Complex Analysis: #11 Monodromy Theorem

• Complex Analysis: #11 The Monodromy Theorem

Why confine our thoughts on analytic continuation to a single continuous path γ? After all, it seems obvious that we can move the path back and forth to some extent — at least when staying within the open discs which we are using — without affecting the arguments of the previous section. This is the idea of the Monodromy Theorem. But first we must define what “moving a path” is supposed to mean.

Definition 9
Let Q = {(x, y) ∈ ℝ² : 0 ≤ x, y ≤ 1} be the unit square, and let H : Q → ℂ be a continuous mapping such that

• H(0, y) = H(0, 0), for all y ∈ [0, 1], 
• H(1, y) = H(1, 0), for all y ∈ [0, 1], 

Let α : [0, 1] → ℂ be the path α(t) = H(t, 0) and let β(t) = H(t, 1). Then we have α(0) = β(0) and α(1) = β(1). The mapping H is said to be a homotopy from α to β. One also says that α and β are homotopic to one another. If G ⊂ ℂ is a region, and H(Q) ⊂ G, then it is a homotopy within the region.

In the more general setting of topology, this idea of homotopy is quite important. But historically, the idea grew out of these applications in complex analysis. If we work with closed paths α, so that α(0) = α(1), then we can define the fundamental group of the topological space. This is dealt with to a greater or lesser degree in all of our textbooks. But I will skip over these things here. Of course, as an additional remark, you should note that the fact that the paths are being parameterized using the unit interval [0, 1] represents no loss of generality. It all works just as well if we use some other interval [t₀, t₁]. Finally, note that if f : G → ℂ is given as an analytic function and α and β are homotopic to one another within G, then theorem 4 shows that

 

Theorem 23 (Monodromy Theorem)
Let α and β be two homotopic paths in ℂ. Assume there is an open disc B₀ centered on α(0) = β(0) and an analytic function f₀ : B₀ → ℂ. For each τ ∈ [0, 1] let h_τ : [0, 1] → ℂ be the path h_τ(t) = H(t, τ). (Thus α = h₀ and β = h₁) Assume that f₀ has an analytic continuation along the path h_τ for each τ. In particular there exists an open disc B₁ centered at α(1) = β(1) such that the analytic continuation along α produces the function f₁ : B₁ → ℂ and the analytic continuation along β produces the function f₁ : B₁ → ℂ. Then f₁ = f₁.

Proof
The proof uses the technique which we have seen in theorem 4. The construction of an analytic continuation along each of the paths h_τ involves some finite chain of open discs. So the set of all such discs covers the compact set H(Q). Take a finite sub-covering. Take the inverse images of the sets in this sub-covering. We obtain a finite covering of Q by open sets. Take a subdivision of Q into sub-squares of length 1/n, for n sufficiently large, so that each of the sub-squares is contained in a single one of these open sets covering Q. Then let γ_j = h_j/n, for j = 0, . . . , n. Now the argument in the proof of the previous theorem (theorem 22) shows that the analytic continuation along γ_j leads to the same function as that along γ_j+1, for each relevant j. In particular, the function is uniquely defined through the power series representation in each of the sub-squares. Therefore it’s values along one segment of the curve γ_j determines uniquely it’s values along the corresponding segment of the next curve γ_j+1. Since the endpoints of all of the curves are identical (that is, γ_j(1) = α(1) = β(1) for all j), we must have the power series expression at this endpoint for each of the analytic continuations of the original function being the same.

• Complex Analysis: #10 Analytic Continuation

We now know that an analytic function f : G → ℂ can be represented as a power series centered at any given point z₀ ∈ G. The function f is equal to the function defined by the power series in the largest possible open disc around z₀ which is contained in G. But of course G is not, in general itself an open disc. Therefore there might be parts of G where f is not given by this power series centered on z₀. Or, (thinking about the logarithm function) we might have the situation that G could be expanded to a larger region G_*, with G ⊂ G_* where the function could be defined. But then perhaps there might be different ways of “continuing” this definition of f from G to G_*.

