METRIC SPACES AND COMPLEX ANALYSIS. 35

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1 METRIC SPACES AND COMPLEX ANALYSIS The Complex Plne: topology nd geometry. For the rest of the course we will study functions on C the complex plne, focusing on those which stisfy the complex nlogue of differentibility. We will thus need the notions of convergence nd limits which C possesses becuse it is metric spce (in fct normed vector spce). In this regrd, the complex plne is just R 2 nd we hve seen tht there re number of norms on R 2 which give us the sme notion of convergence (nd open sets). The dditionl structure of multipliction which we equip R 2 with when we view it s the complex plne however, mkes it nturl to prefer the Eucliden one z = (Re(z) 2 + Im(z) 2. More explicitly, if z = (, b) nd w = (c, d) re vectors in R 2, then we define their product to be z.w = (c bd, d + bc). It is stright-forwrd, though bit tedious, to check tht this defines n ssocitive, commuttive multipliction on R 2 such tht every non-zero element hs multiplictive inverse: if z = (, b) (0, 0) hs z 1 = (, b)/( 2 + b 2 ). The number (1, 0) is the multiplictive identity (nd so is denoted 1) while (0, 1) is denoted i (or j if you re n engineer) nd stisfies i 2 = 1. Since (1, 0) nd (0, 1) form bsis for R 2 we my write ny complex number z uniquely in the form + ib where, b R. We refer to nd b s the rel nd imginry prts of z, nd denote them by R(z) nd I(z) or Re(z) nd Im(z) respectively. Definition If z = (, b) we write z = (, b) for the complex conjugte of z. It is esy to check tht zw = z. w nd z + w = z + w. The Eucliden norm on R 2 is relted to the multipliction of complex numbers by the formul z = z z, which moreover mkes it cler tht zw = z w. (We cll such norm multiplictive). If z 0 then we will lso write rg(z) R/2πZ for the ngle z mkes with the positive hlf of the rel xis. Becuse subsets of the complex plne cn hve much richer structure thn subsets of the rel line, the topologicl mteril we developped in the first hlf of the course will be indespensible in understnding complex differentible functions. We will need the notions of completeness, compctness, nd connectedness, long with the bsic notions of open nd closed sets. Definition A connected open subset D of the complex plne will be clled domin. As we hve lredy seen, n open set in C is connected if nd only if it is pth-connected. We will lso use the nottions of closure, interior nd boundry of subset of the complex plne. The dimeter dim(x) of set X is sup{ z w : z, w X}. A set is bounded if nd only if it hs finite dimeter. Recll tht the Heine-Borel theorem in the cse of R 2 ensures tht subset X C is compct (tht is, every open covering hs finite subcover) if nd only if it is closed nd bounded Circles nd lines. Lemm A line L is the complex plne cn be described s the locus {z C : I(z) = b} where = 1 nd 0 rg() < π, nd b R. A circle C my

2 36 KEVIN MCGERTY. be described s the locus {z C : z c = r} where r R >0 nd c C. The prmeters, b, c, k re uniquely determined by L nd C respectively. Proof. First note tht the rel xis is the locus {z C : I(z) = 0}, thus we cn obtin the eqution of ny line L by using n isometry to move it to the rel xis. More precisely, suppose tht L is line nd t L. Then trnslting by t we obtin new line L 1 through the origin. This line mkes n ngle θ with the positive rel xis, where 0 θ < π. Rotting by n ngle θ thus moves L 1 to the rel xis. It follows tht the composition of trnsltion by t nd rottion by θ moves L to the rel xis. These trnsformtions re given by z z t nd z z respectively, where = e iθ, thus their composition is z z t, nd hence L = {z C : I(z t) = 0}. Tking b = I(zt) we find tht L = {z C : I(z) = b} s required. The vlue of is clerly uniquely determined by L, while if we pick nother point s L, note tht I(s t) = I((s t)) = 0, since s t is in the direction of L, nd hence t n ngle θ with the rel xis, so tht (s t) = e iθ (s t) is rel. For the second prt, if C hs centre c C nd rdius r > 0 then clerly C = {z C : z c = r}. The