 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-1/2f/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

a

is an expansion

f (z) = \sum_{n = - \infty}^{\infty} c_{n} (z - a)^{n}

that converges to

f

on an annulus

r < ∣ z - a ∣ < R

\frac{10}{( z + 2 ) ( z ^{2} + 1 )} = \frac{2}{z + 2} + \frac{4 - 2 z}{z ^{2} + 1} .

For

0 < ∣ z ∣ < 1

f (z) = n = 0 \sum \infty \frac{( - 1 ) ^{n}}{2 ^{n}} z^{n} + (4 - 2 z) k = 0 \sum \infty (- 1)^{k} z^{2 k} .

This is actually a Taylor series at zero, since the apparent puncture at zero contains no singularity.

For

1 < ∣ z ∣ < 2

, the

z + 2

term still expands in nonnegative powers, while the quadratic term must be expanded in negative powers:

\frac{4 - 2 z}{z ^{2} + 1} = (4 z^{- 2} - 2 z^{- 1}) \frac{1}{1 + z ^{- 2}} .

Hence

f (z) = n = 0 \sum \infty \frac{( - 1 ) ^{n}}{2 ^{n}} z^{n} + (4 z^{- 2} - 2 z^{- 1}) k = 0 \sum \infty (- 1)^{k} z^{- 2 k} .

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