 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-1/11/c/solution

For

b \neq = 0

, write

z = b w

. The equation becomes

w^{2} + w + 1 = 0

, whose roots are

(- 1 \pm i 3) /2

, both of modulus one. Hence

∣ z ∣ = ∣ b ∣

. If

b = 0

z = 0

, giving the same relation. Therefore

∣ z ∣ \leq 1 ⟺ - 1 \leq b \leq 1.

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