Complex plane

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 $|z - z_0| = 3$ :

$\begin{aligned} |z - z_0| &= |x + yi - (1 + 2i)| \\ & = |(x-1) + (y - 2) i| \\ & = \sqrt{(x-1)^2 + (y-2) ^2} = 3. \end{aligned}$

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

$\begin{aligned} (1+i)z &= (1+i)(x+yi) \\ &= x + yi + xi - y \\ & = x - y + (x+y) i. \\ \end{aligned}$

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 $2 \sqrt{3}$ .

Latest Tweets

For Those Who Want To Learn More:

Navigation

Latest posts

The bridges of Konigsberg

Prime numbers

Euclidean algorithm

Koch snowflake

Most downloaded worksheets