Geometric representation of complex numbers.

Let $z=a+bi$ , with real $a,b$ . To this complex number we associate the point $P$ and the vector $\overrightarrow{OP}$ , both with coordinates $(a,b)$ .

When used to represent complex numbers, the Euclidean plane is called the Cauchy-Argand plane or Gauss plane. Biographies of Augustin Cauchy (1789-1857) and Jean Argand (1768-1822) can be found in The MacTutor History of Mathematics archive, at the following URL:

$z+z'=(a+c)+(b+d)i$

A biography

Figure 1: Augustin Cauchy.

Let $A$ and $B$ be two points with respective ``complex coordinate'' $z$ and $z'$ . Then $AB= \vert z'-z\vert$ .

Figure 2: Distances and circles.

Example 1.2.1 (A perpendicular bisector)   Represent in the plane the set of complex numbers $z$ such that $\vert z-2i\vert=\vert z-1\vert$ .

Geometrically, the given equation represents the set of all the points $M$ at the equal distances from the points $A(0,2)$ and $B(1,0)$ , i.e. the perpendicular bisector of the segment $AB$ .

Let $z=x+iy$ , where $x,y$ are real. We have:

$$\vert z-2i\vert=\vert z-1\vert \Longleftrightarrow \qquad \vert z-2i\vert^2 =\vert z-1\vert^2$$

$$x^2+(y-2)^2 = (x-1)^2+y^2$$

$$x^2 +y^2-4y+4 = x^2-2x+1+y^2$$

$$2x-4y+3 =0.$$

This is the equation of the perpendicular bisector of $AB$ (see Figure 3).

Figure 3: A perpendicular bisector.

Example 1.2.2 (Circles.)   If $z_0$ is the complex coordinate of a point $A$ in the plane, then the circle whose center is in $A$ and whose radius is equal to $R$ is defined by the equation $\vert z-z_0\vert=R$ (see Figure 2(b)).

Example 1.2.3 (A solved exercise.)   Find the geometric locus of all the points $M(z)$ in the complex plane such that

$$\lvert z-2\rvert =\lvert z\rvert =\lvert\frac 1z\rvert .$$

On the one hand we have:

$$\lvert z\rvert =\lvert\frac 1z\rvert \Longleftrightarrow \lvert z\rvert ^2=1 \Longleftrightarrow \lvert z\rvert =1.$$

Thus, the required geometric locus is contained in the unit circle.

On the other hand, $\lvert z-2\rvert =\lvert z\rvert$ if, and only if, the point $M(z)$ is on the perpendicular bisector of the segment whose endponts are the origin and the point $T(2)$ . Therefore the $x-$ coordinate of $M$ is equal to 1.

It follows that the required geometric locus is made of all the points on the unit circle whose $x-$ coordinate is equal to 1; it contains exactly one point, namely $A(1)$ ; see Figure 4.

Figure 4: Intersection of the unit circle and a line.