uniqueness of r nd c is cler. Lemm Any line or circle cn be described s {z C : z = k z b }, where, b C nd k (0, 1] nd b. If k = 1 one obtins line, while if k < 1 one obtins circle. The prmeters, b, k re not unique. Proof. Let C,b,k = {z C : z = k z b }. First suppose tht k < 1. Then we hve: z = k z b z 2 = k 2 z b 2 z z z āz + ā = k 2 (z z b z bz + b b) (1 k 2 )z z ( k 2 b) z (ā k 2 b)z = ā + k 2 b b z ( k2 b) 1 k k 2 ( b + āb) + k 4 b 2 (1 k 2 ) 2 = k2 b k 2 z k2 b 1 k 2 2 = k2 ( 2 b āb + b 2 ) (1 k 2 ) 2 z k2 b k 2 1 k 2 2 = (1 k 2 ) 2 b 2. k Thus C,b,k is circle of rdius 1 k b nd centre k2 b 2 1 k. If k = 1, then 2 C,b,1 is just the locus of points equidistnt from nd b, which is clerly line (explicitly it is the line through ( + b)/2 perpendiculr to the line through nd b). We hve thus shown tht the loci C,b,k re either lines or circles. Next we show tht ny line or circle my be described in this form. If L is line, picking ny two points, b equidistnt to L we see tht L = C,b,1. Now suppose tht C is circle. If T : C C is the trnsformtion z rz + s (where r 0), then it is esy to check tht C,b,k = T (C ( s)/r,(b s)/r,k ), thus the set of circles of the from C,b,k is preserved under the ction of the group of ffine liner trnsformtions. But since we cn trnsform ny circle in C to ny other circle using such trnsformtions, it follows tht every circle occurs s locus C,b,k for some, b C, k (0, 1).

3 METRIC SPACES AND COMPLEX ANALYSIS. 37 Remrk Let S 1 = {z C : z = 1} be the unit circle in C. The proof of the bove Lemm shows tht if we tke w 0 with 0 < w 0 < 1 nd let w 1 = w 0 / w 0 2 nd k = w 0, then S 1 = C w0,w 1,k. Thus, just s for lines, the set of prmeters (, b, k) such tht C,b,k corresponds to prticulr circle is infinite. The points nd b re sid to be in inversion with respect to the circle C = C,b,k. 12. Complex differentibility If D is n open subset of C nd f : D C is function, we sy tht f is holomorphic (or complex differentible) t z 0 D if the limit f(z) f(z 0 ) lim z z 0 z z 0 exists, nd s usul when it exists we denote it by f (z 0 ). Thus the definition t lest formlly is exctly s in the cse of rel vrible. Notice however tht, unlike in the cse of R, z my tend to z 0 is ny direction, nd in fct it turns out tht the existence of the limit is stronger condition thn it might pper t first sight. It will turn out tht the theory of complex differentible functions is very different to tht of rel vrible, with mny results which re quite flse for the rel cse holding for the complex cse. For exmple, we will see tht if function is complex differentible then it is infinitely differentible, wheres the corresponding sttement for rel functions is esily seen to be flse. Just s in the cse of rel vrible, we hve the following reformultion of differentibility condition: Lemm Let f : D C be function defined on n open subset D of C. Then f is holomorphic t z 0 D with derivtive f (z 0 ) if nd only if there is function ɛ: D C which is continuous t z 0 with ɛ(z 0 ) = 0 stisfying f(z) = f(z 0 ) + f (z 0 )(z z 0 ) + (z z 0 )ɛ(z). Proof. Rerrnging the eqution it is esy to see tht it is equivlent to { f(z) f(z0) ɛ(z) = z z 0 f (z 0 ), z z 0, 0 z = z 0. The continuity of ɛ t z = z 0 is then clerly equivlent to f being holomorphic t z 0. The Lemm shows tht f is holomorphic t z 0 if it hs best liner pproximtion in the sense tht the error f(z) (f(z 0 ) + (z z 0 )f (z)) tends to zero superlinerly. (In the cse of one rel vrible, the grph of z f(z 0 ) + (z z 0 )f (z 0 ) is just the tngent line to the grph of f t z 0.) Although we will see tht the condition of complex differentibility is much stronger thn it is in the rel-vrible cse, the similrity of the two definitions shows us immeditely the stndrd properties of differentiblility still hold in the complex cse: Proposition Let U be n open subset of C nd let f, g be complex-vlued functions on U. (1) If f, g re differentible t z 0 U then f + g nd fg re differentible t z 0 with (f + g) (z 0 ) = f (z 0 ) + g (z 0 ); (f.g) (z 0 ) = f (z 0 ).g(z 0 ) + f(z 0 ).g (z 0 ).

