1. Complex Sequences and Series

Transcription

1 . omplex Sequences and Series Let denote the set {(x, y): x, y real} of complex numbers and i denote the number (, ). For any real number t, identify t with (t, ). For =(x, y) =x+iy, let Re = x, Im = y, = x iy and = x + y. The distance between and w is then given by w. For, arg denotes the polar angle of (x, y) in radian (modulo π). Every nonero complex number has a polar representation = r cis θ, where r =, θ = arg and cis θ = cos θ + i sin θ. Properties. Let, w be complex numbers and n be an integer. Then () + =Re, =i Im, =, = cis( arg ) for, () + w = + w, w = w, w = w, /w = /w for w, (3) + w + w, w w, w = w, /w = / w for w, (4) for, w, arg(w) = arg + arg w, arg( w ) = arg arg w, arg(n )=n arg. In analysis, reasoning involving limits are very common and important. We will begin with the concept of the limit of a sequence. (For convenience, we will abbreviate if and only if by iff or in the sequel.) Definition. A sequence {,, 3,...} (or in short, { n }) converges to (denoted by n ) iff for each ε>, there is N ε such that n N ε implies n ε (in short, lim n =.) Otherwise, the n sequense is said to diverge. converges to if <, Examples. () n = n converges to if =, (For <, n = n asn. For diverges otherwise. >, n = n.for = and, we have = cis θ with θ kπ, n = cis(nθ) spins around the unit circle.) n () lim =, since n n n + n + = n + = x + y asn. (n + x) + y Often the limit of a sequence is difficult or impossible to find. We now introduce a criterion that allows us to conclude a sequence is convergent without having to identify the limit explicitly. Definition. A sequence { n } is a auchy sequence iff for each ε>, there is N ε such that m, n N ε implies m n ε (in short, lim m,n n m = ). Lemma. A auchy sequence {a n } of real numbers must converge to some real number. Proof. For ε =, there is N such that m, n N implies a m a n (i.e. a n a m a n +). Taking n = N and letting p = min{a,...,a N } and q = max{a,...,a N } +, we get p a m q for all m. Next, let S = {x R : x a k for infinitely many k}. Note p S, but q S. Also, if x x and x S, then x S. This implies S is an interval of the form (,a)or(,a], where a is the right endpoint of S. We will show a n a.

2 Given ε>, there is N ε such that m, n N ε implies a m a n ε. Since a + ε S, a + ε a k for only finitely many k. However, a ε S, soa ε a k for infinitely many k. Hence, we can find some m N ε such that a ε a m a + ε (i.e. a m a ε ). Then for all n N ε, we have a n a a n a m + a m a ε + ε = ε. Theorem. A sequence { n } converges if and only if it is a auchy sequence. Proof. Suppose n, then for any ε >, the closed disk with as center and radius ε contains K, K+, K+,... for some K. Let N ε = K, then m, n N ε implies m n = ( m )+( n ) m + n ε + ε = ε. onversely, suppose { n } is a auchy sequence. Since Re m Re n m n, {Re n } is also a auchy sequence. Similarly, {Im n } is also a auchy sequence. By the lemma above, {Re n } converges to some real x and {Im n } converges to some real y. Then { n } converges to x + iy. Definitions. A series k = converges iff the sequence, +, + + 3,...converges k= (iff S n = n is a auchy sequence). Otherwise, the series is said to diverge. There are two simple tests for checking convergence of series, namely the term test and the absolute convergence test. The former provides a necessary condition for convergence and the latter provides a sufficient condition for convergence. Term Test. If k converges, then lim n = lim (S n S n ) = lim S n lim S n =. n n n n k= Absolute onvergence Test. If k converges, then k converges (because T n = n k= is a auchy sequence and for m>n, S m S n = n+ + n m n+ + n m = T m T n, forcing S n to be a auchy sequence.) k= k= Examples. () k + i converges because k + i = k= k= k= ( ) ( k i () diverges because Re =Re k + i k + i k + k= k= ( k4 + ) = k= k k + k k= ) converges. ( k= ) diverges. k Exercises. For what complex values will the following series converge ( ) n n (a) ; (b) + +? n n=. When will equality occur in the triangle inequality? That is, under what conditions on w and will w + = w +? n=

3 n 3. Establish the identity α k β k = k= n n α k β k α k β j α j β k for the case n =. k= (This implies the auchy-schwar inequality k= k<j n n α k β k n α k n β k.) k= k= k= 4. Suppose <a a a n. Prove that the polynomial P () =a n + a n + + a n has no root in the open unit disk D = { : < }. [Hint: onsider ( )P ().] 5. Prove that if +i 9 +i =, then =. [Hint: Solve for 9.] (This problem came from the 989 William Lowell Putnam Mathematical ompetition.) 6. Let P () = n + c n + + c n with c,,c n real. Suppose P (i) <. Prove that there is a root x+iy of P () satisfying (x +y +) 4y <. (This problem came from the 989 USA Mathematical Olympiad.) 3

4 . Set Descriptions and Terminologies For { a, b(, ) the line } L through a in the direction b consists of all = a + tb ( <t< ), so a L = :Im =. The circle with center c and radius r is given by = {: c = r}. b Imaginary axis y=-x y y y=x y O Real axis S -i O x _ i 4 i 4 = ( x,y ) x O (,) x Examples. () S = {: + i } consists of complex numbers that are closer to i than, i.e. the closed half plane below y = x. () To describe the parabola y = x in complex variable, we set x = +, y = and get = ( ) i i +. Alternatively, using the fact that the focus is at (, 4 )= 4 i and the directrix is y = 4, the equation of the parabola is i 4 = Im + 4. { (3) :< arg( i) < π } is the open quarter-plane lying in the first quadrant with vertex at (, ) and edges parallel to the x, y axes. Next, we will introduce some notations and terminologies often used in the sequel. The open disk {: <r} is denoted by B(,r) and is referred to as the r-neighborhood of. The closed disk {: r} is denoted by B(,r). The circle {: = r} is denoted by (,r). The boundary of a set S is denoted by S and consists of all points such that every neighborhood of contains a point in S and a point not in S. In particular, we note that a set S and its complement \S have the same boundary. A set is open iff it does not contain any boundary point. A set is closed iff it contains all boundary points. A set containing some, but not all, boundary points is neither open nor closed. The interior of a set S is S \ S and the closure of S is S S. A set is bounded iff it is contained inside a neighborhood of O. A set is compact iff it is closed and bounded. A set S is disconnected iff it is contained in the union of two disjoint, open sets A, B each of which contains at least one point of S, i.e. there are open sets A and B such that S A B,A B =,S A and S B. A set is connected iff it is not disconnected. A set is a region or a domain iff it is open and connected. Remarks. () S is open every point in S has a neighborhood containing in S. () S is closed \S is open n S, n implies S. Theorem. Every region S is polygonally connected, i.e. any two points in S can be joined by a polygonal line lying totally in S. In fact, the polygonal line can be chosen to consist of horiontal or vertical segments. 4

