Complex Number Conditions

Question

Find all the complex numbers z₁, z₂, z₃ that satisfy the three following conditions simultaneously.

|z₁| = |z₂| = |z₃|

z₁ + z₂ + z₃ = 1

z₁z₂z₃ = 1

I'm a high school student and this was the final question in an exam yesterday. I've tried to consider a conjugate pair and third roots of unity but the algebra isn't working out. I could probably find one solution just by guessing values but am interested in how to develop a general formula for all solutions for conditions like this.

Answer 1

The first and last conditions together require that all of these complex numbers have modulus 1, since we can rewrite this as 1 = |z₁z₂z₃| = |z₁|*|z₂|*|z₃| = |z₁|³, which only happens when 1 = |z₁| = |z₂| = |z₃|. Now, if we add all of the nth roots of unity, they always add to 0, but a neat trick we can do is add the previous n-1th roots of unity in the following sense: if we are to find 3 numbers z₁, z₂, and z₃ that do this trick, we can set z₁ and z₂ equal to the two second roots of unity, and set z₃ equal to 1 itself. Now, we have z₁ + z₂ + z₃ = (z₁ + z₂) + z₃ = 0 + 1 = 1. The only problem is, when you multiply all the nth roots of unity, the value is (-1)ⁿ, and so in general, if we have an odd number for our n (like 3), then we can simply multiply the n-1th roots of unity by (-1)^(1/n-1). This does not change the value of the addition, and rotates the numbers around the unit circle. The product, in general, now becomes z₁z₂...z_n = z₁z₂...z_n-1*1 = (-1)^(n-1/n-1)*(-1)*1 = 1, where in the last equality, the first (-1) term comes from factoring a (-1)^(1/n-1) out of each z_i, the second (-1) term comes from the product z₁z₂...z_n-1 (since n itself is odd), and the last 1 term comes from z_n = 1.