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

1
=
|z − a| 2
1
ρ 2 − |a| 2
2πia
• |a| < ρ |ρ2−|a| 2 |
2πiρ
• |a| > ρ 2
a|ρ2−|a| 2 |
a ρ2
+
z − a ρ2 − az
• n ≤ 0
|z|=1
• n > 0 n
• n > 0 n ≡ 1 4
• n > 0 n ≡ 3 4
• n ≥ 0 m ≥ 0
• n ≥ 0m < 0
• n < 0m ≥ 0
• n < 0 m < 0
|z|=2
z n (1−z) m dz = 2πi
sin z
zn dz = 0
sin z
|z|=1 zn dz = 0
sin z
|z|=1 zn dz = 2πi
(n−1)!
sin z
|z|=1 zn dz = − 2πi
(n−1)!
|z|=2 zn (1 − z) m dz = 0
|z|=2 zn (1−z) m dz = 2πi(−1) m
|z|=2 zn (1−z) m dz = 2πi(−1) n+1
|m| + |n| − 2
|n| − 1
n
|m| − 1
m
|n| − 1
|m| + |n| − 2
+
|m| − 1
z |z| ≤ ρ δ = R − ρ > 0
Bδ(z) ⊂ BR(0)
f
f
f n (γ) = n!
2πi
∂Bδ(z)
f(ζ)
(ζ − z) n+1
x = 1
2