5 Proof. Suppose p S. Let A = { S: can be connected to p by a polygonal line lying totally in S} and B = S \ A. Then A must be open because for each A, there is B(,r) S and every point in B(,r) can be connected to (hence to p) by such a polygonal line, i.e. B(,r) A. Similarly B is open because for each w B, there is B(w, t) S and every point in this B(w, t) can be connected to w (hence not to p) by such a polygonal line, i.e. B(w, t) B. Since S is connected, S = A B and A is nonempty, B must be empty. Therefore A = S. The theorem requires S open! A circle is connected, but not polygonally connected. In advanced courses, it will be showed that a polygonally connected set is connected. So an open and polygonally connected set is a region. y y x x Examples. () Let S = B(, ) \{}. Then S = (, ) {} ( is a boundary point because every neighborhood of contains points in S and, which is not in S). So S is open, bounded, connected, not closed and not compact. () Let S = {:Re }, the right half plane. Then S is the imaginary axis. So S is closed, connected, not open, not bounded and not compact. (3) Let S = (, ), the unit circle. Then S = S. So S is closed, bounded, compact, connected and not open. (4) Let S = {: = n } for n =,, 3,.... Observe that is not in S, but it is a boundary point of S. So { S = S {}. S is not connected because for instance, S : > 3 } { : < 3 }, where the two sets 4 4 }{{}}{{} open open are disjoint. Definitions. () A right-angled polygon is a polygon (without self intersections) whose edges consist of horiontal and vertical segments. () A simply connected region is a region such that if a right-angled polygon is in the region, then the inside of the polygon is also in the region. Remarks. Simply connected regions have no holes. A region is simply connected iff every polygon can be shrunk to a point without leaving the region. Examples. () The following are examples of simply connected regions: (i) open convex sets (in particular, disks, half-planes, infinite strips), (ii) complex plane with nonintersecting infinite slits removed, (iii) open unit disk with a spiral joining the center to the boundary removed. o o o o () The following are examples of non-simply connected regions: (i) punctured disk, (ii) annulus, (iii) complement of a segment in. 5

6 o o Exercises. Describe the sets whose points satisfy the following relations (a) + = ; (b) + < ; (c) = ; (d) arg + i = π 3.. Given distinct complex numbers α,β,γ. Show that they are collinear if and only if Im(αβ+βγ +γα) =. 3. Show that the numbers α,β,γ are the vertices of an equilateral triangle if and only if α + ωβ + ω γ =, where ω is a cube root of unity. 4. Prove that if,, 3 are distinct complex numbers and = = 3, then arg 3 3 = arg. 5. What is the boundary of the set { :Re and Im are rational}? 6

7 3. ontinuity and Uniform onvergence Definitions. A function f: D has a limit c at D D (i.e. lim f(w) =c) iff for every w sequence n D, we have f( n ) c (iff for each ε>, there is δ>such that < w <δ implies f(w) c <ε.) We say f is continuous at D iff lim f(w) =f(); f is continuous on D iff f is w continuous at every point of D. The ε-δ definition of limit at can be rephrased as for each ε-neighborhood B(c, ε) ofc, there is a δ-neighborhood B(,δ) of such that w B(,δ) \{} implies f(w) B(c, ε). Later we will discuss about neighborhoods of and use this version to extend the notion of limit to the case at. Of course, as usual, the sum, difference, product, quotient (where the denomiator is nonero) and composition of continuous functions are continuous. Examples. () Polynomials P () =a n n + a n n a are continuous everywhere on. () Rational functions P () (where P and Q are polynomials) are continuous at, where Q(). Q() (3) If P (x, y) = N m,n= of continuous functions are continuous. a mn x m y n, then f() =P (,) is continuous everywhere on. In fact, compositions For functions defined by taking limits or summations, to check continuity we usually rely on uniform convergence. Definition. A sequence of functions S n : D converges uniformly to a function S: D iff for every ε>, there is N ε such that m N ε implies S() S n () <εfor all D. (For series f k (), consider S n = n f k ().) k= Weierstrass M-test. If for each k, there is a number M k such that f k () M k for all in D and converges, then f k () converges (absolutely and) uniformly on D. k= Proof. (Absolute convergence follows by comparing ε>. Since f k () with k= M k converges, there is N ε such that n N ε implies k= f k () k= n f k () = k= k=n+ f k () 7 k=n+ k= k= M k M k.) For uniform convergence, let k= k=n+ f k () M k <ε. Then for every in D, k=n+ M k <ε.

8 Properties of Uniform onvergence. () If S k s are continuous on D and S k converges uniformly to S on D, then S is continuous on D and lim lim S k(w) = lim S(w) =S() = lim S w k w k() = lim lim S k(w). (For series, if f k s are continuous on k k w D and f k converges uniformly to f on D, then f is continuous on D and lim f k (w) = lim f(w) = w w k= k= f() = f k () = lim f k(w).) w k= k= () If S k s are integrable and converges uniformly, then lim S k = lim S k. (For series, if f k s are k k integrable and f k converges uniformly, then f k = f k.) k= k= (Proof. We will prove () and postpone () until complex integration is defined. (In (), the parenthetical n statement follows from the first statement by taking S n = f k.) Given ε>. Since S k converges uniformly to S, there is N = N ε/3 such that k N ε/3 implies S() S k () < ε for all D. For a fixed w D, 3 since S N is continuous at w, there is δ> such that w <δ S N (w) S N () < ε 3. Then S(w) S() = S(w) S N (w)+s N (w) S N ()+S N () S() So S is continuous at any arbitrary w D.) Example. onsider f() = because lim k R k+ / R k (k + )! k! k= k= S(w) S N (w) + S N (w) S N () + S N () S() < ε 3 + ε 3 + ε 3 = ε. k= k k!. Suppose B(,R). The series = R lim k k + =. Since k= R k k! converges by the ratio test k k! Rk = M k, by Weierstrass M-test, k! converges uniformly on B(,R). Since k are continuous on B(,R), f is continuous on B(,R). Finally, k! since R is arbitrary, f is continuous everywhere on the complex plane. k= k k! Exercises. Show that f() =. Show that f() = k= k + is continuous on the right half plane H = { :Re>}. k k is continuous on D = { : < }. k= 8

