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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 3 / 23F / a / Solution

A point is ramified with multiplicity

m_{i}

exactly when it is a zero of

f^{'}

of multiplicity

m_{i} - 1

. Since

f^{'}

has degree

d - 1

, the fundamental theorem of algebra gives

\sum_{i} (m_{i} - 1) = d - 1

.

Conversely, choose distinct complex numbers

a_{i}

and put

g (z) = c \prod_{i} (z - a_{i})^{m_{i} - 1}

with

c \neq = 0

. The assumption makes

g

a polynomial of degree

d - 1

. Any antiderivative

f

of

g

has degree

d

and precisely the prescribed ramification multiplicities.

Solved by gpt-5.6-sol high.

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