Thursday, January 31, 2013

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

Complex Analysis: #12 Index of a Point with Respect to a Closed Path equation pic 1

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 νγ(z0).


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

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


Theorem 25
Let γ, z0 and νγ(z0) be as in the definition (but this time we assume that γ is continuously differentiable). Then
Complex Analysis: #12 Index of a Point with Respect to a Closed Path equation pic 2

Proof
Obviously the function 1/(z − z0) is analytic in ℂ \ {z0}. For simplicity, assume that z0 = 0 and γ(0) = γ(1) = 1 = ei·0 , where γ : [0, 1] → ℂ \ {0}. Write γ(t) = r(t)e2πiθ(t). Let

hτ(t) = e2πiθ(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

Complex Analysis: #12 Index of a Point with Respect to a Closed Path equation pic 3


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 νγ(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

Complex Analysis: #12 Index of a Point with Respect to a Closed Path equation pic 4

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 νγ(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, νγ(z1) = a and νγ(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, . . . , γn 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 γj separately, and thus we can define the integral over the whole cycle to be simply the sum of the separate integrals.

Complex Analysis: #12 Index of a Point with Respect to a Closed Path equation pic 5

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

Complex Analysis: #12 Index of a Point with Respect to a Closed Path equation pic 6

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

Complex Analysis: #12 Index of a Point with Respect to a Closed Path equation pic 7

Let β(t) = z0 + r · e2πit be the path we are thinking about in this path-integral. (Here t ∈ [0, 1].) Obviously we have νβ(z0) = 1. Now take the cycle consisting of the given path γ, together with −νγ(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 − νγ(z0)β. Thus, according to theorem 26,

Complex Analysis: #12 Index of a Point with Respect to a Closed Path equation pic 8

Monday, January 28, 2013

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

 Complex Analysis: #11 The Monodromy Theorem equation pic 1



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 hτ : [0, 1] → ℂ be the path hτ(t) = H(t, τ). (Thus α = h0 and β = h1) Assume that f0 has an analytic continuation along the path hτ 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 hτ 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 γj = hj/n, for j = 0, . . . , n. Now the argument in the proof of the previous theorem (theorem 22) shows that the analytic continuation along γj leads to the same function as that along γj+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 γj determines uniquely it’s values along the corresponding segment of the next curve γj+1. Since the endpoints of all of the curves are identical (that is, γj(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.

Thursday, January 24, 2013

Complex Analysis: #10 Analytic Continuation

  • 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, . . ., Bn) say, with

Bj = {z ∈ ℂ : |z − pj| < rj


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

Definition 8
The functions fj here are called function elements, and if we have fj(z) = fj+1 (z) for z ∈ Bj ∩ Bj+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 fn 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, . . ., Bn) be a chain of open discs with Bj ∩ Bj+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 fj', the derivative of fj, 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 gj : B1 → ℂ. i.e. g1(z) = f1'(z) for all z ∈ Bj. The assumption is that for each j there is an analytic function gj : Bj → 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 Bn−1, giving gj : Bj → ℂ such that gj : fj' for each j < n. According to theorem 3, gn has an antiderivative, Gn : Bn → ℂ, with Gn' : gn. Now in the region Bn−1 ∩ Bnwe have gn−1 = gn. That is, fn−1' = Gn', or fn−1' − Gn' = 0. Thus fn−1 − Gn = k say, where k is a constant number. But then the function Gn + k is also an antiderivative to gn, and we can take fn = Gn + 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 pn. So each straight segment, say from pj to pj+1 is contained in the union of the two discs Bj ∪ Bj+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 < · · · < τn = t1 be a partition of the interval [t0, t1]. Assume that we have a corresponding set of open discs Bj = {z ∈ ℂ : |z − γ(τj)| < rj}(where rj > 0) such that γ(t) ∈ Bj ∪ Bj+1 , for t ∈ [τj, τj+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 < · · · < τn = t1 and t0 = ρ0 < ρ1 < · · · < ρm = 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.