9 4. Stereographic Projection In many situations, it is best to treat infinity as a point. Stereographic projection explains how this can be done. Let S = {(ξ,η,ζ): ξ +η +ζ =} be the unit sphere and N =(,, ) be the north pole. There is a one-to-one correspondence between S \ N and the complex plane given by Z = (ξ,η,ζ) = x + iy = (x, y, ) as in the figure. The equation of the line through N, implies x ξ = y η = ξ. It follows that x = ζ ζ, y = η ζ. onversely, = ξ +η +ζ =(x +y )( ζ) +ζ x + y = +ζ ζ = ζ. So ξ = x( ζ) = x x + y +, y η = y( ζ) = x + y +, ζ = x + y x + y +. Z N O Z We see that if = x + iy, then x + y =,so(ξ,η,ζ) (,, ). Thus N =(,, ). The set { } is usually referred to as the extended complex plane. Because of the correspondence, some like to use S to denote the same set. If S is used, then the extended complex plane is sometimes also referred to as the Riemann sphere. Now to analye how the neighborhoods of a point are transformed under stereographic projection, we study the correspondence of the boundary circles. A circle on S is the intersection of S with a { plane Aξ + Bη + ζ = D. This corresponds to ( D)(x + y )+Ax +By = + D, which is a circle on if D line on if = D. onversely, a circle on is given by x + y + ax + by = c, which corresponds to +ζ + aξ + bη = c( ζ) ons, i.e. a circle on S. In particular, the circle on S coming from the plane ζ = ζ < corresponds to x + y = +ζ, i.e. ζ the boundary of a neighborhood of N corresponds to a circle centered at the origin. Then neighborhoods of N correspond to sets of the form {: >r}. By abuse of notation, we will let B(,r)={: >r}. The remark following the definition of lim f(w) suggests how limits involving can be defined. w Definitions. () n iff n (iff for any ε>, there is N ε such that n N ε n >ε.) () lim f() =c iff for any ε>, there is δ> such that >δ f() c <ε(iff lim f() = lim f( w w )=c.) (3) lim f() = iff for any ε>, there is δ> such that < c <δ f() >ε(iff lim c c f() =.) (4) lim f() = iff for any ε>, there is δ>such that >δ f() >ε. 9

10 (5) f() iscontinuous at iff g() =f( ) is continuous at. Definitions. () Let f be defined in a neighborhood of. Then f is differentiable at iff f () = f( + h) f() lim h,h h exists; f is differentiable at iff f( ) is differentiable at. () f is holomorphic or analytic (or regular) on a set S iff f is differentiable at every point of some open set containing S. (3) f is univalent (or schlicht) on an open set iff f is differentiable and one-to-one there. (4) f is entire (or integral) ifff is differentiable on the complex plane. Of course, as usual, the sum, difference, product, quotient and chain rules are valid in (complex) differentiation. Some common functions are differentiable, e.g. polynomials are entire functions and ( n ) = n n for integer n. However, some are not, e.g. the conjugate function is not differentiable anywhere because lim h,h + h h = lim h,h h lim h,h R h = h h doesn t exist, as can be seen by the following computations : lim h h,h R h = and lim h h=it,t R h = lim it =. t it Exercises. The chordal metric d(w, ) of two complex numbers w and is the distance in R 3 between the points corresponded to w and under stereographic projection. (a) Express d(w, ) in terms of w and only. (b) Express d(w, ) = lim d(w, ) in terms of w only. (c) Show that n { } if and only if d( n,).. Prove that f() = is not differentiable anywhere, but g() = is differentiable at only.

11 5. Power Series Definition. A power series centered at c is a function of the form f() = f(c) =a, so the series converges for at least the point = c.) a n ( c) n. n= (Note that To understand what the domains of power series (as functions) look like, we introduce the concept of the upper limit of a sequence and recall the root test. Definition. Let {x n } be a sequence of real numbers. The upper limit or limit superior of {x n } (denoted by lim n or limsup x n is n n (a) the number L if L has the properties that for each ε>, (i) x n <L+ ε for all except finitely many n and (ii) x n >L ε for infinitely many n. (b) + if for each real number r, there is x n >r. (c) if for each real number r, only a finite number of the x n s are greater than r. Essentially, by taking ε, items (i) and (ii) imply L is the limit of some subsequence. Furthermore, item (i) implies no larger number is also a limit of a subsequence. So the upper limit of {x n } is the largest limit of any subsequence of {x n }. Now if lim x n exists, then all subsequences converge to the same limit. n In that case, lim x n = lim x n. n n Examples. () lim n n =. () lim [n n +( )n n]= lim {, 4,, 8,,,...} = lim {4, 8,,...} =+. n (3) lim sin nπ n = lim {,,,,,,...} = lim {,,,...} =. n Root Test. Given a series a n, let ρ = lim n n an, then { converges absolutely if ρ<, a n diverges if ρ>, is inconclusive if ρ =. Proof. (ase ρ<) Take x such that ρ<x<. Let ε = x ρ>. Then by property (i), all but finitely many n satisfy n a n <ρ+ ε = x, then a n <x n. Since x<, x n converges, which implies a n converges. (ase ρ>) Take x such that <x<ρ. Let ε = ρ x>. Then by property (ii) n a n >ρ ε = x> for infinitely many n, then lim a n =, so a n diverges. n (ase ρ = ) onsider n and. For both cases, ρ =, but the first series diverges and the second n n= n= series converges. Theorem. The power series f() = a n ( c) n converges absolutely for c <Rand diverges for c >R, where R = n lim n n=. (In case R =, the series converges at = c only.) R is called the an

12 radius of convergence and the disk c < Ris called the disk or domain of convergence. Furthermore, a n ( c) n converges uniformly on any smaller disk c R <R. n= Proof. The first statement follows from the root test. The last statement follows by Weierstrass M-test. Since a n ( c) n a n R n for all c R, so we set M n = a n R n and use the absolute convergence of the power series at = R + c. In a course on real analysis, it is usually shown that if lim This can be used in computing the radius of convergence. n a n+ a n n exists, then lim an = lim n n The convergence on the boundary circle c = R can be arbitrary as the following examples show. Examples. () For () For n n, R =. If =, then n= n, R =. If =, then n= n= n= n= n= n n = n= n diverges because lim n n. n= converges by p-test. n n (3) For n, R =. If =, n diverges. If =, ( ) n converges by alternating series test. n Later we will show that every holomorphic function can be locally represented by power series (i.e. f holomorphic at c f() = a n ( c) n in a neighborhood of c). Therefore, the local properties of n= holomorphic functions can be understood by studying power series. First, we will see that power series can be differentiated (term-by-term) in their domains of convergence. Theorem. If f() = a n ( c) n converges on c <R, then f () = n= a n+ a n. na n ( c) n converges on c < R, too. onsequently, power series are infinitely differentiable on their disk of convergence. Proof. The proof is divided into two steps. The first step is to observe that lim ( ) n n a n = lim (n a n ) n n n n n n= ( ) = lim n n an n n = lim an, n n so the power series have the same disk of convergence. The second step is to obtain an inequality of the f( + h) f() form na n ( c) n A h. h n= Fix such that c <R. Suffices to consider small h (say h <δ<r c.) Then f( + h) f() h Observe that for k, ( ) ( ) n n (n k +) = k k k ( n ( ) n na n ( c) n = a n )h k ( c) n k. k n= n= k= }{{} b n (n k +) k ( ) n n(n ). k

