 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-1/12g/c/solution

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

For

g (z) = \frac{e ^{z} - 1}{z l o g ( 1 + z )},

both

e^{z} - 1

and

lo g (1 + z)

have a simple zero at zero. Hence

g

has a simple pole, and its residue is

Res (g, 0) = lim_{z \to 0} \frac{e ^{z} - 1}{l o g ( 1 + z )} = 1 .

For

h (z) = sin z sin (1/ z)

, multiplication of the two Laurent series shows that all powers are even:

h (z) = \sum_{p, q \geq 0} \frac{( - 1 ) ^{p + q}}{( 2 p + 1 )! ( 2 q + 1 )!} z^{2 (p - q)} .

There are infinitely many negative powers, so zero is an essential singularity. There is no

z^{- 1}

term, and therefore

Res (h, 0) = 0 .

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