Algebraic form.

A complex number is an expression of the form $ z=a+bi$ , where $ a,b$ are real numbers and $ i$ is a ``number'' such that $ i^2=-1$ . The number $ a$ is called the real part of $ z$ ( $ a=\operatorname{Re}\nolimits (z)$ ) and $ b$ is the imaginary part (or imaginary coefficient) of $ z$ ( $ b=\operatorname{Im}\nolimits (z)$ ).

The set of all complex numbers is denoted $ \mathbb{C}$ . We denote $ \mathbb{C}^* = \mathbb{C}-{0}$ .

Properties:

  1. Equality: $ z_1=z_2 \Leftrightarrow \operatorname{Re}\nolimits (z_1)=\operatorname{Re}\nolimits (z_2)$ and $ \operatorname{Im}\nolimits (z_1)=\operatorname{Im}\nolimits (z_)$ .
  2. $ z \in \mathbb{R} \Leftrightarrow$   Im$ (z)=0$ .
  3. If $ \operatorname{Re}\nolimits (z)=0$ , $ z$ is called pure imaginary.

Example 1.1.1       

Definition 1.1.2 (Operations)   Let $ z=a_1+bi$ and $ z'=c+di$ .
  1. Addition: $ z+z'=(a+c)+(b+d)i$ .
  2. Multiplication: $ zz' = (ac-bd)+(ad+bc)i$ .

Example 1.1.3 (Maple writing)   : Z1 := 2 + 5*I:

Z2 := 3 - 7*I:

z[1] = Z1;

z[2] = Z2; ` `;

z[1]+z[2]=Z1+Z2;

z[1]*z[2] = Z1*Z2;

These operations have the same algebraic properties as the corresponding operations in $ \mathbb{R}$ (associativity, commutativity, etc.; please prove ...). Thus, the classical formulas (such as Newton's binomial) are also true in $ \mathbb{C}$ .

  1. $ \forall z_1, z_2 \in \mathbb{C}, \; (z_1+z_2)^2=z_1^2+2z_1z_2+z_2^2$ .
  2. $ \forall z_1, z_2 \in \mathbb{C}, \; (z_1-z_2)^2=z_1^2-2z_1z_2+z_2^2$ .
  3. $ \forall z_1, z_2 \in \mathbb{C}, \; (z_1-z_2)(z_1+z_2)=z_1^2-z_2^2$ .
  4. $ \forall z_1, z_2 \in \mathbb{C}, \forall n \in \mathbb{N}, \;
(z_1+z_2)^n= \underset{k=0}{\overset{n}{\sum}} \begin{pmatrix}n \ k \end{pmatrix} z_1^kz_2^{n-k}$ .
  5. $ \forall z_1, z_2 \in \mathbb{C}, \forall n \in \mathbb{N}, \;
z_1^n-z_2^n= \underset{k=0}{\overset{n-1}{\sum}} z_1^{n-1-k}z_2^k$ .

Example 1.1.4       

It is possible to use Maple to verify these laws (associativity, commutativity, etc.). Here a a few examples:

Definition 1.1.5   Let $ z=a+bi$ , where $ a,b\in \mathbb{R}$ . The complex conjugate of $ z$ is the number $ \overline{z}=a-bi$ .

Example 1.1.6       

Example 1.1.7 (With Maple)   Z1 := 3 + 2*I:

z[1] = Z1;

conjugate(z[1]) = conjugate(Z1); `     `;

Z2 := -7 + 5*I:

z[2] = Z2;

conjugate(z2) = conjugate(Z2);

Proposition 1.1.8   $ \forall z_1, z_2 \in \mathbb{C}$ :
(i)
$ \overline{z_1+z_2}=\overline{z_1}+\overline{z_2}$ .
(ii)
$ \overline{z_1 \cdot z_2}=\overline{z_1} \cdot \overline{z_2}$ .

