 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-1/12f/c/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Put

z_{k} = \frac{1}{a _{k}} = \frac{2}{( 2 k + 1 ) π}, w = z - z_{k} .

Near

a_{k}

δ = 1/ z - a_{k}

gives

tan (a_{k} + δ) = - cot δ = - \frac{1}{δ} + \frac{δ}{3} + O (δ^{3}) .

But

δ = \frac{1}{z} - \frac{1}{z _{k}} = - \frac{w}{z _{k} ( z _{k} + w )},

and hence, exactly,

- \frac{1}{δ} = \frac{z _{k} ( z _{k} + w )}{w} = \frac{z _{k}^{2}}{w} + z_{k} .

Since

δ = O (w)

, the first two Laurent terms are

tan \frac{1}{z} = \frac{z _{k}^{2}}{z - z _{k}} + z_{k} + O (z - z_{k}) .

Equivalently, substitute

z_{k} = 2/ ((2 k + 1) π)

in this expression.

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