13 So, for c, b n k= ( ) n h k c n k n(n ) h k c n n(n ) h c j= ( ) n h j c n j = j n k= ( ) n h k c n k+ k n(n ) h c ( c + h ) n. Therefore, as h, f( + h) f() na n ( c) n = h n= a n b n a n b n n= n= h c n(n ) a n ( c + δ) n. n= }{{} A (Note that the series in A converges because the first step above implies f (w) = n= n(n )a n(w c) n converges absolutely on w c <R, in particular at w = c + c + δ by the condition on δ.) Taylor s Theorem for Power Series. If f() = a n ( c) n has a nonero radius of convergence, then n= the coefficient a n = f (n) (c), i.e. f() equals its Taylor series n! n= f (n) (c) ( c) n. n! Proof. Apply theorem repeatedly (i.e. inductively) and evaluate at = c. Identity Theorem for Power Series. If f() = a n ( c) n =for = k c, (k =,,...) and { k } n= converges to c, then a n = for all n. Proof. We are given a power series f() =a +a ( c)+a ( c) +.Nowa = f(c) = lim f( k)=, k f() a = lim c c = lim f( k ) k k c =,..., a f() n = lim c ( c) = lim f( k ) =,... n k ( k c) n Uniqueness Theorem for Power Series. If a n ( c) n = b n ( c) n for = k c, (k =,,...) n= and { k } converges to c, then a n = b n for all n. Proof. onsider f() = (a n b n )( c) n and apply the identity theorem for power series. n= n= Exercises. Find lim x n, where n ( ) (a) x n = n sin ; (b) x n = cos nπ n 4 ; (c) x n =+( ) n n n +. 3

14 . Find the radius of convergence for the following power series: (a) n! ; n= (b) (n + n ) n. n= 3. Give an example of two power series respectively, such that the power series a n n and n= b n n with radii of convergence R and R, n= (a n + b n ) n has a radius of convergence greater than R +R. n= 4. Show that if <, then n =. Find a power series for with centered at. What ( ) n= is the radius of convergence for this series? 5. Explain why there is no power series f() = f () >. 6. Does there exist a power series f() =,, 3,...? 7. If f() = c n n satisfies f n= f (n) (), n=,, 3... c n n such that f n= c n n such that f n= ( ) = for k =, 3, 4,... and k ( ) = k ( k and f ) = k k 3 for k = ( ) = k k k,k =,, 3,..., compute the values of the derivatives + 4

15 6. auchy Riemann Equations f( + h) f( ) Recall f is differentiable at means lim = f ( ) exists and f is holomorphic at h,h h means f is differentiable in a neighborhood of in which case f (w) exists in some disk B(,r) about. Question. Given a function f: R R with two real variables, say f(x, y) =(u(x, y),v(x, y)) = u(x, y)+ iv(x, y), how can we tell if it is a differentiable function of = x + iy? Example. If f() =, = x + iy, then f(x, y) =(x + iy) =(x y )+i(xy) =(x y, xy). real real {}}{{}}{ Theorem. If f() = f(x, y) =( u(x, y), v(x, y)) = u(x, y) +iv(x, y) =u() +iv() is differentiable at, then u x, u y, v x, v y exist at, u x ( )= v y ( ) and u y ( )= v x ( ). [Notation. If we define f x = f x = u x + i v x and f y = f y = u equations u x = v y and u y = v x are equivalent to f y = if x.] Proof. Taking h along the real axis, we have y + i v, then the above auchy-riemann y f f(x + iy + h) f(x + iy ) f(x + h, y ) f(x,y ) ( ) = lim = lim = f x (x,y ). h h h h Taking h along the imaginary axis, say h = it, we have f f(x + iy + it) f(x + iy ) ( ) = lim = t it i lim f(x,y + t) f(x,y ) = t t i f y(x,y ). Theorem. If f x and f y exist in a neighborhood of, are continuous at and f y ( )=if x ( ), then f is differentiable at. Proof. Write f( )=u(x,y )+iv(x,y ),h= s + it. Then u( + h) u( ) h = u(x + s, y + t) u(x,y ) s + it = u(x + s, y + t) u(x + s, y ) = (by the mean value theorem). Similarly, v( + h) v( ) h t s + it s + it [ ] u y (x + s, y + αt) + u(x + s, y ) u(x,y ) s + it + s [ ] u s + it x (x + βt,y ), <α,β< = t [ ] v s + it y (x + s, y + γt) + s [ ] v s + it x (x + δt,y ), <γ,δ<. 5

16 Now, f y ( )=if x ( ) implies f x ( )= t s + it f y( )+ t s + it, s s + it, it follows that as h = s + it, [ f( + h) f( ) f x ( ) h = t u s + it y (x + s, y + αt)+i v + s s + it s s + it f x( ). Using these equations and the fact ] y (x + s, y + γt) f y ( ) [ u x (x + βt,y )+i v x (x + δt,y ) f x ( )] u y (x + s, y + αt)+i v y (x + s, y + γt) f y ( ) + u x (x + βt,y )+i v x (x + δt,y ) f x ( ). So f is differentiable at and f ( )=f x ( ). Examples. () If a holomorphic function G() = u()+ iv() = u(x, y)+ iv(x, y) has the real part u(x, y) = x y, what can v() be? Solution. v x = u y =y v(x, y) =xy + (y) v y = u x =x v(x, y) =xy + (x) v(x, y) =xy + constant () If f = u + iv is holomorphic on a region D and u constant, show that f constant. Solution. v x = u y = v(x, y) (y) v y = u v(x, y) constant f constant x = v(x, y) = (x) (3) If f is holomorphic on a region D and f() constant, then f constant. Solution. If f, then f. Otherwise u +v constant, so taking partial derivatives, we get u u x + v v v u, u u +v. Since x y y y = v x, v y = u ( u, we get = v ), v ( = u ). x x y x y Then u constant, v constant, so f constant. Examples () and (3) are special cases of the open mapping theorem, to be proved later. Exercises. If f = u + iv is holomorphic on some domain, given u(x, y) below, find the possibilities of v(x, y): (a) u(x, y) =x 3 3xy ; (b) u(x, y) =e y cos x; (c) u(x, y) = ln(x + y ); (d) u(x, y) = y ( x) + y.. Show that there is no holomorphic function f = u + iv with u(x, y) =x + y. 3. Suppose f is an entire function of the form f(x, y) = u(x) + iv(y). Prove that f is a polynomial of degree at most one. 6

17 4. Suppose that f and f are holomorphic on a region D. Show that f is a constant function on D. 5. Let G be a region and G = { : G} is the mirror image of G across the x-axis. If f : G is holomorphic, show that f : G defined by f () =f() is holomorphic. 6. Write in polar coordinates. Then f() = u() + iv() = u(r, θ)+ iv(r, θ). Establish the polar form of the auchy-riemann equations u r = v v and r θ r = u r θ. 7

