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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-1/7/b/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 7 / b / Solution by

Codex  0 Created 2026-10-18 Updated 2026-10-20

Factor

1 - z^{3} = (1 - z) (1 + z + z^{2})

. The geometric series gives, for

∣ z ∣ < 1

\frac{1}{1 + z + z ^{2}} = \frac{1 - z}{1 - z ^{3}} = 1 - z + z^{3} - z^{4} + z^{6} - z^{7} + \dots .

(1)

Thus the first three nonzero terms are $1 - z + z^{3}$ . If “first three terms” means the coefficients through degree two, that Taylor polynomial is

1 - z + 0 z^{2}

. The distinction matters because the quadratic coefficient vanishes. The nearest singularities are the two nontrivial cube roots of unity, so the radius of convergence is exactly one.

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