This is read as: Q is the set of all things of the form a/b where a and b are elements of Z and b is not equal to 0.
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- Amie Stokes
- 5 years ago
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1 1. Some nottion 1.1. Nottion from set theory: We wont worry bout wht set is. If S is set nd x is member of S, we write x S (red s x belongs to S ). If x is not member of S we write x / S. The empty set is the only set tht hs no elements. The empty set is denoted by. A set B is clled subset of set A, written B A, if x B implies x A. A set will be described in one of the two following mnners. for exmple we might sy: N = {1, 2, 3, }, or Z = {, 3, 2, 1, 0, 1, 2, 3, } The first sentence is red s: N equl to the collection 1, 2, 3, dot, dot, dot. Or we might sy: Q = { :, b Z, b 0} b This is red s: Q is the set of ll things of the form /b where nd b re elements of Z nd b is not equl to 0. defining new sets from old: Let A, B be sets. Then one hs the following sets: A B is the union of A nd B. One hs x A B, if nd only if x A or x B. A B is the intersection of A nd B. One hs x A B, if nd only if x A nd x B. A \ B is the set theoretic difference of A nd B (red s A minus B ). One hs x A \ B, if nd only if x A nd x / B. A B is the product of A nd B. The set A B consists of ll ordered pirs of the form (, b) where A nd b B. We write A 2 = A A, A 3 = A A A, etceter. If we hve sets A 1,, A n, we cn define their union n j=1a j = A 1 A n, intersection n j=1a j = A 1 A n nd product A 1 A 2 A n in similr fshion. The product A 1 A n consists of ll ordered tuples of the form ( 1,, n ), where 1 A 1,, n A n. More generlly, let J be set nd {A j : j J} be collection of sets, one for ech element of J. Then we cn define their union j J A j nd intersection j J A j. One hs x j J A j if nd only if x A j for some j J. One hs x j J A j, if nd only if x A j for ll j J Some more nottion: Let N be the set of nturl numbers, Z be the set of integers nd Q be the set of rtionl numbers mentioned bove. Let R be the set of rel numbers ( collection of numbers contining Q stisfying list of xioms, defined in clculus clss). Let C be the set of complex numbers (coming soon). Note tht N Z Q R. Note tht 1 Z, but 1 / N, 1/2 Q but 1/2 / Z, 2 R but 2 / Q. 1 So Z is strictly lrger thn N nd so on: N Z Q R. We write R 2 = R R, R 3 = R R R, etc. The set R d is clled the d-dimensionl Eucliden spce. An element of R d is n ordered tuple ( 1,, d ) where 1,, d re rel numbers. We reserve few more symbols for specific purpose: 1 From the xioms of rel number system one cn show tht there is unique positive rel number whose squre is equl to 2. This number is wht we denote by 2. It requires tricky little rgument to prove tht 2 does not belong to the set Q. In fct for long time the Greeks believed tht ll numbers (or mgnitudes) were rtionl, so finding out tht 2 / Q ws big surprise! 1
2 ϵ nd δ will lwy denote strictly positive rel numbers. The sme goes for ϵ 1, ϵ 2,, δ 1, δ 2,. r will denote rel number in [0, ) or r could be the forml symbol. We shll write r [0, ]. The sme goes for r 1, r 2,. N, n, m will lwys denote non-negtive integers. Sme goes for N 1, N 2,, n 1, n 2,, m 1, m 2, Functions: Let A nd B be two sets. A function or mp F from A to B is rule, which, given ny element of A, produces n element of B, denoted F (). We sy tht F () is the imge of the element under the mp F. If F () = b, we sy tht given input, the function F produces output b. If F is function from A to B, we write F : A B. Let A be ny set. The identity function on A is the function id A : A A defined by id A () = for ll A. If F : A B nd G : B C re functions then one cn define function G F : A C by (G F )(x) = G(F (x)). We sy tht G F is the composition of G nd F. Let F : A B. We sy tht A is the domin of F. Let S A. Then we define F (S) = {y B : y = f(x) for some x S}. So F (S) is subset of B. It is clled the imge of S under the function F. In prticulr F (A) is clled the imge of F. If T B, then we write F 1 (T ) = {x A: f(x) T }. So F 1 (T ) is subset of A. It is clled the preimge of T under the function F. Let F : A B. We sy tht F is invertible if there exists function G : B A such tht F G = id B nd G F = id A. If such G exists, then it is necessrily unique. (check this). In this cse we write G = F 1 nd sy tht G is the inverse of F. Let F : A B nd G : B A. Then G = F 1 if nd only if F = G 1. (check this). We sy tht F : A B is onto if F (A) = B. We sy tht F : A B is one to one if for ll x, y in A, F (x) = F (y) implies x = y. We sy tht F is bijection if F is one to one nd onto. A function F : A B is bijection if nd only if it is invertible. (check this). 2
3 2. The complex number system 2.1. Definition (complex numbers, rel nd imginry prt). A complex number z is n expression of the form z = + bi, where nd b re rel numbers. Sometimes we shll write z = + ib which mens the sme thing. In these notes, when we sy number, we men complex number, unless otherwise stted. If z = + ib, we define Re(z) = nd Im(z) = b. We sy tht is the rel prt of z nd b is the imginry prt of z. So two complex numbers re equl if their rel prts re equl nd their imginry prts re equl. The set of ll complex numbers will be denoted by C. For now, i is just forml symbol. We shll identify the rel numbers s subset of complex numbers by thinking of rel number c s the complex number c + 0.i. In prticulr 0 nd 1 re the complex numbers 0 = i nd 1 = i Definition (ddition, multipliction, multipliction, division). We define two opertions, clled ddition nd multipliction on the set of complex numbers s follows: ( + ib) + (c + id) = ( + c) + i(b + d) nd ( + ib).(c + id) = (c bd) + i(d + bc). In prticulr i 2 = 1. One checks esily tht these opertions stisfy the following rules (known s the field xioms ) z + w = w + z, z + (w + u) = (z + w) + u, z + 0 = z, for every complex number z = + ib there exists complex number, ( z) = ( ) + i( b) such tht z + ( z) = 0, z.w = w.z, z.(w.u) = (z.w).u, z.1 = z, for every non-zero complex number z = + ib there exists complex number, z 1 = ( ) + i( b 2 +b 2 such tht z.z 1 = 1, z.(u + w) = z.u + z.w. Now one cn define subtrction nd division of complex numbers in the uusl mnner. Let z, w be complex numbers. Define z w = z + ( w). If w 0, then define z/w = z.w b 2 ) 2.3. Definition (conjugte, norm). Let z = + ib is complex number. The complex number z = + i( b) is clled the conjugte of z. Observe tht z. z = 2 + b 2 is non-negtive rel number nd z. z = 0 if nd only if z = 0. The non-negtive rel number z z = 2 + b 2 is clled the norm of the complex number z. We denote z = z z 3
