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       

• $2i$ is pure imaginary.
• $4 \pi$ is real.

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       

• $(2+3i)+(4+7i)= 6+ 10i$ .
• $(2+3i)(4+7i)= (2 \cdot 4 - 3 \cdot 7) + i(2 \cdot 7 + 3 \cdot 4)=-13=26i$ .
• $(2+3i)^2=2^2+2 \cdot 2 \cdot 3i + (3i)^2= 4+6i-9=-5+6i$ .
• $(5-2i)^3=5^3-3 \cdot 5^2 \cdot 2i + 3 \cdot 5 \cdot (2i)^2 -(2i)^3=125 -150i-60+8i=65-142i$ .
• $z^3+i=z^3-i^3=(z-i)(z^2+iz+i^2=(z-i)(z^2+iz-1)$ .

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

• : Associativity: Z1 := x[1] + I*y[1]: `z1 ` = Z1;

  Z2 := x[2] + I*y[2]: `z2 ` = Z2;

  Z3 := x[3] + I*y[3]: `z3 ` = Z3;

  w1 := Z1*(Z2 + Z3):

  w2 := Z1*Z2 + Z1*Z3: ` `;

  `z1*(z2 + z3) ` = w1;

  `z1*z2 + z1*z3 ` = w2;

  w1 := expand(w1):

  w2 := expand(w2): ` `;

  `z1*(z2 + z3) ` = w1;

  `z1*z2 + z1*z3 ` = w2; ` `;

  `Does z1*(z2 + z3) = z1*z2 + z1*z3 ¿;

  evalb(w1 = w2);

• Commutativity: Z1:='Z1': Z1 := x[1] + I*y[1]:

  Z2:='Z2': Z2 := x[2] + I*y[2]:

  z[1] = Z1;

  z[2] = Z2; ` `;

  `z1 + z2` = Z1 + Z2;

  `z2 + z1` = Z2 + Z1; ` `;

  `Does z1 + z2 = z2 + z1 ¿;

  Z1+Z2 = Z2+Z1;

  evalb(Z1+Z2 = Z2+Z1);

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       

• $\overline{2+4i}=2-4i$ .
• $\overline{5-2i}=5+2i$ .
• $\overline{4i-3}=-3+4i$ .

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:

$$\overline{z_1+z_2} =\overline{(x_1+x_2)+i(y_1+y_2)}$$

$$\quad =(x_1+x_2)-i(y_1+y_2)$$

$$\quad =(x_1-iy_1)+(x_2-iy_2)$$

$$\quad =\overline{z_1}+\overline{z_2}.$$

(ii)

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

$$\overline{z_1z_2} =\overline{(x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1)}$$

$$\quad =(x_1x_2-y_1y_2)-i(x_1y_2+x_2y_1$$

$$\quad = (x_1-iy_1)(x_2-iy_2).$$

$\qedsymbol$

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:

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

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:

• If $x=-1$ , then $z=iy$ and $z^2=-y^2$ (a negative real number).
• If $y=0$ , then $z$ itself is a real number, and the square of a real number is a real number.

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:

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

Proof.     

$$z^{-1} = \frac 1z = \frac {1}{a+ib}$$

$$\quad = \frac {a-ib}{(a+ib)(a-ib)}$$

$$\quad = \frac {a-ib}{a^2+b^2}$$

$$\quad = \frac {a}{a^2+b^2} - i \frac {b}{a^2+b^2}.$$

$\qedsymbol$

In other words, we have:

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

Example 1.1.12  

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

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

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:

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

(ii)

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

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

$\qedsymbol$

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:

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

Example 1.1.15  

$$\vert 4+3i\vert = \sqrt {4^2+3^2}=\sqrt{25}=5.$$

$$\vert 3+5i\vert = \sqrt{3^2+5^2}=\sqrt{34}.$$

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

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:

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

On the other hand, we have:

$$\vert z_1\vert^2 \cdot \vert z_2\vert^2= (x_1^2+y_1^2)+(x_2^2+y_2^2).$$ (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 $z_2=3-4i$ ,

• $z_1+z_2=4-2i$ .
• $z_1 \cdot z_2 = 11+2i$ .
• $\vert z_1\vert=\sqrt{5}$ and $\vert z_2\vert=5$ .

Then we have:

$$\vert z_1 z_2\vert = \sqrt{11^2+2^2}=\sqrt{125}=5\sqrt{5}=\vert z_1\vert\;\vert z_2\vert.$$

$$\vert z_1 +z_2\vert = \sqrt{4-2i}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5} \leq 5+\sqrt{5}.$$

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

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

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:

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

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:

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

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  

$$\frac {1+2i}{2-3i} =\frac {(1+2i)(2+3i)}{(2-3i)(2+3i)}=\frac {-5+7i}{13}=\frac {1}{13}(-5+7i).$$

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