Problem:
Let ω=−21+21i3. Let S denote all points in the complex plane of the form a+bω+cω2, where 0≤a≤1,0≤b≤1, and 0≤c≤1. What is the area of S ?
Answer Choices:
A. 213
B. 433
C. 233
D. 21π3
E. π Solution:
Notice that ω=e32iπ, which is one of the cube roots of unity. We wish to find the span of (a+bω+cω2) for reals 0≤a,b,c≤1. Observe also that if a,b,c>0, then replacing a,b, and c by a−min(a,b,c),b−min(a,b,c), and c−min(a,b,c) leaves the value of a+bω+cω2 unchanged. Therefore, assume that at least one of a,b,c is equal to 0 . If exactly one of them is 0 , we can form an equilateral triangle of side length 1 using the remaining terms. A similar argument works if exactly two of them are 0 . In total, we get 3+(23)=6 equilateral triangles, whose total area is 6⋅43=(C)233.
Note: A diagram of the six equilateral triangles is shown below.