4 nd cll it the bsolute vlue of z. Some bsic observtions bout the conjugte nd norm re collected in the lemm below. The proof is left s n exercise Lemm. Let z nd w be complex numbers. Then () z = z, (b) Re(z) = (z + z)/2, Im(z) = (z z)/2i, (c) Re( z) = Re(z), Im( z) = Im(z), (d) (e) (f) zw = z w, zw = z w, z + w 2 = z 2 + w Re( zw) Remrk (on division). Note tht z 1 = z z 2 nd z w = z w w w = z w w 2 This lets us write the quotient of two complex numbers gin in the form +ib. For exmple, if z = 2 + 5i nd w = 1 + 2i then 2 + 5i 1 + 2i = z w = z w (2 + 5i)(1 2i) (2.1 5.( 2)) + i(2( 2) + 5.1) = = = 12 w i. 4
5 3. Geometry of complex numbers 3.1. Definition. A complex number z = +ib will be represented geometriclly s the point (, b) in the Eucliden plne R 2. So ddition of complex numbers correspond to the usul ddition on R 2 given by the prllelogrm lw: w z + w z 3.2. Definition (Polr representtion). Let z = + ib be complex number. Note tht z = 2 + b 2 (1) is the length of the segment joining 0 nd z. We sy z is the bsolute vlue of z. So the norm is the squre of the bsolute vlue. If z 0, let rg(z) be the ngle between the positive x-xis nd the ry tht strts t 0 nd ends t z. We sy rg(z) is the rgument of z. So tn(rg(z)) = b/ (2) nd z = z (cos(rg(z)) + i sin(rg(z))) (3) This is clled the polr representtion of the complex number z. For exmple, let z = i. Then z = (2 3) = 4, rg(z) = rctn( 2 2 ) = π, 3 6 nd the polr representtion of z is z = 4(cos π + i sin π ). (see the figure below): 6 6 z z Im(z) rg(z) Re(z) 3.3. Remrk. We mke some commments on the nture of the rg function without being completely precise nd without complete justifiction. Note tht the rgument is only well defined upto ddition of n integer multiple of 2π. For exmple, let z = 4(cos(π/3) + i sin(π/3)). Then we my sy tht rg(z) = π. 3 Equivlently we might sy tht rg(z) = π + 2π. The possible vlues of rg(z) for 3 this z re, π 4π, π 2π, π, π + 2π, π + 4π, There is no wy to define the rgument in continuous mnner on ll non-zero complex numbers. However if we delete ry originting t 0 then the rgument cn be well defined s continuous function on the complement. For exmple, if we delete the positive x-xis, then the rgument cn be defined s continuous function 5
6 on the complement, tking vlues in [0, 2π). If we delete the negtive x-xis, then the rgument cn be defined s continuous function on the complement, tking vlues in [ π, π). Finlly, observe tht if A is contined in connected nd simply connected region (roughly this mens connected region with no holes, see exmples below) in C not contining 0, then the rgument cn be defined s continuous function on A in n un-mbigous mnner. A 1 A 2 A 3 the shded regions A 1 nd A 2 re simply connected, but A 3 is not. These observtions re chrcteristic of wht is known s multi-vlued function. Frequent ppernce of multivlued functions is feture tht is is common in complex nlysis nd is different from rel nlysis. We should hndle them with cre Lemm. () Let z C nd c is positive rel number. Then cz = c z nd rg(cz) = rg(z) (b) A complex numbers hs bsolute vlue is 1, if nd only if it hs the form u = cos θ + i sin θ, where θ = rg(u). Proof. Prt () follows directly from the definition of bsolute vlue nd rgument in eq. (1) nd eq. (2). Prt (b) follows directly from the polr representtion, eq. (3) Lemm. () One hs (cos θ 1 + i sin θ 1 )(cos θ 2 + i sin θ 2 ) = cos(θ 1 + θ 2 ) + i sin(θ 1 + θ 2 ). (b)[de Moivre s foruml] One hs (cos θ + i sin θ) n = (cos nθ + i sin nθ). Proof. Prt () is direct clcultion. Prt (b) follows from prt (): (cos θ 1 + i sin θ 1 )(cos θ 2 + i sin θ 2 ) = (cos θ 1 cos θ 2 sin θ 1 sin θ 2 ) + i(sin θ 1 cos θ 2 + cos θ 1 sin θ 2 ) = cos(θ 1 + θ 2 ) + i sin(θ 1 + θ 2 ) Lemm. Let u 1 nd u 2 be two complex numbers of bsolute vlue 1. Then () u 1 u 2 = 1, nd (b) the ngles (rg(u 1 u 2 )) nd (rg(u 1 ) + rg(u 2 )) differ by n integer multiple of 2π. Proof. Prt () is lredy known 2.4(e). For prt (b), let Let θ 1 = rg(u 1 ), θ 2 = rg(u 2 ) nd θ = rg(u 1 u 2 ). Lemm 3.4(b) implies tht we my write Now lemm 3.5 implies u 1 = cos θ 1 + i sin θ 1, u 2 = cos θ 2 + i sin θ 2 nd u = cos θ + i sin θ. u = u 1 u 2 = cos(θ 1 + θ 2 ) + i sin(θ 1 + θ 2 ). 6
7 So cos(θ) = cos(θ 1 + θ 2 ) nd sin(θ) = sin(θ 1 + θ 2 ). Hence θ differs from θ 1 + θ 2 by integer multiple of 2π Remrk. The circle of rdius 1 centered t the origin will be clled the unit circle. So the complex numbers of bsolute vlue 1 re precisely the points on the unit circle. The lemm bove sys tht when we multiply two complex numbers on the unit circle, we get nother one on the unit circle whose rgument is equl to the sum of the rguments of the two numbers we strted with. We shll often be somewht less precise nd stte prt (b) of lemm (3.6) s rg(u 1 u 2 ) = rg(u 1 ) + rg(u 2 ). (4) However, this eqution should be used with some cution since rg is multivlued function. In other words, one hs to keep in mind tht eqution (4) sttes equlity of two geometric ngles nd not equlity of two numbers. For exmple, suppose we mke the convention (s is common) tht rg(u) [0, 2π). Suppose rg(u 1 ) = 3π/2 nd rg(u 2 ) = π. Then (4) implies tht rg(u) = π/ Lemm. Let z 1 nd z 2 be two complex numbers. Then () z 1 z 2 = z 1 z 2 (b) rg(z 1 + z 2 ) = rg(z 1 ) + rg(z 2 ) if z 1 0, z 2 0, z 1 + z 2 0. The second eqution comes with the sme cvet s (4). Proof. Prt () is lredy known from 2.4(e). Let r 1 = z 1 nd r 2 = z 2. Let u 1 = r1 1 z 1 nd u 2 = r2 1 z 2. Note tht u 1 = r1 1 z 1 = r1 1 z 1 = 1. Similrly u 2 = 1. Note tht z 1 z 2 = r 1 r 2 u 1 u 2. Lemm 3.4 implies rg(z 1 ) = rg(u 1 ), rg(z 2 ) = rg(u 2 ) nd rg(z 1 z 2 ) = rg(u 1 u 2 ). So, using (4), we get rg(z 1 z 2 ) = rg(u 1 u 2 ) = rg(u 1 ) + rg(u 2 ) = rg(z 1 ) + rg(z 2 ) Remrk. Let z 1 nd z 2 be two complex numbers. Let θ 1 = rg(z 1 ) nd θ 2 = rg(z 2 ) so tht z 1 nd z 2 hve polr representtion z 1 = z 1 (cos θ 1 + i sin θ 1 ) nd z 2 = z 2 (cos θ 2 + i sin θ 2 ). Lemm 3.8 implies tht z 1 z 2 hs the polr representtion z 1 z 2 = z 1 z 2 (cos(θ 1 + θ 2 ) + i sin(θ 1 + θ 2 )). So lemm 3.8 mkes it esy to visulize multipliction of complex numbers in polr representtion: 7
