Geometry of complex numbers

 From the definition of complex numbers, we can plot any complex number $z=a+ib\in \mathbb{C}$ with $a,b\in \mathbb{R}$ as a point $(a,b)$ in a two-dimensional plane.

Here the ``$x$'' axis represents the real part of complex numbers, whereas the ``$y$'' axis represents the imaginary part. When we use this ``$x-y$'' plane to plot complex numbers, this plane is called the complex (number) plane, also known as the Argand diagram (Figure 1).

Figure 1. The complex plane or Argand diagram.

Definition (Modulus)

Let $z=a+ib\in \mathbb{C}$ with $a,b\in \mathbb{R}$. The modulus of $z$ is defined by

$$|z|=\sqrt{a^2+b^2}.$$

That is, the modulus $|z|$ is the distance between $z$ and the origin in the complex plane (Figure 1).

In comparison, the modulus of a real number $x\in \mathbb{R}$ is defined (usually) as

$$|x|=\{\begin{array}{cc}x& \text{(if }x\ge 0\text{)},\\ -x& \text{(otherwise)}.\end{array}$$

But this is equivalent to

$$|x|=\sqrt{x^2}.$$

Note the similarity to the above definition. We can see that the modulus of complex numbers is a natural generalization of that of real numbers.

Definition (Complex conjugate)

Let $z=a+ib\in \mathbb{C}$ with $a,b\in \mathbb{R}$. Then the complex conjugate $\overline{z}$ of $z$ is defined by

$$\overline{z}=a-ib.$$

On the complex plane, the complex conjugate $\overline{z}$ is the mirror image of $z$ across the $x$-axis (Figure 1). Also, note that

$$z\overline{z}=a^2+b^2=|z{|}^2.$$

You should try to interpret the following lemma in terms of the complex plane.

Lemma

1. $\mathrm{\forall }z\in \mathbb{C},\overline{\overline{z}}=z$.
2. $\mathrm{\forall }{z}_1,z_2\in \mathbb{C},\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}$.
3. $\overline{z}=z$ if and only if $z\in \mathbb{R}$.
4. $\overline{z}=-z$ if and only if $z$ is purely imaginary.
5. $\mathrm{\forall }z\in \mathbb{C},z+\overline{z}\in \mathbb{R}$.
6. $\mathrm{\forall }z\in \mathbb{C},z-\overline{z}$ is purely imaginary.
7. $\mathrm{\forall }{z}_1,z_2\in \mathbb{C},\overline{z_1{z}_2}=\overline{z_1}\overline{z_2}$
8. $\mathrm{\forall }{z}_1,z_2\in \mathbb{C},|z_1{z}_2|=|z_1|\cdot |z_2|$.
9. $\mathrm{\forall }{z}_1,z_2\in \mathbb{C},|z_1+z_2|\le |z_1|+|z_2|$. (the triangle inequality)

Proof. Let $z=a+ib$ where $a,b\in \mathbb{R}$. Similarly, let $z_1=a_1+i{b}_1$, $z_2=a_2+i{b}_2$.