Thursday, January 17, 2013

Complex Analysis: #9 Simple Consequences

  • 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 · z, for some θ.

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

Complex Analysis: #9 Simple Consequences equation pic 1

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, for some θ.

Tuesday, January 15, 2013

Complex Analysis: #8 Zeros of Analytic Functions

  • 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

Complex Analysis: #8 Zeros of Analytic Functions equation pic 1

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 ck ≠ 0. (If cn = 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 ck ≠ 0.) The easiest case is then that k = 1. In this case, we have

Complex Analysis: #8 Zeros of Analytic Functions equation pic 2

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

Complex Analysis: #8 Zeros of Analytic Functions equation pic 3

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

Complex Analysis: #8 Zeros of Analytic Functions equation pic 4

say, for z ∈ B(z0, ∈ ). Here g : B(z0, ∈ ) → ℂ is analytic, and g(z0) = ck ≠ 0. Thus there are k distinct k-th roots of the number g(z0) = ck. [For any w ≠ 0 in ℂ we have w = re say. Then each of the numbers kr · eiθ/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 ϕk'(z) = zk. We know that ϕk is an entire function, and that ϕk'(z1) = kz1k−1 ≠ 0, since z1 ≠ 0. So there is a neighborhood U1 of z1, and a neighborhood V1 of g(z0), such that ϕk : U1 → V1 is a bijection [ϕk is totally differentiable and non-singular at z1], with ϕk(z1) = g(z0). Let ϕk−1 : V1 → U1 be the inverse mapping. Now choose ∈ > 0 so small that g(B(z0, ∈ )) ⊂ V1. Then take h(z) = (z − z0) · ϕk−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 U0 ⊂ G of z0 with f(U0) = B(0, ∈0). Within U0 ,z0 is the only zero of f, and if w ≠ 0 in B(0, ∈0) then there are precisely k different points v1, . . ., vk in U0 with f(vj) = w, for all j.

Proof
Since f is continuous and B(0, ∈0) is open, it follows that U0 = 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, . . ., uk. They are all in B(0, ∈). Therefore each has a unique inverse under h, namely vj = h−1(uj), for j = 1, . . . , k. Is it possible that some other point, v say, not equal to any of the vj, 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 uj, 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.

Sunday, January 13, 2013

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

Complex Analysis: #7 Some Standard Theorems of Complex Analysis equation pic 1

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
Complex Analysis: #7 Some Standard Theorems of Complex Analysis equation pic 2


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 |cn| ≤ M/rn = 0 for all r > 0. This can only be true if cn = 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) = ∑akzk, (with k = 0, . . .,n), with n ≥ 1 and an ≠ 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
Complex Analysis: #7 Some Standard Theorems of Complex Analysis equation pic 3


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 + · · · + bn| ≥ |a| − |b1| − · · · − |bn|.

Since |an|/2 remains constant, |zn| · |an|/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.

Friday, January 11, 2013

Complex Analysis: #6 Power Series

  • Complex Analysis: #6 Power Series

Definition 4
A power series is a sum of the form

Complex Analysis: #6 Power Series equation pic 1
where (an)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 ∑an(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 = limn→∞ sup n|an|
  4. The function given by f(z) = ∑an(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) = ∑nan(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 = limn→∞ sup n|an|. Thus there exists some N0 ∈ ℕ with √|an| < 1/R for all n ≥ N0. That is, |an| < 1/Rn. Therefore

Complex Analysis: #6 Power Series equation pic 2

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|an|  > 1/R. That is,  |an| > 1/Rn or

Complex Analysis: #6 Power Series equation pic 3

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 limn→∞ sup n|nan| = limn→∞ sup n|an| since limn→∞ sup nn = 1. So let f1(z) = ∑nan(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