8 When we multiply two complex numbers the mgnitudes get multiplied nd the rguments (or ngles with the x-xis) gets dded up (see figure below). z z = z 1 z 2 z 2 θ = θ 1 + θ 2 θ θ 2 z 1 θ Lemm (inequlities). Let z, w, u C. Then: () Re(z) z, Im(z) z. (b) (c) (d) z + w z + w. z w ( z w ). z u z w + w u. Sometimes (b) nd (d) re clled tringle inequlity. Proof. Prt () is cler from z = Re(z) 2 + Im(z) 2. From prt () we get Re( zw) zw = z w. Now prt (b) follows from lemm 2.4(f); since z + w 2 = z 2 + w Re( zw) z 2 + w z w = ( z + w ) 2. The inequlity (b) implies tht z w + w (z w) + w = z So z w z w. Similrly one gets z w = w z w z, so z w lies between z w nd its negtive. This is prt (c). Prt (d) follows from (b) by replcing z, w with (z w) nd (w u) respectively. 8
9 4. Point set topology for R d 4.1. Definition. The bsolute vlue of = ( 1,, d ) R d, denoted = d is the distnce between nd 0 R d. If nd b R d, then b is the Eucliden distnce between nd b. Let B r () = {x R d : x < r}. The set B r () is clled the open bll of rdius r centered t. If d = 1, then B r () is the open intervl ( r, + r) of length 2r. If d = 2, then B r () is the open disc of rdius r round Definition (bounded nd unbounded sets). A subset S R d is clled bounded if there exists rel number R > 0 such tht S B R (0), or in other words, z < R for ll z S. A subset S R d is unbounded if it is not bounded Definition. (Open nd closed sets, limit points) U R d is clled open if for ll w U, there exists r > 0 such tht B r (w) U. A subset C R d is clled closed if R d \ C is n open set. Let U R d. A point p R d is clled limit point of U if B ϵ (p) \ {p} intersects U for ll ϵ > Exercise. () Show tht B r () = {x R d : x < r} is n open set. (b) Show tht {x R d : x r} is closed set Exercise. () Show tht rbitry union of open sets is open. Show tht finite intersection of open sets is open. () Show tht rbitry intersection of closed sets is closed. Show tht finite union of closed sets is closed Definition (limit of sequence). Let ( 1, 2, 3, ) be sequence in R d. Sometimes we denote sequence by writing { n } n=1 or simply s { n }. We sy tht R d is the limit of the sequence { n }, if given ny ϵ > 0, there exists nturl number N such tht n < ϵ for ll n > N. In this cse we write lim n =. n We sy tht sequence converges if it hs limit Definition (Cuchy sequence). A sequence { n } in R d is clled Cuchy sequence if given ny ϵ > 0 there exists nturl number N such tht for ll m, n > N, one hs m n < ϵ Remrk. We recll two bsic fcts from Clculus of rel vribles. (1) Suppose { n } is sequence of rel numbers. Assume tht n s re non-decresing, tht is, n s re bounded bove, tht is, there is some rel number M such tht j < M for ll j. Then { n } converges. (2) A sequence { n } converges if nd only if it is Cuchy. Both these properties re equivlent to the lest upper bound xiom for rel numbers, which is the most importnt property of the rel number system. 9
10 4.9. Exercise (Completeness of R d ). A sequence in R d converges if nd only if it is Cuchy. Proof. Let 1, 2,, n, be sequence in R d. Write n = ( 1 n, 2 n,, d n). Then So for ech j between 1 nd d we hve m n 2 = ( 1 m 1 n) ( d m d n) 2. j m j n m n. So if { n } is Cuchy sequence in R d, then { j n} n=1 is Cuchy sequence in R. So lim n j n exists (see 4.8(2)). lim n j n = j. It follows tht { n } converges to ( 1,, d ). So every Cuchy sequence in R d converges. Conversely, suppose n s n. Fix ϵ > 0. Then there exists N such tht n < ϵ/2 for ll n > N. Now, if m, n > N, then So every convergent sequence is Cuchy. m n m + n < ϵ/2 + ϵ/2 = ϵ Exercise. Let U R d nd p R d. Show tht the following re equivlent: () p is limit point of U. (b) Any disc round p contins infinitely mny elements of U. (c) There is sequence in U \ {p} tht converges to p Definition (closure). Let S be subset of R d. The closure of S, denoted by S, is the intersection of ll the closed sets contining S. So it is the smllest closed set contining S. Exercise.() Check tht the closure of S is the union of S nd ll its limit points. (b) Check tht {x R d : x r} is the closure of B r () Theorem (nested intervl theorem). Let [ 1, b 1 ] [ 2, b 2 ] be nested sequence of closed intervls in R. Then the intersection of these intervls l=1 [ l, b l ] is non-empty. If lim l (b l l ) = 0, then the intersection consists of just one point. sketch of proof. Since [ n+1, b n+1 ] [ n, b n ] for ll n, one esily sees tht n b m b 3 b 2 b 1. Now the sequence { n } is incresing nd bounded bove by b 1, so lim n n = p exists nd p n for ll n. Fix m. Since b m n for ll n, we hve b m p. So p [ m, b m ] for ech m. If p nd p belong to [ m, b m ] then p p (b m m ). So if (b m m ) 0 s m then p p = Theorem (nested rectngle theorem). For l = 1, 2,, let R l = [ l, b l ] [c l, d l ] be closed rectngles in R 2 such tht R 1 R 2. Then their intersection l=1 R l is non-empty. If the size of the rectngles tend to zero, i.e, lim l (b l l ) = 0 nd lim l (d l c l ) = 0, then l=1 R l consists of just one point. 2 sketch of proof. Since R 1 R 2, we get [ 1, b 1 ] [ 2, b 2 ] nd [c 1, d 1 ] [c 2, d 2 ]. Applying the nested intervl theorem twice we get p l [ l, b l ] nd q l [c l, d l ]. Then (p, q) l R l. 2 Ofcourse similr sttement holds in R d in generl. The sme proof works. 10
