Esercizi svolti di analisi reale e complessa - Dipartimento di ...

h −1 e −z θ(z) = h −1 e
−z
(1 + h 2n−1 e z )(1 + h 2n−1 e −z ) =
n≥1
= h −1 e −z (1 + he z )(1 + he −z )
(1 + h 2n−1 e z )(1 + h 2n−1 e −z ) =
n≥2
= (1 + h −1 e −z )(1 + he z )
(1 + h 2n−1 e z )(1 + h 2n−1 e −z ) =
n≥2
= (1 + h −1 e −z )
(1 + h 2n−1 e −z )(1 + he z )
(1 + h 2n−1 e z ) =
n≥2
n = m − 1 n = m + 1
n≥2
h −1 e −z θ(z) = . . . =
(1 + h 2m−3 e −z )
(1 + h 2m+1 e z ) =
=
m≥1
m≥1
m≥1
(1 + h 2m−3 e −z )(1 + h 2m+1 e z ) = θ(z + log h 2 ) .
z =
±n
n 1
n
n 1
n2
h = 1
g(z)
sin πz = ze 1 − z
e
n
z
n
g
π cot πz =
d(sin πz)
sin πz
n=0
1
=
z + g′ (z) +
1 1
+ ;
z − n n
g ′ (z) = 0 g(z) ≡ c
sin πz
lim = π
z→0 z
eg(z) = ec = π
sin πz = πz
1 − z
n
n=0
n=0
e z
n