Complex Analysis: #6 Power Series equation pic 4

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

Complex Analysis: #6 Power Series equation pic 5

(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

Complex Analysis: #6 Power Series equation pic 6

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 Sn is just a polynomial, and thus analytic, we have a δ > 0 such that

Complex Analysis: #6 Power Series equation pic 7

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
Complex Analysis: #6 Power Series equation pic 8



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 ∑an(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
Complex Analysis: #6 Power Series equation pic 9


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 cn is not always zero, still it is always a well defined complex number, for all n. 
  • It looks like cn might vary with r. But this is not the case. Theorem 8 implies that f(n)(0) = n!cn, for all n, and this is certainly independent of r. 
  • The power series converges to f(z) at all points of B(z0, r).

Wednesday, January 9, 2013

Complex Analysis: #5 Cauchy`s Integral Formula

  • 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

Complex Analysis: #5 Cauchy`s Integral Formula equation pic 1

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

Complex Analysis: #5 Cauchy`s Integral Formula equation pic 2

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

Complex Analysis: #5 Cauchy`s Integral Formula equation pic 3

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.

Monday, January 7, 2013

Complex Analysis: #4 Cauchy`s Theorem (simplest version)

  • Complex Analysis: #4 Cauchy`s Theorem (simplest version)


Theorem 1
Let G ⊂ ℂ be a region, and assume that the function f : G → ℂ has an antiderivative (auf deutsch: Stammfunktion) F : G → ℂ with F' = f. Let γ be a closed path in G (that is, a continuous, closed, piecewise continuously differentiable path). (Closed means that γ(t0) = γ(t1).) Then ∫γ f(z)dz = 0.

Proof

Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 1

Since every polynomial has an antiderivative, it follows that the path integral around a closed path for any polynomial is zero.

Of course this is all a bit too trivial. So let’s call the following theorem the simplest version of Cauchy’s integral theorem.


Theorem 2
Let Q be a (solid) triangle in the complex plane. Assume that Q ⊂ G ⊂ ℂ, and take f : G → ℂ to be an analytic function. Let γ be the closed path traveling around the three sides of Q. Then ∫γ f(z)dz = 0.

Proof
We may assume that γ begins and ends in a corner of Q — for example the “lowest” corner γ in the complex plane. If the lower side of Q is parallel to the real number axis, then take the right-hand corner on that side. Let us now divide the sides of Q in half, connecting the half-way points with straight line segments, thus creating four equal sub-triangles, Q1, . . . , Q4. Let γj be the path traveling around the boundary of Qj, for j = 1, . . . , 4. Again we may assume that each γj begins and ends in the bottom right corner of it’s triangle. So we have

Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 2

Assume further that each of these paths is parameterized in the simplest way possible, so that |γ'| = 1 and |γj'| = 1 for all the j. Therefore Lγ is the sum of the lengths of the three sides of the triangle Q, and Lγj = Lγ/2 for each of the j.

Let’s say that γ1 is one of the numbers between one and four such that the value of

Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 3
is the greatest. Then we certainly have

Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 4

The next step is to concentrate on the triangle Qj1. As with Q, we subdivide Qj1 into four equal sub-triangles and we take paths around their boundaries. Choose Qj2 to be one of these sub-triangles of Qj1 which is such that the value of

Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 5
is the greatest. Here γj2 is the path around the boundary of Qj2. Now we have Lγj2= Lγ/4, and
Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 6

This whole process is continued indefinitely, so that we obtain a sequence of triangles, becoming smaller and smaller, converging to a point, z0 ∈ Q say,
Q ⊃ Qj1⊃ Qj2 ⊃ · · · → z0 ∈ Q. 

For each n we have Lγjn= Lγ/2n and
Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 7

But we have assumed that f is analytic, in particular it is differentiable at the point z0. Thus we can write

f(z) = f(z0) + f '(z0)(z − z0) + χ(z),

for points z in G, where χ : G → ℂ is a continuous function with

Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 8

So let ∈ > 0 be arbitrarily given. Then there exists some δ > 0 such that |χ(z)| < |z − z0| for all z with 0 < |z − z0| < δ.