11 4.14. Theorem (Bolzno-Weierstrss theorem). Let A be n infinite bounded subset of R d. Then A hs limit point. Proof. We write the proof for R 2. The generl proof is similr nd should be cler from this. Let R be closed rectngle in R 2. Divide R into four smller closed rectngles by drwing two lines tht join the midpoints of opposite edges of R. We cll these smller rectngles the four qurters of R. Key observtion: If rectngle R contins infinitely mny elements of A, then one of the qurters of R must contin infinitely mny elements of A. Let R 1 = [, b] [c, d] be closed rectngle tht contin A. Pick one of the qurters of R 1 such tht it contins infinitely mny elements of A; cll it R 2 3. Next pick one of the qurters of R 2 tht contins infinitely mny elements of A; cll it R 3. Continuing in this mnner we get sequence of closed rectngles R 1 R 2 R 3 such tht ech of these rectngles contin infinitely mny elements of A. One possible choice of R 1, R 2,, R 5. The rectngle R 2 is shded drker thn R 1, R 3 is drker thn R 2, nd so on. Let M be the length of the digonl of the rectngle R 1. Note tht the length of the digonl of the rectngle R n is M/2 n 1, which goes to 0 s n. The nested rectngle theorem implies tht there is unique c R 2 which belong to ech rectngle R n. Let B ϵ (c) be ny disc round c. Choose n lrge, so tht M/2 n 1 < ϵ. Then R n B ϵ (c). Since R n contins infinitely mny elements of A, so does B ϵ (c). So c is limit point of A Definition. Let K be subset of R d. A collection of non-empty open sets {U j : j J} is clled n open cover of K if j J U j contins K. Let {U j : j J} be n open cover of K. We sy tht this open cover hs finite subcover if there re finitely mny U j s, sy U j1,, U jr such tht K U j1 U jr. In this cse we sy {U j1,, U jr } forms finite subcover of the open cover {U j : j J}. A subset K of R d is clled compct if every open cover of K hs finite subcover Lemm (Closed subset of compct is compct). Suppose K S R d such tht K is closed nd S is compct. Then K is compct. Proof. Note tht U = R d \ K is open. Let {U j : j J} be n open cover of K. Then the U j s together with U form n open cover of S. So we cn find U j1,, U jm such tht S U U j1 U j2 U jm. But then K U j1 U j2 U jm Theorem (Heine Borel theorem). A subset of R d is compct if nd only if it is closed nd bounded. 3 If more thn one qurters contin infinitely mny elements of A then pick ny one of them. 11
12 Proof. compct sets re bounded: Suppose K is compct. For n N, let U n = B n (0). Then n=1u n = R d. So {U n : n N} is n open cover of K. It hs finite subcover, sy B n1 (0),, B nr (0). The lrgest one mong these blls contin K, so K is bounded. compct sets re closed: Suppose K is compct. It is enough to show tht R d \ K is open. Suppose p R d \ K. For n N, let U n be the outside of closed bll of rdius 1/n round p. So ech U n is open nd n=1u n = R d \{p}. So {U n : n N} is n open cover of K. So it hs finite subcover U n1,, U nr. Suppose n t is the lrgest number mong n 1,, n r. Then K U nt. So the open disc of rdius 1/n t round p is contined in R d \ K. closed nd bounded sets re compct: We write the proof for d = 2. The generl proof is similr. Let K be closed nd bounded subset of R 2. Since K is bounded, we cn find rectngle R = [, b] [c, d] such tht K R. By lemm 4.16 it suffices to show tht closed rectngles of the form R re compct. Suppose, if possible, {U j : j J} be n open cover of R such tht it does not hve ny finite sub-cover. Observtion: If rectngle cnnot be covered by finitely mny U j s then t lest one of the qurters of this rectngles cnnot be covered by finitely mny U j s either. We re ssuming tht R 1 = R cnnot be covered by finitely mny U j s. So we cn pick qurter R 2 of R 1 such tht R 2 cnnot be covered by finitely mny U j s either. By repeting this rgument we get sequence of rectngles R 1 R 2 R 3 such tht R n cnnot be covered by finite mny U j s for ny n. By the nested rectngle theorem, we cn pick point p, which belongs to ech R n. Pick U j such tht p U j. Since U j is open, there exists ϵ > 0, such tht B ϵ (p) U j. Let M be the length of the digonl of the rectngle R 1. Then the length of the digonl of R n is M/2 n 1. Choose n lrge so tht M/2 n 1 < ϵ. Then R n B ϵ (p) U j. So R n is covered by single open set U j, which is contrdiction Theorem. Let K be non-empty subset of R d. The following re equivlent: () K is closed nd bounded. (b) Every sequence in K hs convergent subsequence tht converge to some point in K. sketch of proof. Suppose K is not bounded. Then there is sequnce { n } in K such tht n. Such sequence hs no convergent subsequence. Suppose K is not closed. Then there is limit point p of K such tht p / K implies tht there is sequence { n } in K such tht n p. But then ny sub-sequence of { n } lso converge to p, which is outside K. Thus (b) implies (). From Bolzno Weierstrss theorem nd 4.10 it follows tht () implies (b). 12
13 5. Limit nd continuity 5.1. Definition (limit of function). Let S R d. Let f be function from S to R m. Let be limit point of S. We sy tht lim f(z) = l z if, given ny ϵ > 0, there exists δ > 0, such tht for ll z S such tht 0 < z < δ we hve f(z) l < ϵ. If lim z f(z) = l, we sy tht the limit of f(z) is l s z tends to or we sy tht sy tht f(z) converges to l s z tends to, written s f(z) l s z Definition (continuous function). Let S R d nd f : S R m be function. Let S. We sy tht f is continuous t if either is limit point of S nd f(z) f() s z. or is not limit point of S. We sy tht f : S R m is continuous function if f is continuous t for ll S Exercise. Let S R d nd f : S R m be function. Let S. Show tht f is continuous t if nd only if given ny open bll B round f(x), there exists n open bll B round x such tht f(b S) B. We shll often use this chrcteriztion of continuity Lemm. Let S R d nd let f : S R m. Then the following re equivlent: () f is continuous on S. (b) If U R d is open, then f 1 (U) = V S for some open set V R d. Proof. Assume (). Let U be n open subset of R d. If f 1 (U) = we hve nothing to prove. Otherwise, tke ny x f 1 (U). Since U is open nd f(x) U, there exists some open bll B x round f(x) such tht B x U. Since f is continuous t x, there exists some bll B x round x such tht such tht f(b x S) B x. Let V = x f 1 (U)B x. Then V is open, f 1 (U) V nd f(v S) U. So f 1 (U) = V S. Thus () implies (b). The converse is left s n exercise Lemm (Computing limit long sequence or pth). Let S R d. Let f : S R m. Let be limit point of S. () The following re equivlent: (i) lim z f(z) = l. (ii) If { n } is ny sequence lying in S such tht n, then f( n ) l. (b) Suppose lim z f(z) = l. If γ : (, b) S is ny continuous function such tht γ(t) p s t b, then lim t b f(γ(t)) = l Theorem (Continuous imge of compct set is compct). Let K R d is compct nd f : K R m continuous. Then f(k) is compct. Proof. Let {U j : j J} be n open cover of f(k). Since f is continuous, there re open sets V j such tht f 1 (U j ) = V j K. Now {V j : j J} is n open cover of K. so there exists finite subcover V j1,, V jr, tht is, K V j1 V jk. Then f(k) U j1 U jk Theorem (extreme vlue theorem). Let K R d compct nd f : K R continuous. Then there exists p K such tht f(p) f(z) for ll z K. 13