Therefore let us consider a chain of open discs (B₁, . . ., B_n) say, with

B_j = {z ∈ ℂ : |z − p_j| < r_j} 

for a chain of points p_j which are the centers of the discs, and numbers r_j > 0, which are the radii in each case. We assume that it is a connected chain in the sense that B_j ∩ B_j+1 ≠ ∅, in each case. For each B_j let us assume that an analytic function f_j : B_j → ℂ exists, such that in the region of overlap, we have f_j(z) = f_j+1 (z) for z ∈ B_j ∩ B_j+1.

Definition 8
The functions f_j here are called function elements, and if we have f_j(z) = f_j+1 (z) for z ∈ B_j ∩ B_j+1 for all j = 1, . . . , n − 1, then we have an analytic continuation of the function elements through the chain of open discs. Or, if we consider the ordering of the discs, we can say that the final function f_n is obtained by analytic continuation of the initial function f₁ through the chain of discs.

Theorem 17 shows that if, say f₁ is given in B₁, and there exists a chain of open discs allowing some analytic continuation, then this continuation is unique.


Theorem 21
Let (B₁, . . ., B_n) be a chain of open discs with B_j ∩ B_j+1 ≠ ∅, for j = 1, . . . , n − 1. Let f₁ : B₁ → ℂ be some given analytic function. Then there exists an analytic continuation of f₁ throughout the chain [That is to say, there exists a set of function elements forming an analytic continuation, such that the first element in the chain of function elements is f₁.] if and only if there also exists an analytic continuation of f₁' : B₁ → ℂ (the derivative of f₁) throughout the chain.

Proof
“⇒” is trivial. (Just take f_j', the derivative of f_j, for each j.)

As far as “⇐” is concerned, we are assuming that there is an analytic continuation of the function . To avoid confusion, let us call this function g_j : B₁ → ℂ. i.e. g₁(z) = f₁'(z) for all z ∈ B_j. The assumption is that for each j there is an analytic function g_j : B_j → ℂ, providing an analytic continuation, starting with g₁. We now use induction on the number n. For n = 1 there is nothing to prove. So let n > 1, and assume that we have an analytic continuation of g along the chain of open discs from B₁ to B_n−1, giving g_j : B_j → ℂ such that g_j : f_j' for each j < n. According to theorem 3, g_n has an antiderivative, G_n : B_n → ℂ, with G_n' : g_n. Now in the region B_n−1 ∩ B_nwe have g_n−1 = g_n. That is, f_n−1' = G_n', or f_n−1' − G_n' = 0. Thus f_n−1 − G_n = k say, where k is a constant number. But then the function G_n + k is also an antiderivative to g_n, and we can take f_n = G_n + k.

One way to think about these chains of discs is to imagine that they are associated with a path, namely a path starting at p₁ then following a straight line to p₂, then a straight line to p₃, and so forth, finally ending at p_n. So each straight segment, say from p_j to p_j+1 is contained in the union of the two discs B_j ∪ B_j+1 .

Let’s generalize this idea in the following way. Let γ : [t₀, t₁] → ℂ be a continuous path. (It doesn’t have to be differentiable here.) Let t₀ = τ₀ < τ₁ < · · · < τ_n = t₁ be a partition of the interval [t₀, t₁]. Assume that we have a corresponding set of open discs B_j = {z ∈ ℂ : |z − γ(τ_j)| < r_j}(where r_j > 0) such that γ(t) ∈ B_j ∪ B_j+1 , for t ∈ [τ_j, τ_j+1], for all relevant j. Then we will say that we have a chain of discs along the path γ. Furthermore, if we have a sequence of analytic function elements giving an analytic continuation along this chain, then we will say that the function is analytically continued along the path. The next theorem shows that this is a property of the path, independent of the choice of discs along the path.


