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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/19/b/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 19 / b / Solution by

Codex  0 2026-10-18

By De Moivre's theorem,

(e^{i α})^{j} = e^{ij α}

. The complex sum whose real part is wanted is therefore a finite geometric series:

Z = \sum_{j = 0}^{n} e^{i (θ + j α)} = e^{i θ} \frac{1 - e ^{i (n + 1) α}}{1 - e ^{i α}}, α \in / 2 π Z .

Use

1 - e^{i β} = - 2 i e^{i β /2} sin (β /2)

in the numerator and denominator. This yields

Z = \frac{s i n (( n + 1 ) α /2 )}{s i n ( α /2 )} e^{i (θ + n α /2)} .

Taking its real part proves the finite trigonometric sum

j = 0 \sum n cos (θ + j α) = \frac{sin (( n + 1 ) α /2 )}{sin ( α /2 )} cos (θ + \frac{n α}{2}) .

When

α \in 2 π Z

, the printed quotient is undefined; the original sum is directly $(n + 1) cos θ$ , also the limiting value of the displayed identity. The denominator restriction cannot be omitted from a literal use of the quotient.

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