14 Proof. Since f(k) is compct it is closed nd bounded. Since f(k) is bounded, M = sup f(k) exists. Since M is limit point of f(k) nd f(k) closed M f(k). So there is p K such tht f(p) = M. So f(p) f(z) for ll z K Definition. Let A R d. Let f : A R n be continuous function. Sy tht A is uniformly continuous on A if given ny ϵ > 0, there exists δ > 0 such tht whenever x, y A nd x y < δ, then f(x) f(y) < ϵ Lemm. Let K be compct subset of R d. Let {δ x : x K} be collection of strictly positive numbers indexed by K. Then there exists x 1,, x m K such tht K B δx1 (x 1 ) B δx m (x m). Proof. The blls {B δx (x): x K} form n open cover of K. Tke finite subcover Theorem (Continuous functions on compct sets re uniformly continuous). Let K R d compct nd f : K R m continuous. Then f is uniformly continuous. Proof. Fix ϵ > 0. Since f is continuous on K, for every x K, there exists δ x > 0, such tht if y K nd y x < 2δ x, then f(y) f(x) < ϵ/2. By 5.9, we cn find x 1,, x m K such tht K B δx1 (x 1 ) B δxm (x m ). Let δ = min{δ x1, δ xm }. Now let x, y K such tht x y < δ. There exists i between 1 nd m such tht x x i < δ xi. So y x i y x + x x i δ + δ xi 2δ xi. So f(x) f(x i ) < ϵ/2 nd f(y) f(x i ) < ϵ/2 nd hence f(x) f(y) < ϵ Definition. Let C R d. The subset x C B ϵ (x) will be clled the ϵ-neighbourhood of C Theorem. Let U be n open subset of R d. Let K be compct subset of U. Then there exists ϵ > 0 such tht the ϵ-neighbourhood of K is contined in U. Proof. Since U is open, for every x K, we cn pick ϵ x > 0, such tht B 2ϵx (x) U. By 5.9, we cn find x 1,, x m K such tht K B ϵx1 (x 1 ) B ϵxm (x m ). Let ϵ = min{ϵ x1,, ϵ xm }. Suppose x K nd y B ϵ (x). Then there is some i between 1 nd m such tht x x i < ϵ xi. So y x i y x + x x i ϵ+ϵ xi < 2ϵ xi. So y B 2ϵi (x i ) U Lemm. Let U be n open subset of R d. Let γ : [, b] U be continuous function. Then there exists ϵ > 0 such tht the ϵ-neighbourhood of γ([, b]) is contined in U nd there eixsts prtition = 0 < 1 < < n = b of [, b] such tht γ([ 0, 1 ]) B ϵ (γ( 0 )), γ([ n 1, n ]) B ϵ (γ( n )) nd γ([ i 1, i+1 ]) B ϵ (γ( i )) for i = 1, 2,, n 1. Proof. Since γ[, b] is compct, there is ϵ > 0 such tht B ϵ (x) U for ll x γ[, b]. Since γ is uniformly continuous, we cn find δ > 0 such tht t 1 t 2 < δ implies γ(t 1 ) γ(t 2 ) < ϵ. Now choose ny prtition = 0 < 1 < < n = b such tht i i+1 < δ. The lemm follows. 14
15 6. Complex nlytic functions: first properties 6.1. Definition. Let U C be n open set. Let f : U C be function. Let U. Sy f(+w) f() tht f is differentible t if lim w 0 exists. If this limit exists we write w f () = lim w 0 f( + w) f() w = lim z f(z) f() z nd sy tht f () is the derivtive of f t. We sy tht f : U C is differentible function if f is differentible t every U. The lemms below give us some exmples of differentible functions Lemm. Define f : C C by f(z) = z n, where n 0 is n integer. Then f is differentible nd f (z) = nz n 1. Proof. Let C. We know, if z, then z n n z = zn 1 + z n z n 3 + n 1. As z, ech term of the bove sum tends to n 1. So lim z z n n z = n n Lemm. Let U be open subset of C nd f, g : U C re two complex functions. Let U. () If f nd g re differentible t, then so is (f ± g) nd (f ± g) () = f () ± g (). (b) If f nd g re differentible t, then so is fg nd (fg) () = f ()g() + f()g (). (c) If f nd g re differentible t nd g() 0, then (f/g) is differentible t nd (f/g) () = f ()g() f()g () g() Definition. A polynomil (of one complex vrible) if function f : C C of the form p(z) = c 0 + c 1 z + + c n z n. We sy tht n = deg(p). The quotient of two polynomils is clled rtionl function Definition. Let S C. Sy tht S is pth connected if given ny two points x, y in S, there exists continuous function γ : [0, 1] S such tht γ(0) = x nd γ(1) = y Lemm. Let U be n open subset of C. Let f : U C be differentible function. Suppose f (z) = 0 for ll z U. () If B is is ny open disc contined in U, then f is constnt on B. (b) If U is open nd pth connected, then f is constnt on U. Proof. () Let f = u + iv. If f (z) = 0, then the prtil derivtives of u nd v re 0, so u nd v re constnt long horizontl nd verticl stright line segments. Any two points in B cn be joined by horzontl nd verticl stright line segments stying inside B. So u nd v re constnt on B. Hence so is f. (b) follows from prt (). 15 (5)
16 6.7. Lemm (Chin rule). Suppse U, V re open sets in C. Let f : U C nd g : V C. Let U such tht f is differentible t, f() V nd g is differentible t f(). Then g f is differentible t nd (g f) () = g (f())f () Remrk. If p(z) nd q(z) 0 re polynomils, then the domin of rtionl function p(z)/q(z) is the set of ll z such tht q(z) 0 4 Since z n is differentible for n 0 nd sums products nd quotients of differentible functions re differentible, we find tht ll rtionl functions p(z)/q(z) re differentible on their domin. This gives our first clss of exmples of differentible functions. More interesting exmples re often constructed by some limiting process, for exmple, using infinite sums nd products The Cuchy-Riemnn equtions. Let U C be n open set. Let f : U C be complex function. Let u = Re(f) nd v = Im(f). Then u, v : U R. We think of u nd v re two rel vlued functions of two rel vribles. So f(x + iy) = u(x, y) + iv(x, y). Suppose f is differentible t = 1 + i 2. Becuse of lemm 5.5, the limit in (5) cn be clculted s z pproches 0 long ny line. For exmple, we cn pproch 0 long the x-xis, tht is, we let w = t where t R nd let t 0. We get f f( + t) f() () = lim t 0 t f( 1 + t, 2 ) f( 1, 2 ) = lim t 0 t = f x (). Alterntively we cn pproch 0 long the imginry xis, tht is, we let z = it, where t R nd let t 0. We get f () = lim t 0 f( + it) f() it = 1 i lim f( 1, 2 + t) f( 1, 2 ) t 0 t So if f is differentible t then it stisfies the Cuchy-Riemnn eqution: f x = i f y = i f y () If u nd v re the rel nd imginry prt of f (s bove), then substituting f = u + iv in (6) nd equting rel nd imginry prts, we get u x = v y nd (6) u y = v x. (7) Remrk. Let f(z) = z. Write f = u + iv, so u(x, y) = x nd v(x, y) = y. Observe tht u nd v do not stisfy (7). So f(z) = z is not complex differentible function. More generlly we cn mke the following sttement which is somewht forml but gives useful wy to think bout complex differentible functions. For this, define the following differentil opertors on functions defined on n open subset of the plne: z = 1 2( x i y ) nd z = 1 2( x + i y 4 The fundmentl theorem of lgebr sys tht if q(z) hs tmost deg(q) mny distinct roots. So the rtionl function p(z)/q(z) is defined except t finintely mny points 16 )