Theorem 22
Let γ : [t₀, t₁] → ℂ be continuous and let B = {z ∈ ℂ : |z − γ(t₀) < r} and B_* = {z ∈ ℂ : |z − γ(t₁)| < r_*} (with both r, r_* > 0) be open discs centered on γ(t₀) and γ(t₁), respectively. Let t₀ = τ₀ < τ₁ < · · · < τ_n = t₁ and t₀ = ρ₀ < ρ₁ < · · · < ρ_m = t₁ be two different partitions of the interval [t₀, t₁], giving two different chains of discs along the path, satisfying the conditions listed above, where the first and last discs are B and B_*. Assume that f : B → ℂ is an analytic function at the first disc and it has an analytic continuation with respect to the first chain of discs, finally giving the analytic function g : B_* → ℂ. Then f also has an analytic continuation with respect to the second chain of discs, and it also gives the same function g : B_* → ℂ.

Proof
If the whole path γ is completely contained within the disc B then, using theorem 9 we see that the analytic continuation is simply given by the power series representing the function.

Therefore let M ⊂ [t₀, t₁] be defined to be the set of t_* ∈ [t₀, t₁] such that the theorem is true for the interval [t₀, t_*] (with respect to this path γ). But if t ∈ M, then since γ is continuous, and since the analytic function which has been continued out to the point γ(t_*) has a power series representation in a neighborhood of γ(t_*), we must have some ∈ > 0 such that {t ∈ [t₀, t₁] ; |t − t_*| < ∈ } ⊂ M. Therefore M is an open subset of [t₀, t₁]. If [t₀, t₁] \ M ≠ ∅, then the same argument shows that M is closed in [t₀, t₁]. Yet the interval [t₀, t₁] is connected. Therefore M = [t₀, t₁].

All of these thoughts allow us to perform path integrals along continuous, but not necessarily differentiable paths. To see this, take γ to be some continuous path, allowing an analytic continuation from an open disc centered on the starting point of the path. The discs in the finite chain of open discs describing the analytic continuation have centers at various points along the path γ. But now take γ to be the piecewise linear path connecting those centers. This gives a path integral, namely ∫_γ f(z)dz, where f consists of the function elements along the path. We can now simply define ∫_γ f(z)dz to be this integral along γ. Theorem 22 then shows that the path integral for γ is well defined.

• Complex Analysis: #9 Simple Consequences

Theorem 17
Assume f, g : G → ℂ are two analytic functions defined on a region G such that the set {z ∈ G : f(z) = g(z)} has an accumulation point. Then f = g.

Proof
Let z₀ ∈ G be such an accumulation point. Then z₀ is a zero of the analytic function f − g. But this is not an isolated zero. Therefore f − g = 0, the trivial constant function.


Theorem 18
Again, f : G → ℂ is analytic and it is not a constant function. Then f(G) is also a region (that is, open and connected) in ℂ.

Proof
Since f is continuous, f(G) must be connected. Is f(G) open? Take w₀ ∈ f(G), and z₀ ∈ G with f(z₀) = w₀. So then z₀ is a zero of the analytic function f − w₀. Since f −w₀ is not constant, it follows that z₀ is a zero of some particular finite order. Theorem 16 now shows that w₀ lies in the interior of f(G).


Theorem 19
Let f : G → ℂ be analytic and not constant. (G is a region.) Let z₀ ∈ G. Then there exists another point z₁ with |f(z₁)| > |f(z₀)|.

Proof
For otherwise, f(z₀) would lie on the boundary of f(G) in ℂ, and thus f(G) would not be open in contradiction to theorem 18.


Theorem 20 (The Lemma of Schwarz)
Let D = {z ∈ ℂ : |z| ≤ 1} be the closed unit disc in ℂ. Assume D ⊂ G, a region in ℂ, and f : G → ℂ is analytic with f(D) ⊂ D and f(0) = 0. Then |f '(0)| ≤ 1 as well, and in fact |f(z)| ≤ |z| for all z ∈ D. If either |f '(0)| = 1 or there exists some z₀ with 0 < |z₀| < 1 such that |f(z₀)| = |z₀|, then we must have f being a simple rotation. i.e. f(z) = e^iθ · z, for some θ.

