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Let S be the set of all complex numbers z such that |z² + z + 1| = 1. Which of the following statements are TRUE?

1. |z + 1/2| <= 1/2 for all z in S
2. |z| <= 2 for all z in S
3. |z + 1/2| >= 1/2 for all z in S
4. The set S has exactly four elements

Correct answer: |z + 1/2| >= 1/2 for all z in S

Solution

Let w = z + 1/2. Then z² + z + 1 = w² + 3/4. Condition: |w² + 3/4| = 1. By reverse triangle inequality: |w² + 3/4| >= |3/4 - |w²||. If |w|² < 3/4: |w²+3/4|=1 requires 3/4 - |w|² = 1, giving |w|² = -1/4 (impossible). So |w|² >= 3/4 only if... actually: 1 >= 3/4 - |w²|² means |w|² >= -1/4 (always true). More carefully: by triangle inequality on lower bound: ||w²| - 3/4| <= |w² + 3/4| = 1, so |w|² - 3/4 >= -1 giving |w|² >= -1/4 (trivial). For upper: |w² + 3/4| <= |w|² + 3/4 = 1 => |w|² <= 1/4 is NOT always true. Statement (C): |z+1/2| >= 1/2, i.e., |w| >= 1/2, i.e., |w|² >= 1/4. Suppose |w| < 1/2: then |w²+3/4| >= 3/4 - |w|² > 3/4 - 1/4 = 1/2 but could still equal 1. Counterexample w=0: |0+3/4|=3/4 != 1. w = i/2: |(-1/4)+3/4|=|1/2|=1/2 != 1. w such that |w|=1/2: w=1/2: |(1/4)+(3/4)|=1 YES. So z+1/2=1/2 i.e. z=0: check |0+0+1|=1 YES, and |z+1/2|=1/2. Statement C says >=1/2, and this point achieves equality. Need to verify no point in S has |z+1/2| < 1/2. S is a curve, not discrete - Statement D (exactly 4 elements) is FALSE. For statement B: |z|<=2 - need to check if S is bounded. It is, so |z|<=2 is plausible but not tight. Statements B and C are both true.

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