17 To gin some intuition bout this nottion observe tht z z = 1 2( x i ) (x + iy) = 1 ( x y ) i2 + i ( y y 2 x y 2 x + x ) = 1 y Similrly one verifies tht z z = 1, z z = 0, z z = 0. A complex function defined on the plne is function of the vribles x nd y where z = x+iy. Using x = (z + z)/2 nd y = (z z)/2i we my formlly think of f s function of z nd z. The Cuchy-Riemnn eqution (6) now trnsltes into f z = 0. So f is nlytic if it is function of only z nd not of z! 17
18 7. Liner Frctionl trnsformtions 7.1. Definition (stereogrphic projection). Let S 2 = {(x 1, x 2, x 3 ) R 3 : x x x 2 3 = 1} be the unit sphere in R 3. Let (0, 0, 1) be the north pole. If p is point on the unit sphere S 2 other thn the north pole, then the stright line joining the north pole nd p intersects the (x 1, x 2 ) plne t unique point. Cll this point π(p). p π(p) Thus we get mp π : S 2 \ {(0, 0, 1)} C where we hve identified the (x 1, x 2 ) plne with C by (x 1, x 2, 0) = x 1 + ix 2. Suppose π(x 1, x 2, x 3 ) = z = x + iy. Since (x 1, x 2, x 3 ) is on the line joining (0, 0, 1) nd (x, y, 0), we hve (x 1, x 2, x 3 ) = t(x, y, 0) + (1 t)(0, 0, 1) = (tx, ty, 1 t) for some t R. It follows tht x = x 1 /(1 x 3 ) nd y = x 2 /(1 x 3 ). So π(x 1, x 2, x 3 ) = x 1 + ix 2 1 x 3. The mp π is clerly invertible. We clculte the inverse explicitly. Since (x 1, x 2, x 3 ) S 2, we hve 1 = x x x 2 3 = t 2 x 2 + t 2 y 2 + (1 t) 2, which implies Since (x 1, x 2, x 3 ) (0, 0, 1), we hve t 0, so Thus we obtin the formul 2t = t 2 (x 2 + y 2 + 1) = t 2 ( z 2 + 1). π 1 (z) = t = 2 z ( 2 Re(z) z 2 + 1, 2 Im(z) z 2 + 1, z 2 1 ). (8) z Clerly both π nd π 1 re continuous. The mp π is clled the stereogrphic projection Lemm. Let C be circle on S 2. Show tht if C does not pss through the north pole then π(c) is circle in C nd if C psses through the north pole then π(c \ {(0, 0, 1)}) is line in C. Conversely ech circle nd line in C is the imge of some circle in S 2 under stereogrphic projection. 18
19 sketch of proof. Let C be circle on S 2. Then C = S 2 P where P is some plne in R 3. Suppose the plne P hs eqution x + by + cz + d = 0. The point z C corresponds to the point π 1 (z) S 2. Eq. (8) implies tht π 1 (z) P if nd only if 2 Re(z) z b 2 Im(z) z c z 2 1 z d = 0, or in other words (c + d) z Re(z) + 2b Im(z) + (d c) = 0. (9) This is the eqution of generl line or circle in the complex plne (Homework). Further, the plne P psses through the north pole if nd only if c+d = 0, which is just the condition for (9) to describe line in C Definition (limit t infinity). Let S be n unbounded subset of C nd f : S C. We sy lim z f(z) = L if given ϵ > 0, there exists M > 0, such tht whenever z S nd z > M, we hve f(z) L < ϵ Exercise. Suppose f : C C be function. Then f π is function from S 2 \{(0, 0, 1)} to C. Show tht f π extends to continuous function f : S 2 C if nd only if lim z f(z) exists nd in tht cse f((0, 0, 1)) = lim z f(z) Definition (the Riemnn sphere). We shll sometimes identify C with S 2 \ {(0, 0, 1)} vi the mp π. Motivted by the previous exercise, we introduce the nottion = (0, 0, 1). So we shll write S 2 = C { }. The set C { } is known s the extended complex plne or the Riemnn sphere Definition. Let g = ( b c d ) be 2 2 mtrix whose entries re complex number such tht det(g) = d bc 0. Given such mtix g we define function ϕ g (z) = z + b cz + d. Note tht the function ϕ g is defined everywhere on C except for the point d/c if c 0. These functions re clled liner frctionl trnsformtions Remrk. The function ϕ g extends nturlly to function ϕ g : S 2 S 2 s follows: If c 0, then we define /c if z = ϕ g (z) = if z = d/c (z + b)/(cz + d) otherwise. If c = 0 then we define ϕ g (z) = { (z + b)/d if z = otherwise Exercise. Check tht the extended function ϕ g : S 2 S 2 is continuous nd it hs continuous inverse ϕ g Exercise. Check tht ϕ g1 ϕ g2 = ϕ g1 g 2 for ny two invertible 2 2 mtrices g 1 nd g 2. The next couple of results helps one in picturing liner frctionl trnsformtion. 19
20 7.10. Exercise. () A liner frctionl trnsformtion ϕ tkes circles in S 2 to circles in S 2. In other words ϕ tkes lines nd circles in C to lines nd circles in C. (b) Suppose L be line or circle in C. Suppose H 1 nd H 2 be the two components of C\L. Then ϕ(h 1 ) nd ϕ(h 2 ) re the two components of C \ ϕ(l) Lemm. () Let ϕ be liner frctionl trnsformtion such tht ϕ(0) = 0, ϕ(1) = 1 nd ϕ( ) =. Then ϕ is the identity mp. (b) Given ny three distint points z 0, z 1, z 2 S 2 there exists unique liner frctionl trnsformtion ϕ such tht ϕ(0) = z 0, ϕ(1) = z 1 nd ϕ( ) = z 2. (c) If liner frctionl trnsformtion ϕ fixes three distinct points in S 2, then it is identity. (d) Suppose (z 0, z 1, z 2 ) re three distinct points in S 2. Suppose (z 0, z 1, z 2) re lso three distinct points in S 2. Then there exists unique liner frctionl trnsformtion ϕ such tht ϕ(z j ) = z j for j = 0, 1, Exmple (Cyley trnsformtion). Consider the Liner trnsformtion ϕ(z) = (z i)/(z + i). Note tht if z R then ϕ(z) hs bsolute vlue 1, so ϕ(z) is on the unit circle. So ϕ tkes the x-xis to the unit circle. So the two hlf spces bove nd below the x-xis must mp onto the inside nd the outside of the unit circle. Note tht ϕ(i) = 0, so ϕ mps the upper hlf plne onto the open unit disc. So the lower hlf plne must mp to the outside of the unit circle. 20