4 38 KEVIN MCGERTY. (2) If f, g re differentible t z 0 nd g(z 0 ) 0 nd g (z 0 ) 0 then f/g is differentible t z 0 with (f/g) (z 0 ) = f (z 0 )g(z 0 ) f(z 0 )g (z 0 ) g (z 0 ) 2. (3) If U nd V re open subsets of C nd f : V U nd g : U C where f is complex differentible t z 0 V nd g is complex differentible t f(z 0 ) U the g f is complex differentible t z 0 with (g f) (z 0 ) = g (f(z 0 )).f (z 0 ). Proof. These re proved in exctly the sme wy s they re for function of single rel vrible. Just s for single rel vrible, the bsic rules of differentition llow one to check tht polynomil functions re differentible: Using the product rule nd induction one sees tht z n hs derivtive nz n 1 for ll n 0 (s constnt obviously hs derivtive 0). Then by linerity it follows every polynomil is differentible Differentibility in R 2. Since C is just R 2 with the dditionl structure of multipliction, if D is open in C then function f : D C is in prticulr function on D tking vlues in R 2. There is notion of differentiblility for functions on open subsets of R 2 (nd indeed R n for ny positive integer n) which we touch on here in order to get better understnding of the condition of complex differentibility. Definition Suppose U is n open subset of R 2 nd tht f : U R 2 is function. We sy tht f is differentible t z 0 = (x, y) if there is liner mp L: R 2 R 2 nd function ɛ: U R 2 such tht f(w) = f(z 0 ) + L(w z 0 ) + w z 0 ɛ(w), where ɛ(w) 0 s w z 0 (nd by convention we set ɛ(z 0 ) = 0). If it exists, the liner mp L is unique nd is denoted Df z0 (or sometimes Df(z 0 )). Given ny vector v R 2, we cn consider the line {z 0 + tv : t R} through z 0 in the direction of v. Restricting f to tht line gives function of single rel vrible t. The derivtive of this function t t = 0 is clled the directionl derivtive v f(z 0 ) of f t z 0 in the direction v. Explicitly it is defined to be f(z 0 + tv) f(z 0 ) v f(z 0 ) = lim. t 0 t When f is differentible t z 0 with derivtive Df z0 then for ll t 0 nd v R 2 we hve f(z 0 + tv) f(z 0 ) = Df z0 (v) + sign(t) v ɛ(tv) L(v), t s t 0, (where sign(t) = t /t {±1}) so tht the directionl derivtives exist for every v R 2 nd moreover v (f)(z 0 ) = Df z0 (v). In prticulr, if we tke v to be ech of the stndrd bsis vectors e 1 = (1, 0) t nd e 2 = (0, 1) t, then the directionl derivtives re just the prtil derivtives of f with respect to x nd y: e1 f(z 0 ) = x f(z 0 ); e2 f(z 0 ) = y f(z 0 ).

