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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}.

 

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