, or modulus of $z$ is defined by
\begin{equation}
\begin{aligned}
|z|&=\sqrt{z\bar z}\\
&=\sqrt{(\mathrm{Re}\ z)^2+(\mathrm{Im}\ z)^2}\\
&=\sqrt{x^2+y^2}
\end{aligned}
\end{equation}

Example.
\begin{align*}
|-3+2i|&=\sqrt{(-3+2i)(-3-2i)}\\
&=\sqrt{9+4}=\sqrt{13}
\end{align*}

Proposition [Triangle Inequality].
\begin{equation}
|z_1+z_1|\leq |z_1|+|z_2|
\end{equation}

Proof. Left as an exercise.

Corollary.
\begin{equation}
||z_1|-|z_2||\leq |z_1+z_2|
\end{equation}

Proof. Left as an exercise.