1. $\overline{z}=a-ib$ by definition. Thus, $\overline{\overline{z}}=\overline{a-ib}=a+ib$.
2. On one hand, $z_1+z_2=(a_1+a_2)+i(b_1+b_2)$, and hence $\overline{z_1+z_2}=(a_1+a_2)-i(b_1+b_2)$. On the other hand, ${\overline{z}}_1+{\overline{z}}_2=(a_1-i{b}_1)+(a_2-i{b}_2)=(a_1+a_2)-i(b_1+b_2)$. 
3. Suppose $\overline{z}=z$. Then, $a-ib=a+ib$. Thus, $ib=0$, which implies $b=0$. Therefore, $z=a\in \mathbb{R}$. Conversely, suppose $z\in \mathbb{R}$. Then, $b=0$ (the imaginary part is zero). Thus, $\overline{z}=a=z$. 
4. Suppose $\overline{z}=-z$. Then, $a-ib=-a-ib$. Thus, $a=0$, which implies $z=ib$ is purely imaginary. Conversely, suppose $z$ is purely imaginary. Then, $z=ib$ (the real part is zero). Thus, $\overline{z}=-ib=-z$.
5. $z+\overline{z}=(a+ib)+(a-ib)=2a\in \mathbb{R}$.
6. $z-\overline{z}=(a+ib)-(a-ib)=2ib$ is purely imaginary.
7. On one hand, $\overline{z_1{z}_2}=\overline{(a_1{a}_2-b_1{b}_2)+i(a_1{b}_2+b_1{a}_2)}=(a_1{a}_2-b_1{b}_2)-(a_1{b}_2+b_1{a}_2)$. On the other hand, ${\overline{z}}_1{\overline{z}}_2=(a_1-i{b}_1)(a_2-i{b}_2)=(a_1{a}_2-b_1{b}_2)-i(a_1{b}_2+b_1{a}_2)$.
8. $$\begin{array}{rcl}|z_1{z}_2{|}^2& =& (z_1{z}_2)(\overline{z_1{z}_2})\\ & =& (z_1\overline{z_1})(z_2\overline{z_2})\\ & =& |z_1{|}^2|z_2{|}^2\end{array}$$ Since the modulus is always non-negative, $$|z_1{z}_2|=|z_1|\cdot |z_2|.$$
9. Let $\alpha \in \mathbb{C}$. $\alpha -\overline{\alpha }$ is purely imaginary (Part 6) so $(\alpha -\overline{\alpha }{)}^2$ is a non-positive real number: $$(\alpha -\overline{\alpha }{)}^2\le 0.$$ Adding $4\alpha \overline{\alpha }=4|\alpha {|}^2$ to the both sides, $$(\alpha -\overline{\alpha }{)}^2+4\alpha \overline{\alpha }\le 4|\alpha {|}^2.$$ By rearranging the left-hand side, $$(\alpha +\overline{\alpha }{)}^2\le 4|\alpha {|}^2,$$ and hence $$\alpha +\overline{\alpha }\le 2|\alpha |.$$ Now let $\alpha =z_1\overline{z_2}$ to obtain $$z_1\overline{z_2}+\overline{z_1}{z}_2\le 2|z_1||z_2|.$$ Add $z_1\overline{z_1}+z_2\overline{z_2}=|z_1{|}^2+|z_2{|}^2$ to the both sides, and we have $$z_1\overline{z_1}+z_2\overline{z_2}+z_1\overline{z_2}+\overline{z_1}{z}_2\le |z_1{|}^2+|z_2{|}^2+2|z_1||z_2|$$ which can be rearranged into $$|z_1+z_2{|}^2\le (|z_1|+|z_2|{)}^2$$ from which the result follows. ■

Definition (Argument)

Let $u\in \mathbb{C}$ such that $|u|=1$. Then $u$ is on the unit circle with the center at the origin of the complex plane. Let the paths round the unit circle from 1 to $u$ have signed lengths $\theta +2k\pi$ where $k\in \mathbb{Z}$ is measured in the anti-clockwise direction. We define the argument of $u$ to be

$$\arg (u)=\{\theta +2k\pi \mid k\in \mathbb{Z}\}.$$

The unique element of $\arg (u)$ in the interval $(-\pi ,\pi ]$ is called the principal argument of $u$, and is written capitalized as $\text{Arg}(u)$.

More generally, if $z\in \mathbb{C}\setminus \{0\}$, then we can define $\hat{z}=z/|z|$ so $|\hat{z}|=1$. We define $\arg (z)=\arg (\hat{z})$ and $\text{Arg}(z)=\text{Arg}(\hat{z})$. (See Figure 1)

Example. 

1. $\text{Arg}(1)=0$; $\arg (1)=\{0,\pm 2\pi ,\pm 4\pi ,\cdots \}$.
2. $\text{Arg}(i)=\pi /2$; $\arg (i)=\{\pi /2,\pi /2\pm 2\pi ,\pi /2\pm 4\pi ,\cdots \}$.
3. $\text{Arg}(-1)=\pi$;  $\arg (-1)=\{\pm \pi ,\pm 3\pi ,\pm 5\pi ,\cdots \}$.
4. $\text{Arg}(-i)=-\pi /2$; $\arg (-i)=\{-\pi /2,-\pi /2\pm 2\pi ,-\pi /2\pm 4\pi ,\cdots \}$.

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