Theorem 377.1(Goursat).
38174fIf is an open set in , and a triangle whose interior is also contained in , then
whenever is holomorphic in .
Proof.
Let . Subdivide into 4 congruent triangles. Since integrating the same curve in opposite directions yields zero, we can express the integral of over as the sum of the integral of over the four triangles. Continue the process to obtain a point in the intersection. Use the holomorphicity of at and bound the error term by .□
Corollary 377.2.
If is holomorphic in an open set that contains a rectangle and its interior, then
Local existence of primitives and Cauchy’s theorem in a disc
Lemma 377.3(Stein & Shakarchi (2003) 2.2.1).
870ca5A holomorphic function in an open disc has a primitive in that disc.
Proof.
Define as the integral over a unique rectangular path from the origin to .
To show that is holomorphic in and , use Thm 1 to write
where is the straight line segment from to . Now use the continuity of to write , and obtain
□
The proof of Lem 3 also works when is given to be continuous in an open disk and its integral on any triangle contained in that disk is zero; we use this observation in the proof of Thm 14.
Theorem 377.4(Cauchy's theorem for a disc).
6ba5d4If is holomorphic in a disc, then
for any closed curve in that disc.
Proof.
Follows from Lem 3 and Cor 370.13.□
Toy contours
We call a toy contour any closed curve where the notion of interior is “obvious”, and a construction similar to that in Lem 3 is possible in a neighborhood of the curve and its interior. These are useful in evaluating integrals.
To be precise (ish), let if be a toy contour and be holomorphic in a neighborhood of and its interior. Then, is holomorphic inside of a slightly larger version of whose interior contains and . We fix a point . For , let denote any curve contained inside connecting to which consists of finitely many horizontal and vertical segments; the choice doesn’t matter since the integral of over any two such curves would be equal, courtesy Thm 1. We may thus define unambiguously in .
Thus, for a toy contour , we have
whenever is holomorphic in an open set that contains the contour and its interior.
Cauchy’s integral formulas
Theorem 377.5(Stein & Shakarchi (2003) 2.4.1).
d3c867Suppose is holomorphic in an open set that contains the closure of a disc . If denotes the boundary circle of this disc with the positive orientation, then for any ,
Integrate over the keyhole contour.
The regularity of holomorphic functions arises as a corollary.
Corollary 377.6.
58d2c9Let be holomorphic in an open set . If is a circle whose interior is also contained in , then for all in the interior of ,
Thus, if is holomorphic in an open set , then has infinitely many complex derivatives in .
Proof by induction.
The formulas of Thm 5 and Cor 6 are called the Cauchy integral formulas.
Corollary 377.7(Cauchy inequalities).
If is holomorphic in an open set that contains the closure of a disc centered at and of radius , then
where .
Proof.
An easy application of Cor 6.□
We have seen (Thm 370.8) that a power series defines a holomorphic function in its disc of convergence. The converse arises as another corollary of the Cauchy integral formulas.
Theorem 377.8.
f3eea8Suppose is holomorphic in an open set . If is a disc centered at and whose closure is contained in , then has a power series expansion at
for all , and the coefficients are given by
for all .
Proof.
Since power series define infinitely differentiable functions, Thm 8 gives another proof that a holomorphic function function is infinitely differentiable.
Corollary 377.9.
If is holomorphic on all of , Thm 8 implies that has a power series expansion around that converges in all of .
Corollary 377.10(Liouville).
a603feIf is entire and bounded, then is constant.
Proof.
It suffices to show that by Cor 370.14. For each and all , the Cauchy inequalities yield
where is a bound for . Letting gives the desired result.□
Example 377.11(The fundamental theorem of algebra).
7800a0Every non-constant polynomial with complex coefficients has a root in .
Proof.
If has no roots, then is a bounded holomorphic function. To see this, assume and write
for . Since each term in the parenthesis goes to as we conclude that there exists such that
whenever . Thus, is bounded from below when . Since is continuous and has no roots in the compact disc , it is bounded from below in that disc by Thm 138.8. This proves the claim.□
Analytic continuation
Theorem 377.12.
172decSuppose is a holomorphic in a connected region that vanishes on a sequence of distinct points with a limit point in . Then is identically zero.
Corollary 377.13.
Suppose and are holomorphic in a region and for all in some non-empty open subset of (or more generally for in some sequence of distinct points with limit point in ). Then throughout .
Morera’s theorem
The converse of Cauchy’s theorem.
Theorem 377.14(Morera).
d13e0eSuppose is a continuous function in the open disc such that for any triangle contained in
then is holomorphic.
Proof.
By the proof of Lem 3, has a primitive in that satisfies . By the regularity theorem, we know that is indefinitely (and hence twice) complex differentiable, and therefore is holomorphic.□
Sequences of holomorphic functions
Theorem 377.15(Stein & Shakarchi (2003) 2.5.2-3).
4e9e39If is a sequence of holomorphic functions that converges uniformly to a function in every compact subset of , then
- is holomorphic in .
- the sequence of derivatives converges uniformly to on every compact subset of .
Compare with the real analytic version of the same theorem; Thm 15.2 goes from being a conclusion to being a hypothesis.
Holomorphic functions defined in terms of integrals
Theorem 377.16.
adaf14Let be defined for where is an open set in . Suppose satisfies the following properties:
- is holomorphic in for each .
- is continuous on .
Then the function defined on by
is holomorphic.
Schwarz reflection principle
Lemma 377.17(The symmetry principle).
If and are holomorphic functions in and respectively, that extend continuously to and for all , then defined on by
is holomorphic on all of .
Theorem 377.18(Schwarz).
Suppose is a holomorphic function in that extends continuously to such that is real-valued on . Then there exists a function holomorphic in all of such that on .