18 7. Definitions of ommon Functions In this section, we will enlarge our collection of complex-valued functions by extending the common functions, such as exponential, logarithm and trigonometric functions, to complex domains. To define e, we want it to satisfy () e w+ = e w e, () e x is the same as before for x real, and (3) e is entire. Suppose f(w + ) =f(w)f() and f(x) =e x for x real. Then f() =f(x + iy) =e x f(iy) =e x (A(y) + ib(y)). For f to be entire (differentiable everywhere), the auchy-riemann equations require f y = if x, i.e. e x (A (y) +ib (y)) = ie x (A(y) +ib(y)). Then A (y) = B(y) and B (y) =A(y). So A (y) = A(y) and B(y) = A (y). Now f() =, so A() = and B() =. Therefore, A(y) = cos y and B(y) = sin y. Definition. For = x + iy, e = e x (cos y + i sin y) =e x cis y. (In particular, e iy = cis y.) In passing, we mention the famous Euler s equation e iπ + =, which relates five of the most important constants e, i, π,, in mathematics. Properties. () e w+ = e w e () e = e x (3) e for all (4) d d e = e Since e iy = cos y + i sin y, e iy = cos y i sin y, so cos y = eiy + e iy use these identities to define trigonometric functions of complex numbers. and sin y = eiy e iy. We will i Definitions. cos = ei + e i, sin = ei e i, tan = sin i cos, sec = cos, csc = cos,cot = sin sin. The derivatives of these functions are the same as before. Also, cos + sin for all and the usual trigonometric identities are still valid. However, it is not true that cos and sin for all, e.g. cos i = e + e >. Now we turn our attention to defining logarithm. This is done by inverting exponentiation. Definitions. If U, V are open sets, f: U V is one-to-one and onto, then the inverse of f is f : V U defined by f (v) =u whenever f(u) =v. Also we say g is the inverse of f at V if g is the inverse of f in some neighborhood of in V. onsider the equation e = w with w. Let = x + iy be a solution of e = w. Then e = e x cis y = w e x = w,y = arg w =ln w + i arg w. However, arg w is multivalued. Let Arg w denote the so-called principal branch of argument, which is defined by π < Arg w π. The general solution of e = w is =ln w + i(arg w +kπ), where k is any integer. Definitions. For w, log w =ln w + i arg w, where arg w = Arg w +kπ, k any integer. The choice Log w =ln w + i Arg w is the principal branch of the logarithm function. To make log w continuous, we need to choose arg w continuously. 8

19 Examples. () On the annulus A = {w: < w < }, arg w cannot be defined continuously. This can be seen by looking at the values of arg w on the circle w = 3. At w = 3, arg w =kπ. As w moves on the circle w = 3 counterclockwise, the value of arg w increases. When it get back to w = 3 at the end, the value of arg w have increased by π from the starting kπ. So arg w cannot be made continuously. () For the region on the left, arg w can be defined continuously. We can define arg w by choosing arg = Arg = and for any other w in the region, draw a polygonal path joining to w and define arg w continuously along the path by starting at. It is visually clear that arg w will be well-defined (even if different paths are used) and continuous. It is bad to simply define arg w by taking the principal branch Arg w because the region wraps around the origin a little more than once. On the word branch If arg w is defined continuously on a region, then arg w +kπ (for any integer k) is also continuous on the same region and gives also the argument of points on the region. We refer to both of these functions as different branches of the argument function. Similarly, if log is defined continuously on a domain, then log +πi is also a continuous inverse of e (since e log +πi = = e log ). We simply say that both are different branches of the logarithm function. To differentiate the logarithm function, we will apply the inverse rule dw / d d = dw. Inverse Rule. Let g be the inverse of f at and g be continuous near. If f is differentiable at g( ) and if f (g( )), then g is differentiable at,g ( )= f (g( )). Proof. Since f(g()) = in a neighborhood of, and g() g( )as in that neighborhood, we have g() g( ) = f(g()) f(g( )) f (g( )) by the existence of f (g( )). g() g( ) From the theorem, we see that if arg w can be defined continuously on a region, then log w will be continuous and even differentiable. Just take g() = log, f() =e, then d d log = g () = e log =. Definitions. For α, define α β = e (β log α) and the principal value of α β is e (β Log α). Example. i i = e (i log i) = e i(ln i +i arg i) = e i(+i( π +kπ)) = e π kπ (which is amaingly all real-valued!). The principal value of i i is e π. Remarks. () d d α = d d e( log α) = α log α (for each fixed value of log α.) () If arg can be defined continuously on a region, then d d α = d d e(α log ) = e (α log ) α = αα. Example. We will show that, on Ω = \ ((, ] [, + )), there is a continuous branch of. Observe that =( ) / = e log( ) = e (ln +i arg( )). Since onω, ln is continuous on Ω. Next we will consider arg( ) as the composition of w = f() = on Ω with g(w) = arg w on f(ω), where the argument is to be chosen later. Now f is clearly continuous on Ω and f(ω) = \[, + ). Next we will choose < arg w<π on f(ω) so as to make 9

20 g continuous on f(ω). Therefore, arg( ) = g (f()) is continuous on Ω, log( ) is continuous on Ω and is continuous on Ω. Exercises. Find all = x + iy such that e =. Use this to find all the roots of cos and sin.. Prove that sin = sin cos for all complex number. 3. Discuss ( if it ) is possible to define log( ) continuously on \[, ]. Also, discuss the possibility for + log to be continuously defined on \[, ].

21 8. onformal Mappings Definitions. () f is conformal at iff f is holomorphic in a neighborhood of and f ( ). () f is a conformal mapping from a region U onto a region V iff f is conformal at each point of U, one-to-one on U and onto V. In that case, we also say U and V are conformally equivalent. Observation. If f is conformal at, then f is angle preserving (in direction and magnitude). θ γ ( t ) γ ( t ) (For γ(t) =(x(t),y(t)),γ (t) =(x (t),y (t)). In the figure, we have tan θ = dy dx = y (t) x (t). So, θ = tan y (t) x (t) = arg γ (t). Suppose γ (t )=γ (t )=. Then the angle between γ,γ at is given by arg γ (t ) arg γ (t ). θ γ γ ( t ) w=f ( ) γ f γ f γ θ So, angle between f γ,f γ at w = f( ) = arg(f γ ) (t ) arg(f γ ) (t ) = arg(f ( )γ (t )) arg(f ( )γ (t )) = arg f ( ) + arg γ (t ) arg f ( ) arg γ (t ) = angle between γ,γ at.) The observation has an interesting interpretation: if you lives at and f is a typhoon that blows your home from to f( ), then you would not know that you have been moved because all the streets near your home intersect at the same angles as before! Examples. [It will be convenient to treat lines as circles of infinite radius and a circle will mean either a line or a circle.] () w = a + b, a is a conformal mapping from to. Writing a = Re iα,b = u + iv), we see that a + b = R(e iα )+u + iv rotates (with respect to the origin) by angle α, then expands (or contracts) by a factor of R, then translate by (u, v). This mapping takes a line to a line and a circle to a circle, i.e. takes circles to circles. () w = is a conformal mapping of \{} onto \{}. The equation α(x + y )+βx+ γy + δ = (line if α =, circle if α ) is equivalent to α + D + D + δ =, where D = β iγ. Under the mapping it becomes δww + Dw + Dw + α =. So the map w = sends circles to circles. w=a+b θ. w θ r w= - θ r w