21 8. Power series Before strting on series plese red 4.6 through 4.9 on sequences. Pge This prt ws edited in the lst updte Definition. An infinite series is n expression of the form n = n + (10) n=1 where { n } is sequence of complex numbers, clled the terms of the series (10). Given { n }, define new sequence {S n } by S 1 = 1, S 2 = 1 + 2,. S n = n,. The numbers S 1,, S n, re clled prtil sums of the series (10). We sy tht series (10) converges, or tht it is summble if the sequence {S n } of prtil sums converges. Suppose the sequence {S n } converges nd let S = lim n S n. In this cse we sy tht the sum of the infinite series ( n + ) is equl to S nd we write n = S. n=1 If the sequence of prtil sums {S n } does not converge, then we sy tht the series (10) diverges or tht the series is not-summble. We shll often use the following obvious observtion without mention: A series ( ) converges if nd only if the series ( N + N+1 + N+2 ) converges for some nturl number N. In other words, dding or deleting finitely mny terms does not chnge the convergence or divergence of series Lemm (Cuchy condition). Let n=1 n be series. Then the following re equivlent () The series n=1 n converges. (b) Given ny ϵ > 0, there exists nturl number N such tht m+1 + m n < ϵ for ll n > m N. Proof. Let S n = n. Then the condition in prt (b) simply sys tht {S n } is Cuchy sequence, which is equivlent to {S n } being convergent sequence. (see 4.8(2)) Definition. A series n=1 n is sid to be bsolutely convergent if the series n=1 n of rel numbers is convergent Lemm. If series is bsolutely convergent, then it is convergent. Proof. Suppose the series n=1 n converges. Then by Cuchy condition (() implies (b)), given ny ϵ > 0, there exists N such tht m+1 + m n < ϵ for ll n > m N. 21
22 But then tringle inequlity gives m+1 + m n < m+1 + m n < ϵ for ll n > m N. So pplying the Cuchy condition ((b) implies ()) we see tht the series n=1 n converges. We shll stte few tests to check convergence of series Lemm (comprison test). Let n=1 n be series of complex numbers. Suppose {M n } be sequence of non-negtive rel numbers. Suppose n M n for ll n. If n=1 M n converges then n n is converges bsolutely; in prticulr n n is converges. Proof. Let M = n=1 M n. Since M j s re ll non-negtive, we hve (M M k ) M for ll k. Let S k = k. Then {S k } is non-decresing sequence nd S k (M M k ) M. So the sequence {S k } converges (see 4.8(1)). In other words, n=1 n converges Lemm (rtio test). Let { n } be sequence such tht n 0 for ll n nd lim n+1 n n = L exists. If L < 1, then the series n=1 n converges bsolutely. If L > 1, then the series n=1 n diverges. Sketch of proof. Suppose lim n n+1 / n = L < 1. Pick positive rel number M such tht L < M < 1. Then there exists N such tht n+1 / n < M for ll n N (Why?). So So the terms of the series re bounded by the terms of the series N+k = N N+1 N+2 N N+1 N+k N+k 1 N M k. ( N + N+1 + N N+k + ) ( N + N M + N M N M k + ). The geometric series (1 + M + M 2 + ) converges since 0 < M < 1; in fct the sum of this series is 1/(1 M). So comprison test implies tht the series ( N + N+1 + N+2 + ) converges. So the series ( ) converges too Lemm (root test). Let { n } be sequence such tht lim n n 1/n = L exists. If L < 1, then the series n=1 n converges bsolutely. If L > 1, then the series n=1 n diverges Definition (sequence of functions). Let U C. Let f n : U C be functions for n = 1, 2,. Let f : U C be function. We sy tht the sequence of functions {f n } converge (pointwise) to function f : U C, if lim f n(z) = f(z) for ll z U. n 8.9. Remrk. The notion of point-wise convergence of sequence of functions is too wek for mny purposes. For exmple, even in rel vrible clculus, there exists sequences of functions {f n } defined on closed intervl of R such tht ech f n is infinitely differentible, 22
23 nd f n (x) f(x) for ech x in the closed intervl, but f(x) is not even continuous. For exmple, let f n : [0, 1] R be given by f n (x) = x n. So if we wnt the limit function f to inherit some nice property tht ech of the functions f n possess, then we need the functions f n to converge to f in stronger fshion. This stronger notion is known s uniform convergence. It turns out tht the right notion is to require tht the distnce between the functions f n nd f is uniformly smll for lrge n. This discussion is mde precise by the couple of definitions tht follow Definition. Let f, g : U C be two functions. Define the distnce between f nd g to be by f g = sup{ f(z) g(z) : z U} Definition (Uniform convergence). Let U C. Let f n : U C be functions for n = 1, 2,. We sy tht the sequence of function f n converges uniformly to function f : U C if given ny ϵ > 0, there exists nturl number N such tht f n f < ϵ for for ll n > N Theorem (Uniform limit fo continuous functions is continuous). Let U C. Let f n : U C be continuous functions for n = 1, 2,. Suppose {f n } converges uniformly on U. If z U, let f(z) = lim n f n (z). Then f : U C is continous Definition (series of functions). Let U C. Let f n : U C be functions for n = 1, 2,. An infinite series of functions is n expression of the form f n = f 1 + f f n + n=1 We define the prtil sums S 1, S 2, of the series s before: S n = f 1 + f f n. We sy tht the the series of functions n f n is converges (pointwise) if the sequence {S n (z)} converges for ech z U. In this cse we cn define new function S : U C by S(z) = f n (z) = lim S n (z) n n=1 Sy tht n f n converges uniformly if the sequence of prtil sums {S n } converges uniformly. The following test is extremely useful to check uniform convergence of series of functions Theorem (Weierstrss m-test). Let U C. Let f n : U C be functions for n = 1, 2,. Let {M n } be sequence of non-negtive rel numbers such tht n M n converges. Suppose f n (z) M n for ll z U nd ll n 1. Then n f n converges bsolutely nd uniformly on U. Proof. Since n M n converges, comprison test tells us tht for ech fixed z U, the series n=1 f n(z) converges bsolutely. Define f : U C by f(z) = n f n(z). Let S k (z) = k n=1 f n(z). Since M n > 0 nd n M n converges, given ϵ > 0, choose r such tht (M r+1 + M r+2 + ) < ϵ. Then f(z) S r (z) = f r+1 (z)+f r+2 (z)+ f r+1 (z) + f r+2 (z) + M r+1 +M r+2 + ϵ So the prtil sums {S k (z)} converges uniformly to f(z). 23
