• Complex Analysis: #14 Isolated Singularities

A “singularity” is really nothing more than a point of the complex plane where a given function is not defined. That is, if f is defined in a region G ⊂ ℂ, then any point a ∉ G can be thought of as being a singularity of the function. But this is not really what we are thinking about when we speak of singularities. As an example of what we are thinking about, consider the particular function

Here, except for the special point z₀ = 0 which is not so nice, we have a good example of an analytic function. Since only a single point is causing problems with this function, let us more or less ignore this point and call it an isolated singularity. So in general, an isolated singularity is a point z₀ ∈ ℂ such that z₀ ∉ G, yet there exists some r > 0 with B(z₀, r) \ {z₀} ⊂ G.

Definition 11
Let f : G → ℂ be analytic, but with an isolated singularity at z₀ ∈ ℂ. The residue of f at z₀ is the number

According to Cauchy’s theorem, the residue is a well-defined number, independent of the radius r of the circle of the path used to define it, as long as r is small enough to satisfy our condition.

We identify three different kinds of isolated singularities:

• removable singularities, 
• poles,
• essential singularities. 

Let’s begin with removable singularities. Let f : G → ℂ have a singularity at z₀. This singularity is removable if it is possible to find some number w₀ such that if we simply define f₀ : G ∪ {z₀} → ℂ by

then f₀ is analytic (thus also differentiable at the special point z₀). We see then that a removable singularity is really nothing special. We have simply “forgotten” to put in the correct value of the function f at the isolated point z₀. By putting in the correct value, the singularity disappears.

A pole is somewhat more interesting. For example the function

has a “simple” pole at the point z₀ = 0. But then if we multiply f with the “simple” polynomial g(z) = z, then we get f(z) · g(z) = 1. Of course the function f · g has a removable singularity at the special point z₀ = 0. Furthermore, the function which results does not have a zero at the point 0. The general rule is: let z₀ be an isolated, not removable, singularity of the analytic function f. If there is some n ∈ ℕ such that the function given by f(z) · (z − z₀)ⁿ has a removable singularity at z₀, and the resulting function does not have a zero at z₀, then z₀ is a pole of order n.

Finally, an essential singularity is neither removable, nor is it a pole.

Definition 12
Let f : G → ℂ be analytic, such that all it’s isolated singularities are either removable or else they are poles. Then f is called a meromorphic function (defined in G).

• Complex Analysis: #13 Weierstrass`s Convergence Theorem

This is the analog of the theorem in real analysis which states that a sequence of continuous functions which is uniformly convergent converges to a continuous function. But here we are concerned with analytic functions.


Theorem 28 (Weierstrass)
Let G₁ ⊂ G₂ ⊂ · · · ⊂ G_n ⊂ · · · be an increasing sequence of regions in ℂ and let (f_n)_n∈ℕ be a sequence of analytic functions f_n : G_n → ℂ for each n. Let G = ∪_n=1^∞ G_n (that is, n = 1, . . ., ∞), and assume that in every compact subset of G, the sequence (f_n) converges uniformly. Let f : G → ℂ be defined by f(z) = lim_n→∞ f_n(z) for all z ∈ G. Then f is analytic on G, and furthermore f_n' → f ' uniformly on every compact subset of G.

Proof
Let z₀ ∈ G and take r > 0 such that B(z₀, r) ⊂ G. (That is the closure of the open set B(z₀, r).) Let N ∈ ℕ be sufficiently large that B(z₀, r) ⊂ ∪_n=1^N G_n. Then B(z₀, r) ⊂ G_m, for all m ≥ N. According to theorem 2 we have ∫_γf_n(z)dz = 0 for all triangles in B(z₀, r). Moreover we have

owing to the uniform convergence of the sequence. Therefore, by theorem 10, f is analytic. By Goursat’s theorem, the derivatives are also analytic. Specifically, let ζ be the boundary of the disc B(z₀, r). Then, using Cauchy’s formula (theorem 6), we have

and the fact that the convergence of the f_n is uniform in B(z₀, r) shows that f_n' → f ' uniformly.