Proof. Denote $ z_1=x_1+iy_1$ and $ z_2=x_2+iy_2$ , where $ x_1,x_2,y_1,y_2$ are real numbers. Then:
(i)
$ z_1+z_2=(x_1+x_2)+i(y_1+y_2)$ , thus:

$\displaystyle \overline{z_1+z_2}$ $\displaystyle =\overline{(x_1+x_2)+i(y_1+y_2)}$    
$\displaystyle \quad$ $\displaystyle =(x_1+x_2)-i(y_1+y_2)$    
$\displaystyle \quad$ $\displaystyle =(x_1-iy_1)+(x_2-iy_2)$    
$\displaystyle \quad$ $\displaystyle =\overline{z_1}+\overline{z_2}.$    

(ii)
$ z_1z_2=(x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1)$ , thus:

$\displaystyle \overline{z_1z_2}$ $\displaystyle =\overline{(x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1)}$    
$\displaystyle \quad$ $\displaystyle =(x_1x_2-y_1y_2)-i(x_1y_2+x_2y_1$    
$\displaystyle \quad$ $\displaystyle = (x_1-iy_1)(x_2-iy_2).$    

$ \qedsymbol$

The following proposition is very simple; we leave the task of proving it to the reader.

Proposition 1.1.9   Let $ z \in \mathbb{C}$ . Then:
  1. $ z \in \mathbb{R} \Longleftrightarrow \overline{z}=z$ .
  2. $ z$ is pure imaginary if, and only if, $ \overline{z}= -z$ .

Proof. Denote $ z=x+iy$ , where $ x$ and $ y$ are real numbers. Then we have:
  1. $ z\in \mathbb{R} \Longleftrightarrow y=0 \Longleftrightarrow z=x \Longleftrightarrow \overline{z}=x
\overline{z}=z$ .
  2. $ z$ is pure imaginary if, and only if, $ x=0$ , that is $ z=iy$ . In this case, $ \overline{z}=-iy=-z$ .
$ \qedsymbol$

Example 1.1.10   Find all the complex numbers $ z$ such that $ z+1$ is a real number.

Denote $ z=x+iy$ , where $ x$ and $ y$ are real numbers. Then we have:

$\displaystyle (z+1)^2=(x+iy+1)^2=[(x+1)+iy]^2=(x+1)^2-y^2+2i(x+1)\; y.$    

Now, $ z+1$ is a real number if, and only if, $ y(x+1)=0$ , i.e. $ x=-1$ or $ y=0$ .

Let us consider the two cases separately:

Definition 1.1.11 (Inverse of a complex)   If $ z \neq 0$ , then it has a complex inverse $ z^{-1}$ . Let $ z=a+bi$ , where $ a$ and $ b$ are real numbers; then we have:

$\displaystyle z^{-1}=\frac {a}{a^2+b^2} - i \frac {b}{a^2+b^2}.$    

Proof.     

$\displaystyle z^{-1}$ $\displaystyle = \frac 1z = \frac {1}{a+ib}$    
$\displaystyle \quad$ $\displaystyle = \frac {a-ib}{(a+ib)(a-ib)}$    
$\displaystyle \quad$ $\displaystyle = \frac {a-ib}{a^2+b^2}$    
$\displaystyle \quad$ $\displaystyle = \frac {a}{a^2+b^2} - i \frac {b}{a^2+b^2}.$    

$ \qedsymbol$

In other words, we have:

$\displaystyle \forall z \in \mathbb{C}- \{ 0 \}, \; \frac {1}{z}= \frac {\overline{z}}{z \; \overline{z}}.$    

Example 1.1.12  

$\displaystyle \frac {1}{1+i}$ $\displaystyle = \frac {1-i}{(1+i)(1-i)} = \frac {1-i}{2}= \frac 12 - \frac 12 i.$    
$\displaystyle \frac {1}{2-3i}$ $\displaystyle = \frac {2+3i}{(2-3i)(2+3i)}=\frac {2+3i}{13}=\frac {2}{13} + \frac {3}{13}i.$    

Proposition 1.1.13   For any $ z \in \mathbb{C}$ , we have:
(i)
$ z+ \overline{z} \in \mathbb{R}$ .
(ii)
$ z \cdot \overline{z} \in \mathbb{R}^{+}$ .

Proof.     