Now we need only choose n so large that |z − z0| < δ for all z ∈ Qjn. For such z we have
Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 9
On the other hand, again since the length of Lγjnis Lγ/2n, we have

Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 10

Bearing in mind Theorem 1 (and remembering that f '(z0) is simply a constant complex number), we conclude that

Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 11
Since ∈ was arbitrary and Lγ remains constant, we conclude that ∫γ f(z)dz = 0.

But now Theorem 2 can be turned around, and we obtain (almost) the converse.


Theorem 3
Assume G ⊂ ℂ is a region and f : G → ℂ is a continuous function. Assume furthermore that for any solid triangle Q contained in G we have ∫γ f(z)dz = 0, where γ is the path around the triangle. Then f has an antiderivative in every open disc contained in G. That is, let U = {z ∈ ℂ : |z − z*| < r} be some such disc, where z* is a complex number (the middle point of the disc) and r > 0 is the radius of the disc. Then there exists F : U → ℂ with F'(z) = f(z) for all z ∈ U.

Proof
By replacing f with the function f*, where f*(z) = f(z − z*), we obtain the situation that z* = 0. Clearly, if the theorem is true for f*, then it is also true for f. Therefore, without loss of generality, we may simply assume that z* = 0.

Within U the function F is defined to be

Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 12

Here, αz is the straight line from 0 to z, that is, αz(t) = tz. To show that F really is an antiderivative to f in U, let z0 be some arbitrary point of U and let z be some other point of U. Let β be the straight line connecting z0 to z. That is, β(t) = (1 − t)z0 + tz. Being a triangle, the integral of f around the path from 0 out to z, then from z to z0 then from z0 back to 0 must itself be zero. That is,

Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 13

Looking at the definition of the path integral, we see that

Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 14


Combining this theorem with Theorems 1 and 2, we see that if D ⊂ G is a closed disc, and γ is the circle of it’s boundary, then  ∫γ f(z)dz = 0, for any analytic function defined in the region G. In fact, if γ is any (piecewise continuously differentiable and continuous) closed path contained within this disc-like G, then  ∫γ f(z)dz = 0. For example we can look at a rectangle [a, b] × [c, d] contained within G. Since the rectangle can be taken to be a union of two triangles, attached along one side, we see that also the path integral around the rectangle must be zero.

More generally, the following theorem will prove to be useful.


Theorem 4
Let Q = {x + iy : 0 ≤ x, y ≤ 1} be the standard unit square in ℂ. Take ζ to be the standard closed path, traveling around the boundary of Q once in a counterclockwise direction, beginning and ending at 0. Assume that a continuously differentiable mapping ϕ : Q → ℂ is given, such that ϕ(Q) ⊂ G, a region where an analytic function f : G → ℂ is defined. Let γ = ϕ ◦ ζ be the image of ζ under ϕ. Then ∫γ f(z)dz = 0.

Proof.
Since Q is compact, ϕ(Q) is also compact. Therefore it can be covered by a finite number of γ open discs in G. But Q can now be partitioned into a finite number of sub-squares Q1, . . ., Qn such that ϕ(Qj) is in each case contained in a single one of these open discs. The theorem then follows by observing that the path integral around each of these sub-square images must be zero.

 Note:-
[The inverse images of the open discs in Q form a finite open covering V1, . . ., Vm of Q. A sequence of partitions of Q can be obtained by cutting it along horizontal and vertical lines spaced 1/n apart, for each n ∈ ℕ. Can it be that for each of these partitions, there exists a sub-square which is not contained completely in one of the open sets Vk? But that would mean that there exists a limit point q ∈ Q such that for every ∈ > 0, there are infinitely many of these sub-squares contained within a distance of from q. However q ∈ Vk, for some k, and since Vk is open, there exists an ∈-neighborhood of q contained entirely within Vk, providing us with the necessary contradiction.]