An interesting example of this is Riemann’s Zeta function. Let z = x + iy with x > 1. Then

Thus the infinte sum ζ(z) = ∑ n^−z  (with n =1, . . . , ∞) defines a function which is the limit of a uniformly convergent sequence of analytic functions (the partial sums) for all z with Re(z) > a, where a > 1 is a given constant. Therefore Weierstrass’s convergence theorem implies that the Zeta function ζ(z) is analytic in this region. As a matter of fact, Riemann showed that, with the exception of the obvious isolated singularity at the point z = 1, the zeta function can be analytically continued throughout the whole complex plane. The big question is “Where are the zeros of the zeta function?” Some of them are located at negative even integers. (These are the so-called “trivial zeros”.) They are not particularly interesting. But there are lots along the vertical line Re(z) = 1/2. The famous Riemann conjecture — which is certainly the greatest unsolved problem in mathematics today — is that all of the zeros (apart from the trivial ones) lie on this line. [An American businessman, Mr. Landon T. Clay has put aside one million American dollars each for the solution of a certain collection of outstanding problems in mathematics. The Riemann conjecture is one of them.]

Complex Analysis: #12 Index of Point with Respect to a Closed Path

• Complex Analysis: #12 Index of a Point with Respect to a Closed Path

Let γ : [t₀, t₁] → ℂ be a continuous closed curve. Let z₀ ∈ ℂ be a point not on the path. Then, as we have seen in Exercise 2.2, we can define a continuous path θ : [t₀, t₁] → ℝ with θ(t₀) = 0 and

for all t ∈ [t₀, t₁]


Definition 10
The number θ(t₁) ∈ ℤ is called the index of the point z₀ with respect to the closed path γ. Sometimes it is also called the winding number of γ with respect to z₀. It is denoted ν_γ(z₀).


Theorem 24
Let α and β be homotopic closed paths, homotopic by a homotopy H : Q → ℂ. Assume that z₀ ∉ H(Q). then ν_α(z₀) = ν_β(z₀).

Proof
Let θ_α be the corresponding path in ℝ (corresponding to α) and let θ_β be the path in ℝ corresponding to β. Then the homotopy from α to β induces a homotopy of the path θ_α to θ_β in ℝ. Since the endpoints thus remain fixed, the index remains unchanged.


Theorem 25
Let γ, z₀ and ν_γ(z₀) be as in the definition (but this time we assume that γ is continuously differentiable). Then

Proof
Obviously the function 1/(z − z₀) is analytic in ℂ \ {z₀}. For simplicity, assume that z₀ = 0 and γ(0) = γ(1) = 1 = e^i·0 , where γ : [0, 1] → ℂ \ {0}. Write γ(t) = r(t)e^2πiθ(t). Let

h_τ(t) = e^2πiθ(t)(t + (1 − t)r(t)),

for τ, t ∈ [0, 1]. This defines a homotopy from γ to a new path γ, which lies on the unit circle. But

Theorem 26 (Cauchy’s Theorem: Complicated version)
Let f : G → ℂ be an analytic function defined in a region G of ℂ. Let γ : [0, 1] → G be a continuous closed path in G such that ν_γ(a) = 0 for all a ∉ G. Then ∫_αf(z)dz = 0.

Proof
Since the (image of the) path γ is a compact subset of G, there exists an ∈ > 0 such that for all t ∈ [0, 1], we have B(γ(t), ∈ ) ⊂ G. Let n ∈ ℕ be sufficiently large that 1/n < ∈ /√2. Then split up the whole complex plane ℂ into a system of small squares of the form

where p, q ∈ ℤ. γ meets only finitely many of these small squares, and each of the squares which γ does meet is completely contained in G. Our proof now consists in altering γ, one step after the next, through processes which can be achieved using a homotopy. In the end we get a version of γ which is so simple that the theorem becomes obvious.

• Let Q be one of the squares which γ meets. If γ only runs along the boundary of Q without entering it’s interior, then there is nothing further to do in this step of the proof. On the other hand, if γ does enter the interior of Q, then we perform a homotopy on it, moving it to the boundary of Q, but leaving all points of γ which are not in the interior of Q unchanged. Basically, we take an interior point of Q which is not a point of the path, then we push the part of γ which is in the interior of Q radially from that point out to the boundary. Do this with all the Q’s which are on the path, then finally, if necessary, move the endpoint γ(0) = γ(1) to a vertex of one of these squares. [Alternatively, and without loss of generality, we may assume that the endpoint of γ is already in a corner of the lattice] The end result of all this is a homotopy, moving γ so that at the end of the movement it is contained within the lattice of vertical and horizontal lines which make up the boundaries of all the small squares. For simplicity, let us again call this “simplified” version of the path γ.



