(since ), so is a local diffeomorphism near each point! Globally, is the universal covering map!
Logarithms and powers
What about the logarithm? For , we want to define such that . Such a exists, but isn’t unique: we can add integer multiples of . is well defined though, and equal to (for the usual on ).
In general “” is not well defined and continuous on ; but it’s OK over simply connected subsets of (we can’t go around , so is well defined). This is consistent with what we’ve seen about the lifting problem for
The same issue comes up with defining for : we would like to define it as , but this only works on suitable domains. E.g. is multivalued () and we can’t define a continuous function on a domain that encloses the origin.
There are still power series expressions away from the origin. E.g.:
Path integrals and Cauchy’s theorem
Now we consider path integrals of complex 1-forms: given a continuous function and a (piecewise) differentiable path ,
(or: pick points along the path, with ; then ).
Example 29.1. For a path from to :
For a power series , if is entirely contained in the disc of convergence, it follows that
indeed , and so the equality follows from the fundamental theorem of calculus.
In general, a 1-form on need not be exact, and path integrals need not be path-independent. One of the miracles is that things are much simpler in the analytic setting.
Key result:
Theorem 29.2 (Cauchy’s theorem). Let be a bounded region with piecewise smooth boundary, and analytic on open . Then
Proof assuming is continuous. The 1-form is , and
Stokes’ theorem then gives .
(Let’s check this more carefully, just to be safe: , so
by Cauchy–Riemann.) ∎
We’ll see later how to show that analytic continuous. In the meantime we add the continuity of to our working assumptions.
This holds not just for a simply connected region bounded by a simple closed curve! We can also allow holes in the region , e.g. around points where isn’t defined.
Example 29.3.
Let be analytic on , and enclosing as shown. Then
(by Cauchy’s theorem: ).
Now assume is analytic on and (e.g. it’s enough for to be bounded near ). Then
Since this quantity as , and the path integral is independent of , we get:
Theorem 29.4 (“Improved Cauchy”). Cauchy’s theorem () remains true under the weaker assumption that is defined and analytic in ,, and .
However, we can’t get rid of all assumptions about the behavior of at .
Example 29.5. With :
or the same for any going once around , by Cauchy! (Cf. the fundamental theorem / the multivalued nature of .)
Cauchy’s integral formula
Using this, we get to Cauchy’s integral formula:
Theorem 29.6 (Cauchy’s integral formula). Let be a bounded region with piecewise smooth boundary , analytic on an open domain containing , and . Then
Proof. Since , the formula is equivalent to
The differentiability of at implies: as ,, and in particular
(+ it is analytic for ). The result thus follows from improved Cauchy. ∎
Proof (Alternative proof). Cauchy’s theorem gives
∎
This is magical: the values of at every point inside a closed curve can be determined by calculating path integrals on !! (Assuming is defined and analytic everywhere in the enclosed region, of course.) In this version, to emphasize that we can vary the point of evaluation, one usually rewrites as
Next we’ll do even better ( all derivatives exist!!):
Remark. If is given by a power series near , with , then for a small circle ( radius of convergence), uniform convergence of the series implies
(by the calculation above, the integral is for and for ), and Cauchy implies .
But the problem is … we haven’t shown yet that analytic functions are power series! In fact the proof uses Cauchy’s formula … so instead we have to work.
Proposition 29.7. Suppose is continuous on . Then for all ,
is analytic in the interior of , and
Proof. We first prove that is continuous on . Fix with , and let (so and are farther than away from all points of ). Calculate:
So:
Since each term in the sum has , this implies
Taking , this inequality proves that is continuous at , i.e. is continuous on . Moreover,
The continuity result, now applied to , shows that the terms in the right-hand side are continuous functions of ; hence the right-hand side of is continuous, and its limit at equals
This gives the existence of
This holds for all , hence is analytic as claimed and .∎
Now if is analytic in (+ continuous), then by Cauchy’s integral formula,
is the expression denoted in the proposition, for .
Corollary 29.8. The proposition then shows that is infinitely differentiable, all its derivatives are analytic, and , i.e.
This also lets us lift the extra assumption we’ve made so far in all proofs using Cauchy’s theorem, that is continuous.
Proposition 29.9. If is analytic then is continuous.
Proof. If is analytic in a disc , define
where we choose a path consisting of horizontal and vertical line segments. We don’t have the full strength of Stokes’ theorem (we don’t know that is continuous), but we claim it holds for rectangles: for a rectangle (see below).
Given this, our definition of makes sense and doesn’t depend on the path. We claim is analytic and . Indeed:
where goes horizontally from to , then vertically up to . Using continuity of , as we have , hence , hence .
So now is analytic with continuous derivative , so we can apply Cauchy’s integral formula and the above argument to , so has derivatives to all orders. In particular is continuous. ∎
Cauchy’s theorem on rectangles (without assuming continuous)
Proof. Assume is a rectangle, is analytic, and
Cut into 4 equal rectangles; then is the sum of the 4 path integrals around them, so there exists with such that
Repeat this process: with and
Then (a decreasing sequence of nonempty closed subsets in a compact space has a nonempty intersection: else the complements would be an open cover without a finite subcover).
But now with , so
Contradiction. ∎
Consequences: Cauchy’s bound and entire functions
Returning to Cauchy’s integral formula for derivatives:
Theorem 29.10. analytic on has derivatives to all orders in , all derivatives are analytic, and for ,
From this we get, by bounding the integral in the right-hand side:
Theorem 29.11 (Cauchy’s bound). If is analytic in , then
(By considering ,, the result still holds under the weaker assumption that is continuous on and analytic in .)
Cauchy’s bound has important consequences for entire functions, i.e. functions analytic on all of .
Corollary 29.12. If is analytic on all of (“entire function”) and bounded, then is constant.
Proof. Apply Cauchy’s bound with to get .∎
Corollary 29.13. A nonconstant entire function has dense image: .
Proof. If , then there is such that for all , and then is a bounded entire function, hence constant. ∎