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AMC 12 only

Complex Plane Geometry

Treating points as complex numbers turns rotation, reflection, and regular-polygon problems into algebra — often faster than synthetic geometry.

Multiplication as transformation

Multiplying a complex number by eiθe^{i\theta} rotates it by angle θ\theta about the origin, without changing its distance from the origin. Multiplying by a positive real numberkk scales it by kk. Multiplying by keiθk e^{i\theta} does both at once — this is exactly a spiral similarity, expressed algebraically.

Distances and centers

z1z2=distance between the points,midpoint=z1+z22,centroid=z1+z2+z33|z_1-z_2| = \text{distance between the points}, \qquad \text{midpoint} = \dfrac{z_1+z_2}{2}, \qquad \text{centroid} = \dfrac{z_1+z_2+z_3}{3}

Recognizing loci

  • zz0=r|z-z_0|=r — a circle centered at z0z_0 with radius rr.
  • zz1=zz2|z-z_1|=|z-z_2| — the perpendicular bisector of the segment joining z1z_1 and z2z_2.

Rotating a point about another point

To rotate zz by angle θ\theta about a center zcz_c, shift to the origin, rotate, then shift back:

z=zc+(zzc)eiθz' = z_c + (z-z_c)e^{i\theta}

This single formula is the workhorse for equilateral-triangle and regular-polygon construction problems in the complex plane.

Worked example

Problem: Points A=0A=0 and B=4B=4 are two vertices of an equilateral triangle in the complex plane. Find the two possible locations for the third vertexCC.

Solution: CC is obtained by rotating BB about AAby ±60°\pm60° (the two choices give the two possible orientations of the triangle):

C=0+(40)e±i60°=4(cos60°±isin60°)=2±2i3C = 0 + (4-0)e^{\pm i60°} = 4\left(\cos60° \pm i\sin60°\right) = 2 \pm 2i\sqrt3

So C=2+2i3C = 2+2i\sqrt3 or C=22i3C=2-2i\sqrt3 — the two reflections of each other across the real axis, as expected.

Practice problems

1.Find the midpoint of z1=3+2iz_1=3+2i and z2=1+6iz_2=-1+6i.

2.Describe the locus z3=z+3|z-3|=|z+3|.

3.Rotate the point z=2z=2 by 90°90° about the origin.

4.Find the centroid of the triangle with vertices 1+i,3i,2+4i1+i, 3-i, -2+4i.