By Nicolas Conti

ISBN-10: 2754003568

ISBN-13: 9782754003568

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The Taylor expansion of G(z) implies that G(z) is a polynomial of degree μ ≤ ν = α . 1, F(z) = e G(z) has order μ, whence α = μ ≤ α . Thus α = μ = α ∈ N, and deg G(z) = μ = α. 5 Let z 1 , z 2 , . . → ∞ be any sequence of non-zero complex numbers with exponent of convergence β < +∞. The canonical product z , p = zn E n p 1− n z 1 z exp zn k zn k=1 k is an entire function of order β. 11). 2 we know that β ≤ α. 3), z , p zn E n ε exp(|z|β+ε ) for every ε > 0. 22) with ε/2 in place of ε one gets z , p zn E log n ε |z|β+ε/2 = o(|z|β+ε ) ≤ ϑ|z|β+ε with 0 < ϑ < 1, whence E n If |z/z n | ≤ 1 2 z , p zn ≤ exp(ϑ|z|β+ε ) = o exp(|z|β+ε ) .

We define the exponent of convergence of the sequence (z n ) as the infimum β of the exponents B > 0 such that ∞ |z n |−B < +∞. 7) n=1 ∞ If |z n |−B = +∞ for every B > 0, the exponent of convergence of (z n ) is +∞. 2 If an entire function F(z) satisfying F(0) = 0 has order α, and if the sequence z 1 , z 2 , . . of the zeros of F(z) has exponent of convergence β, then β ≤ α. 6) with r = |z n | we get n |z n |−(α+2ε) n −(α+2ε)/(α+ε) = n −(1+ε1 ) , for a suitable ε1 > 0. Therefore ∞ |z n |−(α+2ε) n=1 ∞ n −(1+ε1 ) < +∞.

Of respective multiplicities m 1 , m 2 , . . Then the product g(z)ϕ(z) has no poles, and therefore is an entire function h(z), whence g(z) = h(z)/ϕ(z) with h(z) and ϕ(z) entire functions. If g(z) has a pole at z = 0 of multiplicity m, then g(z) : = z m g(z) is regular at z = 0, whence g(z) = h(z)/ϕ(z) with entire h(z) and ϕ(z), and g(z) = h(z)/(z m ϕ(z)) with entire h(z) and z m ϕ(z). 2 (Weierstrass) Let F(z) be meromorphic, and regular and = 0 at z = 0. Let z 1 , z 2 , . . be the zeros and poles of F(z) with respective multiplicities |m 1 |, |m 2 |, .

### 300 énigmes by Nicolas Conti

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