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.
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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.
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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:

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

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

Figure 3: A perpendicular bisector.
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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

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

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.

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.
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Noah Dana-Picard 2007-12-24