5 METRIC SPACES AND COMPLEX ANALYSIS. 39 When doing computtions, it cn be useful to brek f up into its components, tht is, we write f(x, y) = (f 1 (x, y), f 2 (x, y)) t where f 1 nd f 2 re rel-vlued functions. It is esy to see tht v f(z 0 ) = ( v f 1 (z 0 ), v f 2 (z 0 )) t. Now the liner mp Df z0 cn of course be written s mtrix with respect to the stndrd bsis. Since the mtrix of liner mp hs columns given by the imges of the bsis vectors, it follows tht the columns of this mtrix re Df z0 (e 1 ) nd Df z0 (e 2 ) respectively, nd since these vectors re the directionl derivitives in the directions e 1 nd e 2 respectively we see tht ( ) x f Df z0 = 1 (z 0 ) y f 1 (z 0 ) x f 2 (z 0 ) y f 2 (z 0 ) The mtrix representing the totl derivtive of f t point is clled the Jcobin mtrix. As we hve seen, it my be clculted by computing the prtil derivtives of the components of f. While it is possible for the prtil derivtives of function to exist without the function being differentible in the sense of Definition 12.3, the following theorem shows tht this does not hppen in good situtions: Theorem Let U be n open subset of R 2 nd f : U R 2. Let f(x) = (f 1 (x), f 2 (x)) t. If ll the prtil derivtives of the f i exist nd re continuous t z 0 U then f is differentible t z 0. The proof of this (lthough it is not hrd one only needs the definitions nd the single-vrible men-vlue theorem) is not prt of this course. For completeness, proof is given in the Appendix The Cuchy-Riemnn equtions. We return now to the cse of complex differentible function f : D C on n open subset D of C. Viewing C just s R 2, function f : D C on n open subset D of C becomes function to R 2 of two rel vribles (the rel nd imginry prts). Explicitly, if we write z = x + iy nd f(z) = u + iv where u nd v re rel, then f(x, y) = (u(x, y), v(x, y)). Lemm If U is n open subset of C nd f : U C is complex differentible t z 0 = x 0 + iy 0 U then the ssocited function to R 2 is rel-differentible t (x 0, y 0 ). Proof. If w C is ny complex number then the opertion of multipliction by w defines n R-liner mp. Explicitly, if w = + ib then the mtrix of this liner mp with respect to the stndrd bsis (corresponding to 1, i C) is: ( ) b b If f is differentible t z 0 then letting L be the liner mp given by f (z 0 ) it follows immeditely tht f stisfies the definition of the totl rel derivtive t (x 0, y 0 ) compre the formultion of complex differentibility given by Lemm 12.1 nd Defintion Now the mtrix for the totl derivtive is the mtrix of prtil derivtives of the components u, v of f, nd by the proof of the previous Lemm this mtrix is given by the rel nd imginry prts of f (z 0 ): indeed if f (z 0 ) = r + is then the mtrix of the totl derivtive t z 0 is (12.1) Df z0 = ( x u(z 0 ) y u(z 0 ) x v(z 0 ) y v(z 0 ) ) ( r s = s r ).

6 40 KEVIN MCGERTY. It follows tht x u(x 0, y 0 ) = y v(x 0, y 0 ) = r nd x v(x 0, y 0 ) = y u(x 0, y 0 ) = s. We record this s theorem. Theorem (Cuchy-Riemnn equtions). Let f : U C be function on n open subset U of C nd let f = u + iv where u, v re rel-vlued. Then wherever f is complex differentible the functions u nd v stisfy x u = y v; x v = y u. Conversely, if f : U C is rel-differentible nd its rel nd imginry prts stisfy the Cuchy-Riemnn equtions, then f is complex differentible, with derivtive f (z 0 ) = x u + i x v. Proof. We hve lredy shown this using the definition of the totl derivtive, but one cn lso work directly from the definition of the complex derivtive: Suppose tht f is complex differentible t z 0 = x 0 + iy 0. We hve f(z 0 + t) f(z 0 ) x f(x 0, y 0 ) = lim = f (z 0 ), t 0 t since we re simply tking the limit defining f (z 0 ) in prticulr direction (long the rel xis). On the other hnd for the prtil derivtive with respect to y we hve f(z 0 + it) f(z 0 ) f(z 0 + it) f(z 0 ) y f(x 0, y 0 ) = lim = i lim = if (z 0 ), t 0 t t 0 it so tht f (z 0 ) = 1 i yf(z 0 ). Tking components, we see tht x f = ( x u, x v) nd y f = ( y u, y v), so tht x f(z 0 ) = f (z 0 ) = 1 i yf(z 0 ) becomes ( x u, x v) = ( y v, y u). s required. For the converse, observe tht if f stisfies the Cuchy-Riemnn equtions t z 0 U, then the liner mp ssocited to the mtrix of prtil derivtives t z 0 coincides with tht given by multipliction by the complex number = x u + i y v. But then it follows from the definition of the rel derivtive tht stisfies the conditions of Lemm 12.1 so tht f (z 0 ) = nd f is complex differentible s required. Remrk Since the opertion of multipliction by complex number w is composition of rottion (by the rgument of w) nd diltion (by the modulus of w) the mtrix of the corresponding liner mp is, up to sclr, rottion mtrix. The Cuchy-Riemnn equtions just cpture this fct for the mtrix of the totl (rel) derivtive of complex differentible function. Definition The Cuchy-Riemnn equtions cn be rewritten using certin prtil differentil opertors: Set z = 1 2 ( x i y ); z = 1 2 ( x + i y ). Theorem Let f : U C be function on n open subset of C. If f is complex differentible t z 0 then z f(z 0 ) = 0 nd f (z 0 ) = z f(z 0 ). Moreover if J(z 0 ) is the mtrix of the totl derivtive of f t z 0 U then det(j(z 0 )) = f (z 0 ) 2.

7 METRIC SPACES AND COMPLEX ANALYSIS. 41 Proof. The condition z (f)(z 0 ) = 0 is esily seen to be equivlent to the Cuchy- Riemnn equtions by tking rel nd imginry prts. On the other hnd, we sw in the proof of Theorem 12.6 tht f (z 0 ) = x f(z 0 ) nd y f(z 0 ) = if (z 0 ), so tht z f(z 0 ) = 1 2 ( xf(z 0 ) i y f(z 0 )) = f (z 0 ) s required. The lst prt follows from Eqution (12.1) nd the identities ( ) b det = 2 + b 2 = + ib 2, b for ny, b R. Exercise Show tht if T : C C is ny rel liner mp (tht is, viewing C s R 2 we hve T : R 2 R 2 is liner mp) then there re unique, b C such tht T (z) = z + b z. (Hint: note tht the mp z z + b z is R-liner. Wht mtrix does it correspond to s mp from R 2 to itself?) We finish this section with result which gives useful sufficient condition for function to be complex differentible. Theorem Suppose tht U is n open subset of C nd let f : U C be function. If f is differentible s function of two rel vribles with continuous prtil derivtives stisfying the Cuchy-Riemnn equtions on U, then f is complex differentible on U. Proof. Since the prtil derivtives re continuous, Theorem 12.4 shows tht f is differentible s function of two rel vribles, with totl derivtive given by the mtrix of prtil derivtives. If f lso stisfies the Cuchy-Riemnn equtions, then by Theorem 12.6 it follows it is complex differentible s required. Exmple The previous theorem llows us to show tht the complex logrithm is holomorphic function up to the issue tht we cnnot define it continuously on the whole complex plne! The function z e z is not injective, since e z+2nπi = e z for ll n Z thus it cnnot hve n inverse defined on ll of C. However, since e x+iy = e x (cos(y) + i sin(y)), it follows tht if we pick ry through the origin, sy B = {z C : I(z) = 0, R(z) 0}, then we my define Log : C\B C by setting Log(z) = log( z ) + iθ where θ ( π, π] is the rgument of z. Clerly e Log(z) = z, while Log(e z ) differs from z by n integer multiple of 2πi. We clim tht Log is complex differentible: To show this we use Theorem Indeed the function L(x, y) = (log( x 2 + y 2 ), θ) = (L 1, L 2 ) hs x x L 1 = x 2 + y 2, y yl 1 = x 2 + y 2, x L 2 = y x 2 + y 2, x yl 2 = x 2 + y 2. where in clculting the prtil derivtives of L 2 we used tht it is equl to rctn(y/x) in ( π/2, π/2) (nd other similr expressions in the other two qudrnts). Exmining the formule we see tht the prtil derivtives re ll continuous, nd obey the Cuchy-Riemnn equtions, so tht Log is indeed complex differentible.

