Geometric representation of complex numbers

Geometric representation of complex numbers

Geometric representation of complex numbers.

Let , with real . To this complex number we associate the point and the vector $\overrightarrow{OP}$ , both with coordinates

.

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.
$\begin{figure}\mbox{ \epsfig{file=Cauchy.eps,height=5cm}}\end{figure}$

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

**Figure 2:** Distances and circles.
$\begin{figure}\mbox{ \subfigure[]{\epsfig{file=vector.eps,height=3cm}} \qquad \qquad \subfigure[]{\epsfig{file=circle.eps,height=3cm}} }\end{figure}$

Example 1.2.1 (A perpendicular bisector) Represent in the plane the set of complex numbers

such that $\vert z-2i\vert=\vert z-1\vert$ .

Geometrically, the given equation represents the set of all the points at the equal distances from the points and , i.e. the perpendicular bisector of the segment .

Let , where are real. We have:

$\displaystyle \vert z-2i\vert=\vert z-1\vert \Longleftrightarrow \qquad \vert z-2i\vert^2$	$\displaystyle =\vert z-1\vert^2$
$\displaystyle x^2+(y-2)^2$	$\displaystyle = (x-1)^2+y^2$
$\displaystyle x^2 +y^2-4y+4$	$\displaystyle = x^2-2x+1+y^2$
$\displaystyle 2x-4y+3$	$\displaystyle =0.$

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

**Figure 3:** A perpendicular bisector.
$\begin{figure}\mbox{\epsfig{file=Line-1.eps,height=4cm}}\end{figure}$

Example 1.2.2 (Circles.) If

is the complex coordinate of a point

in the plane, then the circle whose center is in

and whose radius is equal to

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

in the complex plane such that

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

On the one hand we have:

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

Thus, the required geometric locus is contained in the unit circle. $\lvert z-2\rvert =\lvert z\rvert$ if, and only if, the point is on the perpendicular bisector of the segment whose endponts are the origin and the point . Therefore the coordinate of is equal to 1.

It follows that the required geometric locus is made of all the points on the unit circle whose coordinate is equal to 1; it contains exactly one point, namely ; see Figure 44.

**Figure 4:** Intersection of the unit circle and a line.
$\begin{figure}\mbox{\epsfig{file=GeometricLocus-1.eps,height=4cm}}\end{figure}$