Complex Analysis/Cauchy's theorem and Cauchy's integral formula

The Goursat–Pringsheim lemma edit

Cauchy's theorem states that if   is holomorphic on   (  a star-shaped domain; we'll make precise in the next section what that means) and   is a closed contour with values in  , then


As a first step of proving the theorem, we will prove it in the special case where   is a triangle; this is the Goursat–Pringsheim lemma. Goursat had the idea first, but Pringsheim then provided the idea to use triangles (instead of squares, as Goursat had done).

Lemma 6.1 (Goursat–Pringsheim):

Let   be a triangle, where   is holomorphic and the interior of the triangle is contained within  . Then we have



Any triangle   may be divided into four small triangles of equal side length as indicated in the picture. Each of the small triangles will have half the side length of the original triangle; this is clear from the formulae one would assign to the smaller triangle if one were to make things (even more) rigorous. We shall denote the four triangles as  ,  ,   and  . But one of the four non-negative numbers


will have to be the largest. We then set  ,   the first of the four smaller triangles arising from   such that the absolute value as above is largest, and so on; ie. once   is defined for general  , we decompose   into triangles   as above and then define   the one of   where the absolute value of the four ones indicated above is largest.

In this fashion, we obtain a sequence of ever smaller triangles  . We now have the following lemma:

Lemma 6.2 (Cantor's intersection theorem):

Let   be a decreasing sequence of compact sets situated in a Hausdorff topological space  . Then



In a Hausdorff topological space, compact sets are closed; indeed, let   be such a compact set, and assume that the neighbourhood filter of a certain   is such that all its elements have nonempty intersection with  .

Hence, if we assume that


we get by deMorgan's rules, taking complements in  ,


Hence, if we regard   as a topological space with subspace topology, the sets   form an open cover of  . By compactness of  , there is a finite subcover  , where without loss we assume that  . Due to monoticity, we then obtain that in fact,   and thus  , which is a contradiction. 

Thus, we can obtain a point


where   shall denote the filled triangle (i.e. the convex hull) of  .

Now by assumption,   was differentiable, in particular at  . Therefore, we may write


where   as  ; indeed, we may set


Furthermore, if we take any triangle   in the complex plane and form the curve integral of a polynomial over it, we get zero; indeed, every polynomial   has a primitive   which is formed in the obvious way, according to chapter 2; namely it's exactly the same formula as in the real case, and the differentiation formulae confirm that. We may then decompose the triangle   into three lines; say that   are the corners of that triangle, then   (  denoting concatenation, even though we will later, when considering chains, allow for more general curves as arguments on the left and right of  ), and we get


as desired. In particular, we get


and by the fundamental estimate


However, this value is actually greater than the absolute value of the original integral, since we will have


since the integrals cancel out, as indicated in the picture, and hence by the triangle inequality


then use induction to get this "greaterness" for each  . But this shows the theorem, since if the absolute value is less or equal than any positive value, it is zero. 

Cauchy's theorem on starshaped domains edit

Now we are ready to prove Cauchy's theorem on starshaped domains. This theorem and Cauchy's integral formula (which follows from it) are the working horses of the theory; from these two we will deduce the local theory of holomorphic functions, and the global theory will then follow as well.

Definition 6.3:

Let   be open.   is called star-shaped if and only if there exists a designated point   such that for all  , the line   lies completely in  .

Informally,   will "look like a star" with center  .

Regarding these domains, we have the following lemma:

Lemma 6.4:

Let   be holomorphic, where   is star-shaped. Then   will have a primitive in  , i.e. a holomorphic function   such that   at every point  .


We begin by just defining


this will be well-defined just by the star-shapedness of  , i.e.   must be and is defined on all of the path  . Our claim now is that   thus defined is indeed a primitive of  .

Indeed, pick any fixed  . Since   is open, we find a (possibly small) radius   such that   is completely contained in  .

Cauchy's integral formula edit

Cauchy's integral formula is a formula which looks at first glance a bit strange:

Theorem 6.6:

Let   be holomorphic, and suppose that   is a point in  , and   is a radius such that  . Then

For all  , we have  ,

where the closed contour   is traversed counterclockwise, for instance using the parametrisation


Now this formula may seem strange at first, but considering the special case   and inserting the definition of curve integrals, we obtain that a special case of it is actually a mean value formula:


In what follows, we shall always assume that the path   is traversed counterclockwise.


We first note that the function


is continuous; this follows from dominated convergence by the boundedness of   on compact sets. Then we note that for any  , the above expression coincides with


Furthermore, given  , we have


where   is the curve


where we define


  denoting concatenation of curves. The curve is depicted in the following picture:

This picture depicts the curve which is used in the proof I gave for Cauchy's integral formula.

However, one finds star-shaped domains (eg. half-planes) which contain each of the cycles   of which   is composed, whence, in fact


by Cauchy's theorem on star-shaped domains. Hence,   is constant on   and by continuity and  , this constant must be exactly  . 

Remark: Using a similar construction like the curve   above, we in fact obtain the following much more general version of Cauchy's integral formula:

Now note that in our situation, we may differentiate under the integral sign, which gives us formulae for   (the higher derivatives) for   in the respective ball. Indeed, we get


Later, we will deduce the very same formulae from the Taylor expansion which we will compute; in fact, all holomorphic functions are equal to their Taylor series, as we will show.