A special case is the following.


Theorem 5
Let D1 and D2 be closed discs in ℂ such that D2 is contained in the interior of D1. Let γj be the closed path going once, counterclockwise, around the boundary of Dj, j = 1, 2. Let G ⊂ ℂ be a region containing D1\D2 and also containing the boundary of D2. Assume that f : G → ℂ is analytic. Then ∫γ1 f(z)dz = ∫γ2 f(z)dz.

Proof
The annulus between D2 and D1 can be taken to be the image of the unit square under a continuously differentiable mapping.

A convenient notation for this situation is the following. Let γ : [0, 1] → ℂ be the path γ(t) = z0 + re2πit . Then we simply write

Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 15

With this notation, we can say that if the analytic function f is defined in a region containing the annulus {z ∈ ℂ : r ≤ |z| ≤ R}, then we must have

Complex Analysis: #4 Cauchy`s Theorem (simplest version) equation pic 16

Saturday, January 5, 2013

Complex Analysis: #3 Path Integrals

  • Complex Analysis: #3 Path Integrals

Let t0 < t1 be two real numbers. Then a continuous mapping γ : [t0, t1] → G ⊂ ℂ is a path in the region G of ℂ. In the analysis lecture we learned that γ is rectifiable if a number L exists such that for all ∈ > 0, a δ > 0 exists such that for every partition t0 = a0 < a1 < · · · < an = t1 which is such that aj+1 − aj < δ for all j, we have

Complex Analysis: #3 Path Integrals equation pic 1

Let γ(t) = γr(t) + iγi(t), where γr, γi: [t0, t1] → ℜ are real-valued functions. Then we say that the path is continuously differentiable if both the functions γr and γi are continuously differentiable. In this case, γ' = γr' + iγi' is also a path in ℂ. [Thinking in terms of 2-dimensional real geometry, we can say that γ'(t) is the "tangent vector" to γ(t).] We also learned that continuously differential paths are always rectifiable, and we have

Complex Analysis: #3 Path Integrals equation pic 2

All of this has already been dealt with in the analysis lecture. For us now, the interesting thing is to think about path integrals through a region where a complex-valued function is given.


Definition 3
Let G ⊂ ℂ be a region, and let f : G → ℂ be a function. Furthermore, let γ : [t0, t1] → G be a differentiable path. Then the path integral of f along γ is

Complex Analysis: #3 Path Integrals equation pic 3
assuming it exists.

The integral here is simply the sum of the integrals over the real and the imaginary parts. It is not necessary to assume that γ is continuously differentiable, but we will assume that it is piecewise continuously differentiable. That is, there is a partition of the interval [t0, t1] such that it is continuously differentiable along the pieces of the partition. So from now on, we will (almost) always assume that all paths considered are piecewise continuously differentiable.

As an exercise (using the substitution rule for integrals), one sees that the path integral does not depend on the way the path is parameterized. The simplest case is that, say γ(t) = t. Then (taking t from 0 to 1) we just have ∫γ f(z)dz = ∫01 f(t)dt. Almost equally simple is the case that γ(t) = it. Then we have ∫γ f(z)dz = i∫01 f(it)dt.

Increasing the complexity of our thoughts ever so slightly, we arrive at the first version of Cauchy’s integral theorem.

Thursday, January 3, 2013

Complex Analysis: #2 Analytic Functions

  • Complex Analysis: #2 Analytic Functions

Definition 1
Any non-empty connected  open set G ⊂ ℂ will be called a region.

(Recall that in ℜ2, every open connected subset is also path-connected.) So we will generally be interested in functions f : G → ℂ defined in regions.

Definition 2
Let f : G → ℂ be given, and let z0 ∈ G. If

Complex Analysis: #2 Analytic Functions screenshot 1
exists, then it is the derivative of f at z0. The function f will be called analytic in G if it is defined, and has a continuous derivative everywhere in G. The word holomorphic is also used, and it is synonymous with the word analytic.

