 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/11/f/solution

For

θ \in / 2 π Z

e^{i θ}

\sum_{n = 0}^{N - 1} e^{in θ} = \frac{1 - e ^{i Nθ}}{1 - e ^{i θ}} = e^{i (N - 1) θ /2} \frac{s i n ( Nθ /2 )}{s i n ( θ /2 )} .

The second equality follows by factoring

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

n = 0 \sum N - 1 cos (n θ) = \frac{cos (( N - 1 ) θ /2 ) sin ( Nθ /2 )}{sin ( θ /2 )} .

When

θ = 2 p π

, every cosine is

1

, so the sum is

N

. This also gives the continuous limiting value at the excluded arguments.

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