2007年Peter Milley教授复分析笔记：解析函数与Cauchy-Riemann方程
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"这份名为《ComAna-ReviewNotes.pdf》的文档是2007年悉尼大学第二学期Peter Milley教授的复分析复习笔记。该笔记针对课程第一至第六周的内容，主要关注复分析的基本概念和理论。在复分析中，一个函数如果在其定义域D上的每一个点都存在导数，那么称这个函数在该点是解析的。解析性的一个关键性质是Cauchy-Riemann方程，它涉及到函数的实部u(x,y)和虚部v(x,y)之间的关系：若f(z) = u(x,y) + iv(x,y)，则解析性条件为∂u/∂x = ∂v/∂y且∂u/∂y = -∂v/∂x。
解析函数的一个重要特性是，如果u和v是两阶可微的，那么它们同时也是调和函数，意味着它们满足二阶偏导数相等且和为零的条件：∂²u/∂x² + ∂²u/∂y² = ∂²v/∂x² + ∂²v/∂y² = 0。这意味着解析函数不仅在局部表现出了特殊结构，而且其几何特性也受到严格限制。
此外，笔记强调了开放连通集（domain）的概念，通常不包括集合的边界，尽管在某些情况下会考虑那些可以连续扩展到边界上的函数。这对于理解和分析解析函数的性质以及它们在复平面上的行为至关重要。
整个文档涵盖了复分析的基本理论，如解析函数的定义、性质、以及与实函数如调和函数的关系，为学生准备期中考试或深入理解复分析提供了详尽的复习资料。对于那些希望在复变函数领域深化学习的人来说，这份笔记是一份宝贵的参考资料。"
2.2 Existence of antiderivatives
Note that Green’s Theorem is one of the multivariable generalizations of
the Fundamental Theorem of Calculus. This suggests that there may be
a “Fundamental Theorem of Calculus” for complex path integrals too, and
there is, sort of. If f (z) has an anti-derivative F (z) on D, also called a
primitive, then
R
γ
f(z) dz = F (b) − F (a) for all paths γ in D from a to b.
However, unlike in the real case f(z) may not have such a primitive even if
f is analytic; it depends both on f and on the shape of D. For example,
f(z) = 1/z has no primitive on the punctured plane C \ {0}, but does have
a primitive on the slit plane C \ {x ∈ R, x ≤ 0}. Once again, the notion of
“simply connected” rears its head.
2.3 Cauchy’s Integral Formula
To move from Cauchy’s Theorem to Cauchy’s Integral formula, we first note
the Mean Value Property for harmonic functions. If u(x, y) is a harmonic
function defined on a disk of radius r centred at z
0
, then
1
2π
Z
2π
0
u(z
0
+ re
iθ
) dθ = u(z
0
).
In other words, the average value of u on the boundary of the disk is equal
to the value of u at the centre. This is an important property of harmonic
functions, and a consequence of Green’s Theorem. Now suppose that f(z)
is analytic in a neighbourhood of z
0
and that γ is the boundary of the disk
{|z−z
0
| ≤ r}, oriented counter-clockwise. Then by substituting z = z
0
+re
iθ
,
taking real and imaginary parts, and applying the Mean Value Property, we
get
1
2πi
Z
γ
f(z)
z − z
0
dz = f(z
0
).
Now suppose that f is analytic on a domain D and extends continously to
∂D, and fix a point z
0
in the interior of D. Let D
0
be an open disk of radius r
around z
0
, where r > 0 is chosen small enough that D
0
⊂ D, and let D
1
⊂ D
be the complement of the closure of D
0
. Since z
0
6∈ D
1
, f(z)/(z − z
0
) is
analytic in D
1
, and by Cauchy’s Theorem
1
2πi
Z
∂D
1
f(z)
z − z
0
dz = 0.
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