As in real analysis, we have the simple rules for combining the derivatives of two functions f and g:

Complex Analysis: #2 Analytic Functions screenshot 2

Nevertheless, there is a very big difference between the idea of a derivative in complex analysis, and the familiar derivative in real analysis. The reason for this is that a common limit must exist, regardless of the direction with which we approach the point z0 in the complex plane. This leads to the Cauchy-Riemann differential equations.

Looking at the definition of the complex derivative, one immediately sees that it is really a special version of the total derivative (as in analysis 2) in ℜ2. Thus, for ξ ∈ ℂ sufficiently small (that is |ξ| small), we have

f(z0 + ξ) = f(z0) + Aξ + |ξ|ψ(ξ), 

where A is a 2 ×2 real matrix, and limξ→0 ψ(ξ) = 0. But what is A? It represents multiplication with the complex number f '(z0) = a + ib, say. That is,
Complex Analysis: #2 Analytic Functions screenshot 3

So what are these real numbers a and b? Let f(z) = u(z) + iv(z), where u, v : G → ℜ are real functions. Then writing z = x + iy, we have f(x + iy) = u(x + iy) + iv(x + iy). Identifying ℂ with ℜ2, we can consider the partial derivatives of u and v. Since A is simply the Jacobi matrix of the mapping f at the point z0, we must have

Complex Analysis: #2 Analytic Functions screenshot 4

These are the Cauchy-Riemann equations. Another way to express this is to simply say that we must have

Complex Analysis: #2 Analytic Functions screenshot 5

Thinking in geometrical terms, we see that if f is analytic, then it is a conformal mapping, at least at the points where f ' is not zero. That means that, locally, the mapping preserves angles. Looked at up close, the mapping is

Complex Analysis: #2 Analytic Functions screenshot 6


Of course, as we have already seen, the rotation preserves orientation. Thus it is an element of the group SL2 (R).

Another interesting detail is that the real and imaginary parts of an analytic function are themselves harmonic functions. Anticipating a later conclusion, let us assume that the parts of the analytic function f = u + iv are twice continuously differentiable. Since ∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x, we have

Complex Analysis: #2 Analytic Functions screenshot 7

Or, expressed in another standard form of notation,

uxx + uyy  = ∆u = 0. 

Here, ∆ is the Laplace operator. Similarly, we see that ∆v = 0.

All of this shows that we cannot simply choose any old smooth function f : G → ℂ and expect it to be analytic. On the contrary, there is a very great “rigidity”, which means that most smooth functions — even though they may be partially differentiable when considered as mappings of 2-dimensional Euclidean space — are not complex differentiable.


Examples
  1. The first example is the nice and smooth function f(z) = f(x + iy) = x2 + y2. Here ux = 2x and vy = 0. But according to the Cauchy-Riemann equations, we must have ux = vy; that is, x = 0. This only holds along a single line in the complex plane ℂ. Therefore it certainly can’t hold in any region of ℂ (since regions are defined to be open), and thus f, despite all appearances of being a nice function, is definitely not analytic.


  2. Having been cautioned by the previous example, let us try to construct an analytic function. For example, let us assume that u(x + iy) = x. What possibilities are there for v(x + iy)? Since ux = 1 = vy and vy = 0 = vx, it is clear that the only possibility is v(x + iy) = y + constant. So this is just the rather boring function f(z) = z + constant.


  3. Thinking more positively, we have just seen that the simplest non-trivial polynomial, namely f(z) = z, is analytic throughout ℂ. Of course the simplest polynomial, f(z) = constant, is also analytic. But then, noting that we can use the sum and product rules for differentiation in complex analysis, we see that any arbitrary complex polynomial is analytic throughout ℂ. Indeed, z−n is also analytic (in ℂ \ {0}) for any n ∈ ℕ.
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