Proof
Since f(0) = 0, we can write

say, where g is an analytic function, defined in some neighborhood of D. So f '(0) = g'(0) and |f(z)| = |z| · |g(z)| ≤ 1. Thus |g(z)| ≤ 1/|z|. This holds in particular for |z| = 1.

On the other hand, theorem 18 says that g(B(0, 1)) ⊂ ℂ is open in ℂ. [Here B(0, 1) = {z ∈ ℂ : |z| < 1} is the open disc centered at zero, with radius 1.] If there were some point z_* ∈ B(0, 1) with |g(z_*)| > 1 then we could choose it to be a point such that this value is maximal. However that would then be a boundary point of f(B(0, 1)), contradicting the fact that f(B(0, 1)) is open. Thus |g(z)| ≤ 1 for all z ∈ D. In particular, |f '(0)| = |g(0)| ≤ 1 and |f(z)| = |z| · |g(z)| ≤ |z| for all z ∈ D.

Finally, let us assume that a z₀ exists with 0 < |z₀| < 1 such that |f(z₀)| = |z₀|. That means |g(z₀)| = 1. But according to theorem 19, this can only be true if g is a constant function. Since it is a constant number with absolute value 1, it must be of the form e^iθ, for some θ.

• Complex Analysis: #8 Zeros of Analytic Functions

Definition 7
Let f : G → ℂ be an analytic function defined in a region G. A point z₀ ∈ G with f(z₀) = 0 is called a zero of the function.

So let z₀ ∈ G be a zero of the analytic function f : G → ℂ. As usual, without loss of generality, we may assume that z₀ = 0. As we have seen, we can choose some r > 0 such that B(0, r) = {z : |z| < r} ⊂ G and

for all such z ∈ B(0, r).

The fact that f(0) = 0 means that c₀ = 0. Let k > 0 be the smallest integer such that c_k ≠ 0. (If c_n = 0 for all n, then f is simply the constant function which is zero everywhere. This is not what we are interested in here so we will assume that some k exists with c_k ≠ 0.) The easiest case is then that k = 1. In this case, we have

and in particular f '(0) = c₁ ≠ 0. Could it be that for every ∈ > 0 there exists a complex number z_∈ with 0 < |z_∈| < ∈ and yet f(z_∈) = 0? But that would imply that

That is impossible, since f '(0) ≠ 0. Therefore we have:


Theorem 14
Let f : G → ℂ be analytic and let z₀ ∈ G be such that f(z₀) = 0 while f '(z₀) ≠ 0. Then there exists an ∈ > 0 such that B(z₀, ∈ ) ⊂ G and the only zero of f in B(z₀, ∈ ) is the single number z₀.

Of course, another way of thinking of these things — and remembering what was done in Analysis II — is to consider f to be a continuously differentiable mapping of G into ℂ, represented as ℝ². The mapping f is then totally differentiable, and the derivative at z₀ is not singular; thus it is a local bijection around z₀.

More generally, we might have k being greater than 1. In any case, the number k is called the order of the zero. A zero of order 1 is also called a simple zero.


Theorem 15
Let f : G → ℂ be analytic in the region G, and let z₀ ∈ G be a zero of f of order k. Then there exists an ∈ > 0 such that in the open disc B(z₀, ∈ ) of radius ∈ around z₀ we have f(z) = (h(z))^k where h : B(z₀, ∈ ) → ℂ is analytic with a simple zero at z₀.