24 8.15. Definition (power series). Let z 0 C. A power series centered t z 0 is series of functions of the form c n (z z 0 ) n = c 0 + c 1 (z z 0 ) + c 2 (z z 0 ) c n (z z 0 ) n + n where c 0, c 1, c 2, re numbers Theorem. () Let n c n(z z 0 ) n be power series. Then there exists unique r [0, ] such tht n c n(z z 0 ) n converges bsolutely for ll z with z z 0 < r nd n c n(z z 0 ) n diverges for ll z with z z 0 > r. 5 (b) If r is positive number strictly less tht r, then n c n(z z 0 ) n convereges uniformly in the disc B r (z 0 ). Proof. It suffices to prove the theorem under the ssumption z 0 = 0. The generl cse then follows immeditely. (why?). () Clim: Suppose the power series n c nz n converges for z =. Then the power series n c nz n converges bsolutely for every z in the open disc of rdius round 0. Proof of clim: Since n c n n converges, lim n c n n = 0, so lim n c n n = 0. So we cn choose rel number M > 0 such tht c n n M for ll n 1. Now suppose w is some complex number such tht w <. Then c n w n = c n n w n M w n. Since w/ < 1, the geometric series n w n converges. So by comprison test n c nw n converges bsolutely. This proves the clim. Let r be the supremum of the bsolute vlues of ll the complex numbers for which the power series n c nz n converges. Then the clim shows tht the power series converges bsolutely for z < r nd diverges for z > r. (b) Let r < r. Pick such tht r < < r. If w B r (0) then, s bove c n w n M. r n. Let M n = M. r n. Then n M n converges. So prt (b) follows from Weierstrss m-test Definition. The r tht ppers in the bove lemm is clled the rdius of convergence of the power series n c n(z z 0 ) n. Suppose n c n(z z 0 ) n is power series hving rdius of convergence r. Then the sum of this series defines function f : B r (z 0 ) C. f(z) = c n (z z 0 ) n = c 0 + c 1 (z z 0 ) + c 2 (z z 0 ) c n (z z 0 ) n + (11) n=0 Formlly differentiting the power series f(z) term-by-term we obtin the power series nc n (z z 0 ) n 1 = c 1 + 2c 2 (z z 0 ) + 3c 3 (z z 0 ) 3 + (12) n=1 Formlly integrting the power series f(z) term by term we obtin the power series c n n + 1 (z z 0) n+1 = c 0 z + c 1 2 (z z 0) 2 + c 2 3 (z z 0) 3 + (13) n=0 5 Allowing r = is just convenient nottionl device. If r =, then the sentence simply sys tht the power series bsolutely converges for ll z C. 24
25 8.18. Theorem. () The power series (12) nd (13) hs the sme rdius of convergence s f given in (11). (b) Suppose r is the rdius of convergence of the power series f. Then f is nlytic on B r (U). The derivtive of f is given by the power series (12). Proof. () As before, we my ssume z 0 = 0. Let r be the rdius of convergence of (11). Let z be complex number such tht z < r. Choose w such tht z < w < r. Then n c nw n converges. Note tht since z/w < 1, lim n n z/w n = 0. So we my choose N such tht n z/w n < 1 for ll n N. So for every n > N we hve nc n z n = n z w n c n w n c n w n. So n nc nz n converges by comprison test with n c nw n. So n nc nz n 1 converges for ll z with z < r. It follows tht the rdius of convergence of the power series obtined by differentiting term by term is tlest s big s the rdius of convergence of the originl power series. But the reverse inequlity holds obviously. So (12) nd (11) hve the sme rdius of convergence. The rgument for the series (13) is similr. (b) Without loss ssume z 0 = 0. Let r be the rdius of convergence of f. Let z, w be two points in B r (0). Pick K < r such tht z < K nd w < K. Then from prt () we know tht the rdius of convergence of (12) is lso r. so g(w) = n nc nw n 1 converges. We need to show tht f (z) = g(z). Now we estimte Now f(z) f(w) z w zn w n z w g(z) = n c n ( z n w n z w ) nwn 1 c n zn w n z w nwn 1 n 1 nwn 1 = (z j w n j 1 w n 1 ) n 1 w n j 1 (z j w j ) j=0 The formul z j w j = (z w) j 1 t=0 zt w j t 1 gives us the estimte z j w j jk j 1 z w. So zn w n z w n 1 nwn 1 K n j 1 jk j 1 z w = j=0 n j=0 n(n 1) z w K n 2. 2 So f(z) f(w) g(z) z w n(n 1) c n K n 2. z w 2 n The power series n n(n 1)c nz n 2 is obtined from n c nz n by twice differentiting term by term. So prt () implies tht n(n 1) n c 2 n z n 2 lso hs rdius of convergence equl to r. Since K < r the sum n c n n(n 1) K n 2 converges. Let M be the sum of this series. Then 2 f(z) f(w) g(z) M z w z w Tking limit s z w, we see tht f (z) = g(z) Exercise. Let f(z) = n c n(z z 0 ) n be power series hving rdius of convergence r. Suppose lim n c n 1 n exists. Show tht 1 r = lim n c n 1 n. 25
26 8.20. Remrk. If you know bout lim sup then there is more generl formul known s the Cuchy-Hdmrd theorem: It sys tht the rdius of convergence of power series n c n(z z 0 ) n is given by ( ) 1. lim sup c n 1 n n 26
27 9. Complex exponentil nd trigonometric functions 9.1. Lemm. The power series z n n=0 converges for ll z C, tht is, its rdius of n! convergence is equl to. Proof. Fix ny z C. Let n = z n /n!. Then n+1 n = zn+1 /(n+1)! = z z n /n! n+1 0 s n. So the rtio test implies tht the power series converges bsolutely for ll complex number Definition. Since the rdius of convergence of the power series z n n=0 converges for ll n! z C is, theorem 8.18 implies tht the sum of this power series defines n entire function. This function is clled the (complex) exponentil function nd is denoted by exp(z): exp(z) = n=0 n=0 z n n! = 1 + z + z2 2 + z3 6 + z In similr mnner one defines the (complex) trigonometric functions sin nd cos s the sum of the following infiinte series: ( 1) n z 2n+1 sin(z) = = z z3 (2n + 1)! 3! + z5 5! z7 7! + cos(z) = n=0 ( 1) n z 2n (2n)! = 1 z2 2! + z4 4! z6 6! + In mnner similr to the exponentil series one cn verify tht the power series on the right converges for ll z. So sin(z) nd cos(z) re entire functions Remrk. If z hppens to be rel number then the bove definition of sin, cos nd exp mtches with the usul definition in rel vrible clculus. One cn verify this, for exmple, by clculting the power series expnsion of the rel sine nd cosine function by using Tylor series Definition (complex power of positive rel number). Let c be strictly positive rel number nd z C. Define c z = exp(z log c). Recll tht there is trnscendentl number e (the bse of nturl logrithm) such tht log e = 1. It follows tht e z = exp(z) Lemm. One hs de z dz = d sin z d cos z ez, = cos z, = sin z. dz dz Proof implies tht the derivtive of power series cn be clculted by differentiting the power series term by term Lemm. One hs e z+w = e z e w for ll z, w C. In ptriculr e z e z = e 0 = 1, so e z 0 for ll z C. 27
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