22 (3) w = α,α> is conformal at all. w= α ϕ θ r αϕ αθ w r α (4) w = Log is conformal at \(, ]. w= Log πi w θ R θi LogR (5) Möbius transformations (or linear fractional transformations or bilinear transformations) are functions of the form w = T () = a + b, where ad bc. (Note if ad bc =, then T () constant.) c + d Properties of Möbius Transformations. () (Basic Property) T () = a + b (ad bc ) is a one-to-one map from { } onto { }, where c + d { { a T ( ) = if c c, = T ( d c ) if c ; T () = d b if c = T ( ) if c = c + a,t(t ()) = = T (T ()); T ad bc d () = for. So for c =,T() is a conformal mapping from (c + d) onto and for c { c,t() is a conformal mapping from \ d } { a } onto \. c c () (Algebraic Property) If T () = a + b (a d b c ) and T () = a + b (a d b c ), c + d c + d then T T (x) =T (T ()) = (a a + b c ) +(a b + b d ) (c a + d c ) +(c b + d d ). ( )( ) ( ) a b (Observe that a b a a = + b c a b + b d. This shows that there is a group c d c d c a + d c c b + d d a d + b d homomorphism from the Möbius transformations to the two-by-two matrices.) ( ) a ad bc (3) (Geometric Property) T () = a + b c + d = c if c c c + d is a composition of map- if c = pings of the form A + B and. From examples () and (), we conclude that Möbius transformations take circles to circles. (4) (Analytic Property) A fixed point of T () is a point such that T ( )=. The only Möbius transformation having more than two fixed points is the identity mapping. (Proof. T () = a + b c + d = c +(d a) b =, which has at most two roots, unless c =, d a =,b =.) Hence, if two Möbius transformations take the same values on three points of { }, then they must be identical.

23 (Proof. S( )=T ( ), S( )=T ( ), S( 3 )=T ( 3 ) S T has three fixed points T ( ), T ( ), T ( 3 ) S T is the identity mapping S() T ().) Definitions. For distinct, 3, 4 { }, let S, 3, 4 () = / 3 3. (Then S, 4 3, 4 is a Möbius 4 transformation that takes the circle through, 3, 4 to the real axis because S, 3, 4 sends, 3, 4 to,,, respectively.) The cross ratio of,, 3, 4 is (,, 3, 4 )=S, 3, 4 ( )= 3 4. (If one of,, 3, is, then we take limit to get the cross ratio.) Properties of ross Ratio. ( w) () (,,, ) = lim (,,,w) = lim w w ( w) =. () (,, 3, 4 )=(,, 3, 4 ). (3) (T,T,T 3,T 4 )=(,, 3, 4 ) for any Möbius transformation T. T (Proof. S, 3, 4 T : T 3. So by the holomorphic property, S, 3, 4 T = S T,T 3,T 4. Hence, T 4 evaluating both sides at T ( ), we get S, 3, 4 ( )=S T,T 3,T 4 (T ).) (4),, 3, 4 are distinct points on a circle if and only if (,, 3, 4 ) is real. (This is because S, 3, 4 takes the circle through, 3, 4 to the circle through,,, which is the real axis.) Below we shall show that cross ratio is a useful device that allows us to transform geometrical informations (such as, symmetry of points and inside/outside of circles) into algebraic quantities for computation and derivation of important principles. Γ c R b w w* Definition. For a given circle Γ: c = R, points w, w { } are symmetric with respect to Γiff(w c)(w c) =R. This is denoted by w Γ w. Equivalently,w Γ w w = R + c. (Observe w c that c, w, w are collinear because w R c = (w c). Also, w c w c R = R w c implies cbw =9. ) In the case Γ is a line, w is taken to be the mirror image of w with respect to Γ. Properties of Symmetric Points. () The line ww is orthogonal to Γ (because c, w, w are collinear). () A point is the center of Γ if and only if it is symmetric to with respect to Γ, i.e. w = c w =. (3) w Γ w for any distinct, 3, 4 on Γ, (w,, 3, 4 )=(w,, 3, 4 ). R (Proof. onsider the Möbius transformation T () = + c, then T ( c w c + c) =w. Since on Γ implies ( c)( c) =R, it follows that T ( )=. Similarly, T ( 3 )= 3 and T ( 4 )= 4. ( ) Ifw Γ w, then (w,, 3, 4 )=(T(w ),T( ),T( 3 ),T( 4 )) = (w,, 3, 4 ). ( ) We have (w,, 3, 4 )=(w,, 3, 4 )=(T ( R w c + c),t( ),T( 3 ),T( 4 )) = ( R w c + c,, 3, 4 ). 3 R

