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Exercise 1229Algebra › Complex numbersOlympiadAdded 4 Nov 2026

Three unit complex numbers with zero sum

#complex numbers#unit circle#equilateral triangle#conjugate trick

Statement

  1. Let z,wz,w be complex numbers of modulus 11 with zw≠−1zw\ne-1. Show that z+w1+zw\dfrac{z+w}{1+zw} is real.
  2. Let z1,z2,z3z_1,z_2,z_3 be complex numbers of modulus 11 with z1+z2+z3=0z_1+z_2+z_3=0. Show that they form an equilateral triangle.
Hint

For unit complex numbers, zˉ=1z\bar z=\frac1z. For part 2, take ∣z1+z2∣=∣z3∣|z_1+z_2|=|z_3| and expand the square of the modulus.

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Solution

1. Let u=z+w1+zwu=\dfrac{z+w}{1+zw}. Using zˉ=1z\bar z=\frac1z and wˉ=1w\bar w=\frac1w:
uˉ=1z+1w1+1zw=z+w1+zw=u.\bar u=\frac{\frac1z+\frac1w}{1+\frac1{zw}}=\frac{z+w}{1+zw}=u.
So u=uˉu=\bar u, i.e. uu is real.

2. From z1+z2=−z3z_1+z_2=-z_3 we get ∣z1+z2∣2=1|z_1+z_2|^2=1. Expanding,
∣z1+z2∣2=2+2 Re(z1zˉ2)=1,|z_1+z_2|^2=2+2\,\mathrm{Re}(z_1\bar z_2)=1,
so Re(z1zˉ2)=−12\mathrm{Re}(z_1\bar z_2)=-\frac12. Then
∣z1−z2∣2=2−2 Re(z1zˉ2)=3.|z_1-z_2|^2=2-2\,\mathrm{Re}(z_1\bar z_2)=3.
By symmetry, ∣z2−z3∣2=∣z3−z1∣2=3|z_2-z_3|^2=|z_3-z_1|^2=3. The three sides all have length 3\sqrt3: equilateral.

Lesson. For points on the unit circle use zˉ=1/z\bar z=1/z: it turns geometric conditions into algebraic identities.