Proof
For sufficiently small ∈ > 0, we can write

say, for z ∈ B(z₀, ∈ ). Here g : B(z₀, ∈ ) → ℂ is analytic, and g(z₀) = c_k ≠ 0. Thus there are k distinct k-th roots of the number g(z₀) = c_k. [For any w ≠ 0 in ℂ we have w = re^iθ say. Then each of the numbers ^k√r · e^{iθ/k + 2πil/k} , for l = 0, . . . , k − 1, is a different k-th root of w.] Let z₁ be one of these k-th roots of g(z₀). Now consider the particular polynomial function ϕ_k'(z) = z^k. We know that ϕ_k is an entire function, and that ϕ_k'(z₁) = kz₁^k−1 ≠ 0, since z₁ ≠ 0. So there is a neighborhood U₁ of z₁, and a neighborhood V₁ of g(z₀), such that ϕ_k : U₁ → V₁ is a bijection [ϕ_k is totally differentiable and non-singular at z₁], with ϕ_k(z₁) = g(z₀). Let ϕ_k⁻¹ : V₁ → U₁ be the inverse mapping. Now choose ∈ > 0 so small that g(B(z₀, ∈ )) ⊂ V₁. Then take h(z) = (z − z₀) · ϕ_k⁻¹(g(z)). This defines a function h : B(z₀, ∈ ) → ℂ which satisfies our conditions.[As in real analysis (the proof is the same here in complex analysis) we have the rule that if ϕ is an invertible differentiable function (with non-vanishing derivative), then ϕ⁻¹ is also differentiable, with derivative (ϕ⁻¹) '(z) = 1/ϕ '(ϕ⁻¹(z)).]


Theorem 16
Again let f : G → ℂ be analytic in the region G, and let z₀ ∈ G be a zero of order k. Then there exists an ∈₀ > 0 and an open neighborhood U_∈0 ⊂ G of z₀ with f(U_∈0) = B(0, ∈₀). Within U_∈0 ,z₀ is the only zero of f, and if w ≠ 0 in B(0, ∈₀) then there are precisely k different points v₁, . . ., v_k in U_∈0 with f(v_j) = w, for all j.

Proof
Since f is continuous and B(0, ∈₀) is open, it follows that U_∈0 = f ⁻¹(B(0, ∈₀)) is also open, regardless of how the number ∈₀ > 0 is chosen. So we begin by choosing an ∈₁ > 0 sufficiently small that we can use theorem 15 and write f(z) = (h(z))^k, for all z ∈ B(0, ∈₁). Since h has a simple zero at z ₀, and therefore the derivative at z ₀ is not zero (h '(z ₀) ≠ 0), there exists a neighborhood of z ₀ such that h is a bijection when restricted to the neighborhood. So let ∈ > 0 be chosen sufficiently small that B(0, ∈) is contained within the corresponding neighborhood of 0. Finally, with this ∈ , we take ∈₀ = ∈^k . Then if w ≠ 0 in B(0, ∈₀), we have k different k-th roots of w, lets call them u₁, . . ., u_k. They are all in B(0, ∈). Therefore each has a unique inverse under h, namely v_j = h⁻¹(u_j), for j = 1, . . . , k. Is it possible that some other point, v say, not equal to any of the v_j, also is such that f(v) = (h(v))^k = w? But then h(v) would also be a k-th root of w, not equal to any of the u_j, since after all, h⁻¹ is a bijection when restricted to B(z₀, ∈₀). This is impossible, owing to the fact that there are only k different k-th roots of w.

Complex Analysis: #7 Standard Theorems of Complex Analysis

• Complex Analysis: #7 Some Standard Theorems of Complex Analysis

Combining the last two theorems, we have:

Corollary (Goursat’s Theorem)
The derivative of every analytic function is again analytic. Thus every analytic function has arbitrarily many continuous derivatives.

We can also complete the statement of theorem 3


Theorem 10 (Morera’s Theorem)
Let G ⊂ ℂ be a region and let f : G → ℂ be continuous such that ∫_γ f(z)dz = 0, for all closed paths which are the boundaries of triangles completely contained within G. Then f is analytic.

Proof
According to theorem 3, there exists an antiderivative F : G → ℂ, with F' = f. Thus, by Goursat’s Theorem, f is also analytic.



Theorem 11 (Cauchy’s estimate for the Taylor coefficients)
Again, let f : G → ℂ be analytic, z₀ ∈ G, r > 0 is such that D(z₀, r) = {z : |z − z₀| ≤ r} ⊂ G, and

for all z ∈ D(z₀, r). Since f is continuous and D(z₀, r) is compact, we must have |f| being bounded in D(z₀, r). Let M > 0 be such that |f(z)| ≤ M for all z ∈ D(z₀, r). Then we have

Definition 5
Let the function f : ℂ → ℂ be defined throughout the whole complex plane, and let it be analytic everywhere. Then we say that f is an entire function.