24 Since S, 3, 4 is one-to-one, the definition of cross ratio implies w = R w c + c, i.e. w Γ w.) Möbius transformations send pairs of symmetric points to pairs of symmetric points as the following shows. Symmetry Principle. w Γ w T (w) T (Γ) T (w ) for every Möbius transformation T. Proof. This follows easily by property (3) above because (w,, 3, 4 )=(w,, 3, 4 ) if and only if (T (w ),T( ),T( 3 ),T( 4 )) = (T (w),t( ),T( 3 ),T( 4 )). Möbius transformations need not send the inside of circles to the inside of the image circles (e.g. consider w = inside the unit circle). However, there is an orientation principle which says Möbius transformations do send the right sides of oriented circles to the right sides of their image circles. Definition. An orientation of a circle Γ is an ordered triple points (, 3, 4 ) on Γ. By property (4) of cross ratio, Γ = {:Im(,, 3, 4 )=}. Now, the right side of Γ with respect to (, 3, 4 ) (i.e. the side the right hand touches as one walks on Γ from to 3 to 4 ) is given by the set R, 3, 4 Γ={:Im(,, 3, 4 ) > }. To see this, take the case (, 3, 4 ) is counterclockwise, then the right side is the outside of Γ, which includes. Since S, 3, 4 () =(,, 3, 4 ) is one-to-one and onto and sends Γ to the real axis, it must send the outside of Γ onto either the upper or the lower half plane because of connectivity. (To be precise, suppose p,p are outside Γ such that S, 3, 4 sends them to q,q, one on the upper half plane and the other on the lower half plane. Take a (continuous) curve joining p and p without intersecting Γ, then the image of the curve under S, 3, 4 will not intersect the real axis. However, all curves joining q and q must intersect the real axis, which yields a contradiction.) By definition, R, 3, 4 Γ is the side of Γ that is sent to the upper half plane. We intend to show, for the counterclockwise case, that this is the outside of Γ. For this, it suffices to see R, 3, 4 Γ. o θ 4 3 Γ Now observe that (,, 3, 4 )= 4 = 4 = Re iθ, <θ<π.so, 3 3 Im(,, 3, 4 ) >, which implies R, 3, 4 Γ. The clockwise case is similar. Also, the case Γ is a line can be checked, e.g. R,, R = {:Im>} is the upper half plane) Similarly, the left side of Γ with respect to (, 3, 4 ) is given by L, 3, 4 Γ={:Im(,, 3, 4 ) < }. Orientation Principle. For an orientation (, 3, 4 ) on a circle Γ and T a Möbius transformation, we have T (R, 3, 4 Γ) = R T (),T ( 3),T ( 4)T (Γ) and T (L, 3, 4 Γ) = L T (),T ( 3),T ( 4)T (Γ). In particular, T must send each side of Γ onto a side of T Γ. Proof. R, 3, 4 Γ Im(,, 3, 4 )=Im(T,T,T 3,T 4 ) > T R T,T 3,T 4 T (Γ). Similarly, L, 3, 4 Γ T L T,T 3,T 4 T (Γ). ( ) a Theorem. For real θ, a <, T () =e iθ is a conformal map of the open unit disk D onto itself. a (Later we will show these are the only conformal maps from D onto D.) Proof. Observe that T takes the unit circle D to itself because for =, T () = a eiθ a = a a = =. Since T (a) =, the orientation principle implies T maps D onto D. a a 4

25 In an advanced course, a theorem called the Riemann Mapping Theorem is usually proved, which asserts that for every simply connected region Ω (i.e. Ω has no holes in its interior), there is a conformal mapping f from the open unit disk D onto Ω. That is, every proper simply connected region is conformally equivalent to D. The mapping between D and Ω is called a Riemann mapping. Remarks. By composing with another conformal mapping from D onto D, we see that Riemann mappings are not unique. In fact, if there is one Riemann mapping, there are infinitely many. However, if the values f() Ω and f () > are prescribed, then there is a second half of the Riemann mapping theorem, which asserts that there is exactly one Riemann mapping satisfying these additional conditions. Later on we will prove many theorems about the open unit disk or functions on it. Because of the Riemann mapping theorem, they then can be viewed as theorems about arbitrary proper simply connected regions or functions on them. So, the open unit disk is a canonical object in complex analysis. Let us compute some conformal mappings between D and some regions. Examples (),(),(3),(4) and (6) are Riemmann mappings between the open unit disk D and some common regions. Example (8) shows an interesting application of the symmetry principle and the orientation principle in conformal mappings. Examples. i - D T o UHP Upper Half Plane () T () =S i,, () =i T () = i i +. ( ), + D D S S - R RHP Right Half Plane \ (-oo,] () S() = T () = i +, S () = +. (3) R() =(S()) = ( ). + D Q st Quadrant (4) Q() = ( ) T () = i. + (the principal square root) - i - i P P ϑ i D w T - (5) P () =i ( ),P: + i, i P (w) = i w.(c.f. Example ()) i + w (6) ϑ() =T (P () )=T ( = ( ( + ) i + ) i = i +i + i + ( ) ) + 5

26 N (7) N() =P ( ) ( ( = ) ) ( +) +( ) = ( ). - i P( ) -i - - Γ Γ w. ẇ* (8) Let U be the interior of the annulus bounded by Γ : + =9, Γ : +6 = 6. Find a conformal map from U onto a concentric annulus V centred at the origin. Key. Find a pair of points w and w symmetric with respect to Γ and Γ. (Such a pair must be collinear with the centers, so they must be real.) For such a pair, (w + )(w + ) = 8, (w + 6)(w + 6) = 56. Solving we get w =,w = 6 (or converse). onsider T () =. This is a Möbius transformation, hence takes circles to circles and right 6 sides to right sides. Since w = Γ,Γ w =6, the symmetry principle implies T (w) = T Γ,T Γ T (w )=. It follows T (w) = is the center of T Γ,TΓ. Since T (8) = 3 and T () =, the orientation principle implies that T sends the inside of Γ to the inside of T Γ and the outside of Γ to the outside of T Γ. So, V = { w: 3 < w < }. Exercises. Find conformal mappings from the open unit disk D = { : < } onto the following regions: (a) the infinite strip { :< Im <}; i i (b) the upper semidisk {: < and Im >}. - [Hint: Find a map from st quadrant onto the upper half semidisk.] (c) the slit disk D \ [, ) (d) the finite-slit plane { } \ [, ] - [Hint: Find a map from \(, ] to ( { } \ [, ].] 6

27 . What is the range of e if we take to lie in the infinite strip Im < π? What are the images of the horiontal lines and vertical segments in Im < π under the e mapping? Give an example of a holomorphic function from the open unit disk D onto \{} with derivative never equal to ero. 3. Find r such that there is a one-to-one conformal mapping from { :Im< and +i > } onto {w : r< w < }. 4. Let a<band T () = ia ib. Define L = {:Im = b}, L = {:Im = a},l 3 = {:Re =}. Determine which of the regions A, B,, D, E, F in Figure, are mapped by T onto the regions U, V, W, X, Y, Z in Figure. A B L 3 ib D L E ia F L [Hint: Orient L 3 by (, ia, ib). ] T Y X Z Figure Figure U V W 7

28 9. ontour Integrals Definition. If γ:[a, b] is continuous and is given by γ(t) =u(t)+iv(t), then we define b γ(t) dt = b u(t) dt + i b a a a v(t) dt. Example. For γ(t) =e it, t π, π γ(t) dt = π π cos tdt+ i sin tdt=+i. Definitions. A continuous curve :[a, b], (t) = x(t) + iy(t), is piecewise smooth iff except for finitely many points, x (t) and y (t) are continuous on [a, b] and (t) =x (t)+iy (t) on[a, b]. The curve is closed iff (a) =(b). A closed curve is simple iff the function is one-to-one on [a, b), i.e. it has no self-intersection. Henceforth all curves will be assumed piecewise smooth and have finite lengths. Definitions. If is a smooth curve given by :[a, b] and f is continuous on the curve, then we b b define f() d = f((t)) (t) dt and f() d = f((t)) (t) dt. a Examples. () Let be the unit circle given by (t) =e it ( t π). Now f() = is continuous on. π So d = e it (ieit ) dt =πi and π d = dt =π. () Let be the curve (t) =+it ( t ) and f() =. Then f() d = ( + it ) (it dt) = (it 4t 3 8it 5 ) dt =(it t 4 4i 3 t6 ) = i 3. Properties of ontour Integrals. () If, given by :[a, b], :[c, d] are smoothly equivalent (in the sense that = λ, where λ:[c, d] [a, b] is one-to-one with continuous derivative and λ(c) =a, λ(d) =b), then by the change of variable s = λ(t), we have () (3) f() d = t=d t=c a s=b f( (t)) (t) dt = f( (s)) (s) ds = f() d. f() d = f() d, where is given by :[a, b], (t) =(a + b t). [f() +g()] d = f() d + g() d, (4) Fundamental Theorem of alculus. s=a αf() d = α f() d. f () d = f((b)) f((a)), where is given by :[a, b]. 8

