Problem:
How many integers n≥2 are there such that whenever z1​,z2​,…,zn​ are complex numbers such that
∣z1​∣=∣z2​∣=…=∣zn​∣=1 and z1​+z2​+…+zn​=0
then the numbers z1​,z2​,…,zn​ are equally spaced on the unit circle in the complex plane?
Answer Choices:
A. 1
B. 2
C. 3
D. 4
E. 5
Solution:
For n=2, we see that if z1​+z2​=0, then z1​=−z2​, so they are evenly spaced along the unit circle.
For n=3, WLOG, we can set z1​=1. Notice that now ℜ(z2​+z3​)=−1 and ℑ{z2​}=−ℑ{z3​}. This forces z2​ and z3​ to be equal to ei32π​ and e−i32π​, meaning that all three are equally spaced along the unit circle.
We can now show that we can construct complex numbers when n≥4 that do not satisfy the conditions in the problem.
Suppose that the condition in the problem holds for some n=k. We can now add two points zk+1​ and zk+2​ anywhere on the unit circle such that zk+1​=−zk+2​, which will break the condition. Now that we have shown that n=2 and n=3 works, by this construction, any n≥4 does not work, making the answer (B)2​ .
The problems on this page are the property of the MAA's American Mathematics Competitions