A complex number $z= x + yi$ can be written as the ordered pair $(x,y)$ of real numbers. Therefore, to the complex numbers we can join points in the coordinate plane. The $x$ axis of the coordinate plane is called the real axis and it contains only real numbers. The $y$ axis is called the imaginary axis and it contains only imaginary numbers. The coordinate plane which contains all complex numbers is called the complex plane or $z$-plane.
Every complex number $z=x +y i$ corresponds to the point $M=(x,y)$.
The modulus of the complex number $z = x + yi$ is the positive real number $|z| = \sqrt{x^2 + y^2}$. Geometrically, the modulus of the complex number $z = x + yi$ we interpret as the distance from the point $M = (x, y)$ to the origin:
Let $z_1 = x_1 + y_1 i$ and $z_2 = x_2 + y_2 i$ are the given complex numbers. Then is valid:
$$|z_1 – z_2| = | (x_1 – x_2) + (y_1 – y_2) i| = \sqrt{(x_1 – x_2) ^2 + (y_1 – y_2)^2}.$$
The expression above we recognized as the formula for distance between two points in the coordinate plane. Therefore, $|z_1 – z_2|$ is the distance between points $z_1$ and $z_2$ in the complex plane:
Properties of the modulus of complex numbers
$$|z_1 \cdot z_2| = |z_1| \cdot |z_2|$$
$$ \left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$$
$$|z^n| = |z|^n$$
$$|z_1 + z_2| \le |z_1| + |z_2|$$
$$z\cdot \overline{z} = |z|^2$$
$$z^{-1} = \frac{\overline{z}}{|z|^2}, z \neq 0$$
Example 1. Specify the set of all complex numbers $z$ in the complex plane for which $|z-z_0| =3$ is valid, where $z_0 = 1 + 2i$.
Solution:
Let $z=x + yi$ be the solution. Include $z$ in the equality $|z – z_0| = 3$:
$$|z – z_0| = |x + yi – (1 + 2i)|$$
$$= |(x-1) + (y – 2) i|$$
$$ = \sqrt{(x-1)^2 + (y-2) ^2} = 3.$$
This means that the distance from the point $ T = (x, y)$ to the point $S = ( 1, 2)$ is equal to $3$. A set of points $T$ which are from the point $S$ remote for $3$ is the circle with the center at point $S$ and with the radius $3$.
Example 2.
In the complex plane sketch all complex numbers $z$ for which is valid:
$$-1 \le Im [(1+i)z]\le 1.$$
Solution:
Let $z = x + yi$ be the solution.
Then
$$(1+i)z = (1+i)(x+yi)$$
$$= x + yi + xi – y$$
$$= x – y + (x+y) i. $$
It follows that the imaginary part of a complex number $(1+i)z$ is
$$Im[(1+i)z] = x+y.$$
Now we have inequalities:
$$ -1 \le x+y$$
and
$$ x + y \le 1.$$
This means that we have parts of a plane for which is valid
$$-1 – x \le y$$
and
$$y \le 1-x $$.
Example 3. Determine all numbers $z$ for which is:
$$ 1 \le |z+i| \le 4.$$
Solution:
Let $z = x + yi$ be the solution. Then
$$ 1 \le | x +y i +i| \le 4,$$
$$ 1 \le | x+ (y +1)i| \le 4.$$
We have two inequalities:
$$ | x + (y+1)i| \ge 1$$
and
$$ | x + (y+1)i| \le 4,$$
that is, from the definition of the modulus of complex numbers, we have
$$ \sqrt{x^2 + (y+1)^2} \ge 1$$
and
$$\sqrt{x^2 + (y+1)^2} \le 4.$$
We obtained the annulus between two concentric circles with the center at point $S(0,1)$ and radii $1$ and $4$
Example 4.
Show in the complex plane all complex numbers $z$ for which the following is valid:
$$| z+2| + |z-2| = 8.$$
Solution:
We need to find all complex numbers $z$ for which is the sum of distances from the points which are join to numbers $-2$ and $2$ is constant and equal to $8$.
Without calculating, we can see that this would be an ellipse with focii $F_1(-2, 0)$ and $F_2(2,0)$ and the length of the semi-major axis $a=4$.
Analytically:
Let $z = x+yi$ be the solution. Then
$$ | x+ yi +2 | + |x+yi -2| =8$$
$$|(x+2) + yi | + |(x-2) + yi| =8$$
$$ \sqrt{(x+2)^2 + y^2 } + \sqrt{(x-2)^2 + y^2} = 8$$
$$\sqrt{(x+2)^2 + y^2} = 8 – \sqrt{(x-2)^2 + y^2} / ^2$$
$$(x+2)^2 + y^2 = 64 – 16 \sqrt{(x-2)^2 + y^2} + (x-2)^2 + y^2$$
$$ x^2 + 4x + 4 +y^2 = 64 – 16 \sqrt{(x-2)^2 + y^2} + x^2 – 4x +4 + y^2$$
$$16 \sqrt{(x-2)^2 + y^2} = 64 – 8x / :8$$
$$ 2 \sqrt{(x-2)^2 + y^2} = 8 -x /^2$$
$$ 4 \cdot [(x-2)^2 + y^2] = 64 – 16x + x^2$$
$$ 4 \cdot ( x^2 – 4x + 4 + y^2) = 64 – 16x + x^2$$
$$ 4 x^2 – 16 x +16 + 4 y^2 = 64 – 16x + x^2$$
$$ 3 x^2 + 4 y^2 = 48 \ : 48$$
$$ \frac{3x^2}{48} + \frac{4 y^2}{48} = 1 $$
$$\frac{x^2}{16} + \frac{y^2}{12} = 1$$
We obtained an equation of the ellipse which the length of the semi-major axis is equal to $4$, and the length of the semi-minor axis is equal to $2 \sqrt{3}$.