29 losed urve Theorem. If is a closed curve and f has an antiderivative (say F ()) in a region containing, then f() d = F () d = F ((b)) F ((a)) =. (In particular, f() d = f() d for curves, that have the same initial and terminal points because followed by is a closed curve.) orollary. If is a closed curve, then P () d = for every polynomial P (). (5) M-L Inequality. If f() M for every on and has length L, then b b f() d = f((t)) (t) dt f((t)) (t) dt = f() d M a a b a (t) dt = ML. (6) If f n () converges uniformly to f() on, then lim f n () d = f() d. n (Proof. Let L be the length of. For any ε>, because f n () converges uniformly to f() on, there is N such that n N f n () f() <εfor all on. Then for n N, f n () d f() d = (f n () f()) d εl.). Find Exercises d, where is a smooth curve from to not passing through the origin. f (). Suppose f() is holomorphic and f() < in a region G. Show that d = for every f() closed curve in G, assuming f is continuous. 3. (a) Find d, where is the simple closed curve that goes from to along the real axis, then goes in the clockwise direction on the unit circle to, then goes from to along the real axis and finally goes back to in the counterclockwise direction on the circle =. (b) Find d, where the unit circle = is given the counterclockwise orientation. = 4. Evaluate the integral [Hint: = + + d, where is the curve shown below ] i i If Re a and Re b, show that e a e b a b. 9

30 6. Suppose f is holomorphic on B(,r) and Re f () > for all B(,r). Show that f is one-to-one. [Hint: Express f(a) f(b) as an integral.] 7. If f is a continuous real-valued function and f(), show that [Hint: Let I be the value of the integral, then I = I e iθ.] = π f() d sin t dt =4. 3

31 . auchy Theory auchy theory deals with the integration of holomorphic functions and its consequences as developed by Augustine auchy. First, we deal with the local auchy theory. The word local refers to the situation in a neighborhood of a point. We begin with a fundamental property that leads to the whole development. Definition. (Nonstandard Terminology!) A continuous function f has the rectangle property in a region Ω iff f() d = for every rectangle Γ such that Γ and its inside are contained in Ω and its edges are parallel Γ to the real or imaginary axes. Such a rectangle will be referred to as a rectangle inω. Rectangle Theorem (or Goursat s Theorem). If f is holomorphic on a region Ω, then f has the rectangle property in Ω. Integral Theorem. If f has the rectangle property in a disk B(c, R) ( < R ), then f has an antiderivative there. auchy s Theorem on Disks. If f is holomorphic on a disk B(c, R), then f() d = for every closed curve in B(c, R) (by the rectangle, integral and closed curve theorems.) Γ Proof of the Rectangle Theorem. Let Γ be a rectangle in Ω. Let s be the length of the larger side of Γ. Suppose f() d = I. Divide the Γ interior of Γ into four congruent subrectangular regions with boundaries Γ, Ω s 4 Γ,Γ 3,Γ 4. Since f() d = f() d, one of the integral must satisfy Γ j= Γ j f() d Γ j I 4. Let us define Γ() =Γ j. Repeat this subdivision to the interior of Γ () to obtain a Γ () and so on. We obtain a sequence of rectangles Γ (), Γ (), Γ (3),...in Ω such that f() d I. The rectangular regions bounded by the Γ 4k (k) Γ (k) s shrink to a point in Ω. Since f is holomorphic at,sog() = f() f( ) f ( ) as. So given ε>, there is δ> such that δ g() ε. Observe that the sides of Γ (k) are s s,so k k s for all in Γ (k). hoose k large so that k <δ, then I 4 k f() d = (f( )+f ( )( ) +g()( )) d = Γ (k) Γ }{{} g()( ) d. (k) Γ (k) polynomial ( ε ) s ( k 4 s ) k = 4 s ε 4 k I 4 s ε. 3

32 Since ε is arbitrary, f() d = I =. Γ a b Re b + i Im a Proof of the Integral Theorem. Without loss of generality, let c =. For w in the disk, define F (w) = f() d, where a,b denotes the curve from a to,w Re b + i Im a then to b. By the rectangle property, f() d = for the boundary Γ Γ of every rectangle inside the disk. We will show F = f. Fix w in the disk. For any ε>, since f is continuous at w, there is δ>such that w δ f() f(w) ε. For h δ, length of w,w+h h. Observe that f(w) = h w,w+h f(w) d. So, w+h w F (w + h) F (w) f(w) h = (f() f(w)) d h w,w+h ε h =ε. h O F (w + h) F (w) Therefore, lim = f(w). h h Occasionally, we come across some continuous functions that are almost holomorphic in their domains (with exceptions only on some thin subsets), it is remarkable that such functions still have the rectangle property on their domains, as the following theorem asserts. Extension Theorem. If f is continuous on a region D and holomorphic on D \ L, where L is a (horiontal) line segment (or in the extreme case, a point), then f has the rectangle property on D. y Γ upper lower Γ e ε e Γ x x Γ L L L Proof. Let Γ be a rectangle in D. There are three cases to consider. () If Γ and its interior do not contain any point of L, then Γ is a rectangle in D \ L and f() d = by the holomorphicity of f. Γ () If Γ has a (horiontal) edge e containing points of L, let Γ ε be the rectangle that has the same edges as Γ except the edge e is replaced by an edge e ε, which is ε unit from L on the same side as Γ. By (), f() d =. Now as ε, Γ ε the (uniform) continuity of f on e implies x x f() d = f(x + i(y + ε)) dx f(x + iy ) dx = f() d. e ε x x e Since = f() d f() d as ε, we get f() d =. Γ ε Γ Γ (3) If the interior or a vertical edge of Γ contains a point of L, then f() d = Γ f() d + Γ upper f() d =, where we used the (horiontal) line through Γ lower L to form two new rectangles and applied (). Now we turn to the (global) auchy theory, which deals with the integration of holomorphic functions on (closed) curves in (simply connected) domains. The curves need not be restricted to lie inside disks. Definition. A continuous function f defined on a region D has the polygon property iff f() d = for Γ every right-angled polygon Γ such that Γ and its inside are contained in D. 3