(i)
If $ z=x+iy$ , with real $ x$ and $ y$ , we have:

$\displaystyle z+ \overline{z}=(x+iy)+(x-iy)=2x=2 \operatorname{Re}\nolimits (z).$    

(ii)
If $ z=x+iy$ , with real $ x$ and $ y$ , we have:

$\displaystyle z+ \overline{z}=(x+iy)+(x-iy)=2yi=2i \operatorname{Im}\nolimits (z).$    

$ \qedsymbol$

The property in Proposition 1.13 (ii) justifies the following definition:

Definition 1.1.14 (Absolute value)   $ \forall z \in \mathbb{C}, \; \vert z\vert=\sqrt{z \cdot \overline{z}}$ .

For $ z=x+iy$ , where $ x$ and $ y$ are real numbers, we have:

$\displaystyle \vert z\vert= \sqrt{x^2+y^2}$    

Example 1.1.15  

$\displaystyle \vert 4+3i\vert$ $\displaystyle = \sqrt {4^2+3^2}=\sqrt{25}=5.$    
$\displaystyle \vert 3+5i\vert$ $\displaystyle = \sqrt{3^2+5^2}=\sqrt{34}.$    
$\displaystyle \vert 4-6i\vert$ $\displaystyle = \sqrt{4^2+(-6)^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}.$    

Proposition 1.1.16   For any $ z_1, z_2 \in \mathbb{C}$ , we have:
(i)
$ \vert z_1 \cdot z_2\vert=\vert z_1\vert \cdot \vert z_2\vert$ .
(ii)
$ \vert z_1+z_2\vert \leq \vert z_1\vert + \vert z_2\vert$ .

Proof. We denote $ z=x_1+iy_1$ and $ z_2=x_2+iy_2$ , where $ x_1,x_2,y_1,y_2$ are real numbers.
(i)
$ z_1z_2=(x_1+iy_1)(x_2+iy_2)=(x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1)$ , thus on the one hand, we have:

$\displaystyle \vert z_1z_2\vert^2=(x_1x_2-y_1y_2)^2+(x_1y_2+x_2y_1)^2.$ (1.1)

On the other hand, we have:

$\displaystyle \vert z_1\vert^2 \cdot \vert z_2\vert^2= (x_1^2+y_1^2)+(x_2^2+y_2^2).$ (1.2)

We develop the right-hand sides of Equation 1 and of Equation 2 and the required equality follows.
(ii)
We can leave it to the reader. Section 2 explains why it is possible here to use results from analytic geometry.
$ \qedsymbol$

Example 1.1.17   If $ z_1=1+2i$ and $ z_2=3-4i$ , Then we have:

$\displaystyle \vert z_1 z_2\vert$ $\displaystyle = \sqrt{11^2+2^2}=\sqrt{125}=5\sqrt{5}=\vert z_1\vert\;\vert z_2\vert.$    
$\displaystyle \vert z_1 +z_2\vert$ $\displaystyle = \sqrt{4-2i}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5} \leq 5+\sqrt{5}.$    

Corollary 1.1.18   For any $ z \in \mathbb{C} - \{ 0 \}$ , we have:

$\displaystyle \begin{vmatrix}\frac 1z \end{vmatrix} = \frac {1}{\vert z\vert}.$    

The proof is an easy consequence of Proposition 1.16.

Corollary 1.1.19   For any $ z_1 \in \mathbb{C}$ and any $ z_2 \in \mathbb{C} - \{ 0 \}$ we have:

$\displaystyle \begin{vmatrix}\frac {z_1}{z_2} \end{vmatrix} = \frac {\vert z_1\vert}{\vert z_2\vert}.$    

Here the proof is an easy consequence of Corollary 1.18 and Proposition 1.16.

Remark 1.1.20   For any $ z \in \mathbb{C}^{*}$ , we have:

$\displaystyle z^{-1}= \frac {\overline{z}}{\vert z\vert^2}$    

It follows that the best way to compute a quotient is to multiply both the numerator and the denominator by the conjugate of the denominator, and then simplify.

Example 1.1.21  

$\displaystyle \frac {1+2i}{2-3i}$ $\displaystyle =\frac {(1+2i)(2+3i)}{(2-3i)(2+3i)}=\frac {-5+7i}{13}=\frac {1}{13}(-5+7i).$    
$\displaystyle \frac {5+7i}{3+4i}$ $\displaystyle =\frac {(5+7i)(3+4i)}{(3-4i)(3-4i)}=\frac {43+i}{9+16}=\frac {1}{25}(43+i).$    

Noah Dana-Picard 2007-12-24