Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 137 / 1 / b

Solution

Put

τ = x + i y

. The modular transformation law and

Im (γ τ) = \frac{y}{∣ c τ + d ∣ ^{2}}

y^{k /2} ∣ f (τ) ∣

is invariant under

Γ (1)

. On the region from part (a), it is bounded: it is continuous on every truncated region, while the cusp form condition makes it tend to zero as

y \to \infty

. Thus

∣ f (x + i y) ∣ \leq C_{0} y^{- k /2}

for all

x \in R

and

y > 0

a_{n} = e^{2 πn y} \int_{0}^{1} f (x + i y) e^{- 2 πin x} d x,

and hence

∣ a_{n} ∣ \leq C_{0} e^{2 πn y} y^{- k /2} .

Choosing

y = 1/ n

yields

∣ a_{n} ∣ \leq C_{0} e^{2 π} n^{k /2} .

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