Theorem 12
A bounded entire function is constant.

Proof
Assume that the entire function f : ℂ → ℂ is bounded with |f(z)| ≤ M say, for all z ∈ ℂ, where M > 0 is fixed. Thus |c_n| ≤ M/rⁿ = 0 for all r > 0. This can only be true if c_n = 0 for all n > 0.

Definition 6
A field is called algebraically closed if every polynomial within the field of degree greater than or equal to one has a root.


Theorem 13 (The Fundamental Theorem of Algebra)
ℂ is algebraically closed.

Proof
Let f(z) = ∑a_kz^k, (with k = 0, . . .,n), with n ≥ 1 and a_n ≠ 0 be a polynomial of degree n. Looking for a contradiction, we assume that there is no root, that is, f(z) ≠ 0 for all z ∈ ℂ.

For z ≠ 0, we have

Note that for a and b arbitrary numbers, we have |a| = |a + b − b| ≤ |a + b| + |b| or |a + b| ≥ |a| − |b|, and more generally, |a + b₁ + · · · + b_n| ≥ |a| − |b₁| − · · · − |b_n|.

Since |a_n|/2 remains constant, |zⁿ| · |a_n|/2 becomes arbitrarily large, as |z| → ∞. Therefore |f(z)| → ∞ when |z| → ∞. That is to say, if M > 0 is given, then there exists an r > 0 such that |f(z)| > M for all z with |z| > r. That is, |1/f(z)| < 1/M for |z| > r. Now, since f(z) ≠ 0 always, and f (being a polynomial) is an entire function, we have that 1/f is also an entire function. It is bounded outside the closed disc D(0, r), but since the function is continuous, and D(0, r) is compact, it is also bounded on D(0, r). Thus it is bounded throughout ℂ, and is therefore constant, by theorem 12. Therefore, the polynomial f itself is a constant function. This contradicts the assumption that f is of degree greater than zero.

• Complex Analysis: #6 Power Series

Definition 4
A power series is a sum of the form

where (a_n)_n∈ℕ0 is some arbitrary sequence of complex numbers and z₀ is a given complex number. (ℕ₀ is the set of non-negative integers)

So the question is, for which z does the power series converge? Well it obviously converges for z = z₀. But more generally, we can say the following.


Theorem 8
Let the power series ∑a_n(z − z₀)ⁿ, (with n = 0, . . .,∞) be given. Then there exists 0 ≤ R ≤ ∞, the radius of convergence, such that

1. The series is absolutely convergent for |z − z₀| < R, and uniformly convergent for |z − z₀| ≤ ρ, for 0 ≤ ρ < R fixed. 
2. It diverges for |z − z₀| > R.
3. The radius of convergence is given by 1/R = lim_n→∞ sup ⁿ√|a_n|. 
4. The function given by f(z) = ∑a_n(z − z₀)ⁿ, (with n = 0, . . .,∞) is analytic in the region |z − z₀| < R. For each such z, the derivative is given by the series f '(z) = ∑na_n(z − z₀)^n-1, (with n = 1, . . .,∞), and the radius of convergence of this derivative series is also R. 

Proof
Parts 1 and 2 are proved in the analysis lecture. For 3, let |z − z₀| < ρ < R with 1/R = lim_n→∞ sup ⁿ√|a_n|. Thus there exists some N₀ ∈ ℕ with √|a_n| < 1/R for all n ≥ N₀. That is, |a_n| < 1/Rⁿ. Therefore

with ρ/R < 1. This is a geometric series which, as is well known, converges. On the other hand, if |z − z₀| ≥ ρ > R then there exist arbitrarily large n with ⁿ√|a_n|  > 1/R. That is,  |a_n| > 1/Rⁿ or

So the series cannot possibly converge, since the terms of the series do not converge to zero.

