 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/7/d/ii/solution

For

∣ z ∣ < 1

, multiply the absolutely convergent Taylor series of the sine by the geometric series:

\frac{s i n z}{1 - z ^{2}} = [\sum_{j = 0}^{\infty} \frac{( - 1 ) ^{j} z ^{2 j + 1}}{( 2 j + 1 )!}] [\sum_{k = 0}^{\infty} z^{2 k}] .

The product is odd, so all even coefficients vanish. Collecting the coefficient of

z^{2 m + 1}

gives

a_{2 m} = 0, a_{2 m + 1} = j = 0 \sum m \frac{( - 1 ) ^{j}}{( 2 j + 1 )!}, m = 0, 1, 2, \dots .

For example, the first terms are

z + 5 z^{3} /6 + 101 z^{5} /120 + \dots

. The denominator vanishes at

z = \pm 1

, and the numerator is nonzero there, so these are genuine simple poles. Therefore

R = 1

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