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Topos Uranos
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Geometry with complex numbers

Summary

This lesson uses complex numbers as coordinates in the plane and translates the notions of elementary geometry into algebraic operations: distance is the modulus of a difference, the midpoint and the centroid are averages, and the motions of the plane are short formulas. It proves that R(z)=c+eiφ(z−c)R(z) = c + e^{i\varphi}(z - c) is the rotation about cc through the angle φ\varphi, that the reflection in a line through the origin with angle α\alpha is S(z)=e2iαz‾S(z) = e^{2i\alpha}\conj{z} and that the composition of two reflections is a rotation. It characterizes collinearity and perpendicularity by means of a quotient, defines the orientation of a triangle by the sign of the imaginary part of that quotient, proves that three points form an equilateral triangle if and only if z1+ωz2+ω2z3=0z_1 + \omega z_2 + \omega^2 z_3 = 0, with ω=e2πi3\omega = e^{\frac{2\pi i}{3}}, and shows that the points whose distances to two fixed points are in a constant ratio form a circle, the circle of Apollonius. The problems culminate in Napoleon’s theorem.

Learning objectives

  1. Represent points, distances, midpoints and centroids with complex numbers, and express translations, rotations, dilations and reflections as functions of a complex variable.
  2. Prove the formulas for the rotation about a point and for the reflection in a line, and study the compositions of reflections and of rotations.
  3. Decide collinearity, perpendicularity and the angle between two line segments by means of the quotient of two differences, determine the orientation of a triangle by the sign of the imaginary part of that quotient, and characterize the equilateral triangle with the cube root of unity ω\omega.
  4. Describe circles with the modulus, prove the theorem on the circle of Apollonius and use these tools in classical problems, such as those of Napoleon and of Van Aubel.

Points, distances and centers

Since the Algebra and Geometry I lesson on operations with complex numbers, each complex number z=a+biz = a + bi is identified with the point (a,b)(a, b) of the plane, and the difference z−wz - w is the vector that goes from the point ww to the point zz; by the Pythagorean theorem, the distance between the two points is ∣z−w∣\abs{z - w}. In this lesson the point of view is reversed: the numbers are not studied through the plane, but the plane through the numbers. A figure is described by the complex numbers of its points, and a geometric property by an equality between them.

The main tool is the complex exponential from the lesson on the complex exponential and Euler’s formula. For every real φ\varphi, the number eiφe^{i\varphi} is the point of the circle of radius 11 whose argument is φ\varphi, and it has the following properties; the last one holds because cos⁡(−φ)=cos⁡φ\cos(-\varphi) = \cos \varphi and sin⁡(−φ)=−sin⁡φ\sin(-\varphi) = -\sin \varphi:

eiφ=cos⁡φ+isin⁡φ,∣eiφ∣=1,eiαeiβ=ei(α+β),eiφ‾=e−iφe^{i\varphi} = \cos \varphi + i \sin \varphi, \qquad \abs{e^{i\varphi}} = 1, \qquad e^{i\alpha} e^{i\beta} = e^{i(\alpha + \beta)}, \qquad \conj{e^{i\varphi}} = e^{-i\varphi}

According to the lesson on the polar form of complex numbers, every nonzero complex number can be written z=reiθz = re^{i\theta}, with r=∣z∣r = \abs{z} and θ\theta an argument of zz; when two complex numbers are multiplied, their moduli are multiplied and their arguments are added, and when they are divided, their moduli are divided and their arguments are subtracted. In particular, multiplying zz by eiφe^{i\varphi} preserves its modulus and increases its argument by φ\varphi: it rotates the point zz through the angle φ\varphi about the origin, counterclockwise if φ>0\varphi > 0. The product by i=eiπ2i = e^{\frac{i\pi}{2}}, studied in the Course I lesson on the complex numbers as a quarter-turn rotation, is the case φ=π2\varphi = \frac{\pi}{2}.

The points of a line segment can also be described with complex numbers. According to the Course I lesson on vectors and parametric equations of a line, the points of the line through z1z_1 and z2z_2 are those of the form z1+t(z2−z1)z_1 + t(z_2 - z_1), with tt real, and those of the line segment joining them correspond to 0≤t≤10 \leq t \leq 1.

LemmaMidpoint and centroid in the complex plane

The midpoint of the line segment with endpoints z1z_1 and z2z_2 is half the sum m=z1+z22m = \frac{z_1 + z_2}{2}. The three medians of a triangle with vertices z1z_1, z2z_2 and z3z_3 pass through the point g=z1+z2+z33g = \frac{z_1 + z_2 + z_3}{3}, called the centroid, which lies on each median two thirds of the way from the vertex to the midpoint of the opposite side. The lemma was proved with vectors in the solved problem on the medians in the Course I lesson on vectors and parametric equations of a line; the proof below translates it into complex numbers. The figure follows it with the vertices z1=−2−iz_1 = -2 - i, z2=3−iz_2 = 3 - i and z3=2iz_3 = 2i.

Proof

  1. m=z1+12(z2−z1)=z1+z22\dato{m}{m = z_1 + \resaltar{\frac{1}{2}}(z_2 - z_1) = \resaltar{\frac{z_1 + z_2}{2}}}

    The point of the line segment that corresponds to t=12t = \frac{1}{2} is half the sum of the endpoints.

  2. m−z1=z2−z12=z2−m→∣m−z1∣=∣z2−m∣m - z_1 = \resaltar{\frac{z_2 - z_1}{2}} = z_2 - m \rightarrow \resaltar{\abs{m - z_1} = \abs{z_2 - m}}

    It is at the same distance from both endpoints, because the two vectors joining it to them are equal: it is the midpoint.

  3. z3+23(z1+z22−z3)=z1+z2+z33z_3 + \resaltar{\frac{2}{3}} \left( \frac{z_1 + z_2}{2} - z_3 \right) = \dato{g}{\resaltar{\frac{z_1 + z_2 + z_3}{3}}}

    On the median joining z3z_3 to the midpoint of the opposite side, the point that corresponds to t=23t = \frac{2}{3} is the following one.

  4. z1+23(z2+z32−z1)=g=z2+23(z1+z32−z2)z_1 + \frac{2}{3} \left( \frac{z_2 + z_3}{2} - z_1 \right) = \resaltar{g} = z_2 + \frac{2}{3} \left( \frac{z_1 + z_3}{2} - z_2 \right)

    The expression obtained is symmetric in the three vertices; this being so, the same computation with either of the other two medians gives the same point, and all three pass through gg. With the vertices of the figure, g=13g = \frac{1}{3}.

Thus the midpoint is the average of two complex numbers, and the centroid, that of three.