As far as part 4 is concerned, it is clear that lim_n→∞ sup ⁿ√|na_n| = lim_n→∞ sup ⁿ√|a_n| since lim_n→∞ sup ⁿ√n = 1. So let f₁(z) = ∑na_n(z − z₀)^n-1, (with n = 1, . . .,∞), be the function which is defined in the region |z − z₀| < R. We must show that f is analytic here, with f ' = f₁. To simplify the notation, let us assume from now on that z₀ = 0. Choose some complex number w with |w| < R. We must show that the derivative of f exists at w, and it equals f₁(w).

To begin with, we write

So take some ρ with |w| < ρ < R and we restrict ourselves to examining complex numbers z with |z| < ρ. Furthermore, choose ∈ > 0. We must show that there exists a δ > 0 such that if 0 < |z −w| < δ then

(Remember that the series is absolutely and uniformly convergent in the closed disc with radius ρ.) Thus for some N₁ ∈ ℕ, the “tail” of the series beyond N₁ sums to something less than ∈ /3. Similarly, the series defining f₁ is absolutely and uniformly convergent in this disc. Therefore take N₂ to be sufficiently large that

for all n ≥ N₂. Let N be the larger of N₁ and N₂. Finally we must determine the number δ. For this, we note that since S_n is just a polynomial, and thus analytic, we have a δ > 0 such that

for all z with |z − w| < δ. In particular, if necessary, we can choose a smaller δ to ensure that such z are in our disc of radius ρ. The fact that

Theorem 9
Let f : G → ℂ be an analytic function defined in a region G, and let z₀ ∈ G be given. Then there exists a unique power series ∑a_n(z − z₀)ⁿ, (with n = 0, . . .,∞) whose radius of convergence is greater than zero, and which converges to f(z) in a neighborhood of z₀.

Proof
Let r > 0 be sufficiently small that B(z₀, r) = {z ∈ ℂ : |z − z₀| < r} ⊂ G. In fact, we will also assume the r is sufficiently small that z ∈ G for all z with |z − z₀| = r. Once again, in order to simplify the notation, we will assume that z₀ = 0. That is to say, we will imagine that we are dealing with the function f(z − z₀) rather than the function f(z). But obviously if the theorem is true for this simplified function, then it is also true for the original function. According to theorem 7, for |z| < r we then have

Here are a few points to think about in this proof.

• The third equation is true since |z/ζ| < 1, and thus the sum is absolutely convergent. 
• The fifth equation is true since the partial sums are uniformly convergent, thus the sum and integral operations can be exchanged. 
• Although the function f(ζ)/ζⁿ⁺¹ is not differentiable at zero, it is defined and continuous on the (compact) circle |ζ| = r. Thus, although c_n is not always zero, still it is always a well defined complex number, for all n. 
• It looks like c_n might vary with r. But this is not the case. Theorem 8 implies that f⁽ⁿ⁾(0) = n!c_n, for all n, and this is certainly independent of r. 
• The power series converges to f(z) at all points of B(z₀, r).

• Complex Analysis: #5 Cauchy`s Integral Formula

Theorem 6
Let G ⊂ ℂ be a region and let f : G → ℂ be analytic. Take z₀ ∈ G and r > 0 so small that {z ∈ ℂ : |z − z₀| ≤ r} ⊂ G. Furthermore, let |a − z₀| < r. Then

converges to the constant number f '(a) in the limit as ∈ → 0. On the other hand, the path length around the circle, and the tangent vector to this path, approach zero as ∈ → 0. Thus in the limit, the first integral is zero. As far as the second integral is concerned, we have

A relatively trivial implication is the following theorem.


Theorem 7
The same assumptions as in Theorem 6. But this time take z₀ to be the central point of the circle. Then

So this is just a kind of “mean value theorem” for analytic functions. It shows quite clearly the difference between real analysis and complex analysis. In real analysis, we can make a smooth change in a function, leaving everything far away unchanged, and the function remains nicely differentiable. But in complex analysis, the precise value of the function is determined by the values on a circle, perhaps far away from the point we are looking at. So a change at one place implies that the whole function must change everywhere in order to remain analytic.