Latest Tweets

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 $|z – z_0| = 3$:

$$|z – z_0| = |x + yi – (1 + 2i)|$$

$$=  |(x-1) + (y – 2) i|$$

$$ = \sqrt{(x-1)^2 + (y-2) ^2} = 3.$$

 

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 $2 \sqrt{3}$.