• Now it may be that this simplified γ is still too complicated. Looking at each individual segment of the lattice, it may be that during a traverse of some particular segment, we see that γ moves back and forth in some irregular way. So again using a homotopy, we can replace γ with a new version which traverses each segment of the lattice in a simple linear path.



• We are still not completely happy with the version of γ which has been obtained, for it might be that γ has a winding number which is not zero with respect to one of the squares Q. That is, let q be an interior point of Q. What is ν_γ(q)? If it is not zero, then we can alter γ using a homotopy, changing the winding number with respect to this particular square to zero and not altering the winding number with respect to any of the other squares. The idea is simply to add a bit on to the end of γ, going out to a corner of the offending square along the lattice, then around the square a sufficient number of times to reduce the winding number to zero, returning to that corner, then returning back along the same path through the lattice to the starting point. Since γ is compact, there can be at most finitely many such offending squares. At the end of this operation, we have ensured that γ has winding number zero with respect to all points of ℂ which do not lie on the path γ.

Our curve is now sufficiently well simplified for our purposes. The reason for this is that for every segment of the lattice which γ traverses, it must be that it traverses the segment the same number of times in the one direction as it does in the other direction. To see this, let S be some segment of the lattice. To visualize the situation, imagine that it is a vertical segment. Assume that γ traverses the segment u times in the upwards direction, and d times in the downwards direction. Now S adjoins two squares, say Q₁ on the left-hand side, and Q₂ on the right-hand side. Let us now alter γ, moving the downwards moving parts which traverse S leftwards across Q₁ to the other three sides of Q₁. On the other hand, the upwards moving parts of γ are moved rightwards across Q₂ to the other three sides of Q₂. Let z₁ be the middle point of the square Q₁, and let z₂ be the middle point of Q₂. Before this movement, the winding number of γ with respect to both points was zero. However, afterwards, ν_γ(z₁) = a and ν_γ(z₂) = b. Yet they are in the same region of ℂ\ γ. Thus the points must have the same winding numbers and so a = b.

Finally, after all this fiddling, we see that we must have ∫_γf(z)dz = 0 since the total of the contributions from the path integrals along each of the segments of the lattice adds up to zero in γ each case.

Up to now we have always been performing path integrals around single closed paths. This is quite sensible. But sometimes it is also convenient to imagine two or more closed paths. So let say γ₁, . . . , γ_n be n paths, each of which are piecewise continuously differentiable and closed in some region G. We can think of them together, and call them a cycle, denoted by the letter Ω say. [More generally, we might allow paths which are not necessarily closed. In this case one speaks of “chains”, but we will not persue this idea further in this here.] Then if f : G → ℂ is a function, we might be able to perform the path integrals over each of the γ_j separately, and thus we can define the integral over the whole cycle to be simply the sum of the separate integrals.

Furthermore, if Ω is a cycle and z₀ is a point not on any path of the cycle, then we can define the index of z₀ with respect to the cycle to be the sum of the indices of z₀ with respect to the individual closed paths in the cycle. On the other hand, if we want, we can connect the endpoints of the paths in a cycle together to make a single larger closed path, thus showing that theorem 26 is also true for cycles.


Theorem 27
Again, let G ⊂ ℂ be a region, f : G → ℂ be analytic, γ a closed curve in G with winding number zero with respect to all points of ℂ not in G. Let z₀ ∈ G be a point which does not lie on the path γ, then

Proof
This is a straight-forward generalization of theorem 6. For r > 0 sufficiently small, we have

Let β(t) = z₀ + r · e^2πit be the path we are thinking about in this path-integral. (Here t ∈ [0, 1].) Obviously we have ν_β(z₀) = 1. Now take the cycle consisting of the given path γ, together with −ν_γ(z₀) copies of the path β. (If this is a negative number, then we should travel around β in the reverse direction.) We assume that r is so small that the closed disc D(z₀, r) is contained in G and furthermore the path γ does not meet D(z₀, r). Then z₀ has index zero with respect to the cycle consisting of γ − ν_γ(z₀)β. Thus, according to theorem 26,

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.