8 42 KEVIN MCGERTY Hrmonic functions. Recll tht the two-dimensionl Lplce opertor is the differentil opertor 2 x + 2 y (defined on functions f : R 2 R which re twice differentible in the sense tht their prtil derivtives re gin differentible). A function which is in the kernel of the Lplce opertor is sid to be hrmonic, tht is, function u: D R defined on n open subset D of R 2 is hrmonic if (u) = 2 xu+ 2 yu = 0. There is strong connection between complex differentible functions nd hrmonic functions, s the next result shows. Lemm Suppose tht U is n open subset of C nd f : U C is complex differentible nd f(z) = u(z) + iv(z) re its rel nd imginry prts. If u nd v re twice continuously 21 differentible then they re hrmonic on U. Proof. We hve lredy seen tht u nd v stisfy the Cuchy-Riemnn equtions. Thus we hve 2 x(u) = x ( y v) = y ( x v) = y ( y u) = 2 yu, so tht ( x 2 + y)(u) 2 = 0, tht is u is hrmonic. Note tht in the second equlity we used the symmetry of mixed prtil derivitives x y u = y x u which holds provided u is twice continuously differentible. To see tht v is hrmonic one cn rgue similrly, or note tht v is the rel prt of if, which is clerly complex differentible. Remrk We will shortly see tht if f = u+iv is complex differentible then it is in fct infinitely complex differentible. Since we hve seen tht f = x f = 1 i yf it follows tht u nd v re in fct infinitely differentible so the condition in the previous lemm on the existence nd continuity of their second derivtives in fct holds utomticlly. For proof of the fct tht the mixed prtil derivtives of twice continuously differentible function re equl, see the Appendix. Lemm motivtes the following definition: Definition If u: R 2 R is hrmonic function, we sy tht v : R 2 R is hrmonic conjugte of u if f(z) = u + iv is holomorphic. Notice tht if u is hrmonic, it is twice differentible so tht its prtil derivtives re continuously differentible. It follows tht function v is hrmonic conjugte precisely if the pir (u, v) stisfy the Cuchy-Riemnn equtions. Provided we cn integrte these equtions, hrmonic conjugte will exist, nd we will show lter tht, t lest when the second prtil derivtives re continuous, this cn lwys been done loclly in the plne Power series. Another importnt fmily of exmples re the functions which rise from power series. We review here the min results bout complex power series which were proved in Anlysis II lst yer: Definition Let ( n ) n 0 be sequence of complex numbers. Then we hve n ssocited sequence of polynomils s n (z) = n k=0 kz k. Let S be the set on which this sequence converges pointwise, tht is S = {z C : lim n s n(z) exists}. Note tht since s n (0) = 0 we hve 0 S so in prticulr S is nonempty. On the set S, we cn define function s(z) = lim n s n (z) = k=0 kz k which we cll 21 Tht is, ll of their second prtil derivitives exist nd re continuous.

9 METRIC SPACES AND COMPLEX ANALYSIS. 43 power series. We define the rdius of convergence R of the power series k 0 kz k to be sup{ z : z S} (or if S is unbounded). By convention, given ny sequence of complex numbers (c n ) n 0 we write k=0 c kz k for the corresponding power series (even though it my be tht it converges only for z = 0). We cn give n explicit formul for the rdius of convergence using the notion of lim sup which we now recll: Definition If ( n ) n 0 is sequence of rel numbers, set s n = sup{ k : k n} R { } (where we tke s n = if { k : k n} is not bounded bove). Then the sequence (s n ) is either constnt nd equl to or eventully becomes decresing sequence of rel numbers. In the first cse we set lim sup n n =, wheres in the second cse we set lim sup n n = lim n s n (which is finite if (s n ) is bounded below, nd equl to otherwise). Lemm Let k 0 kz k be power series, let S be the subset of C on which it converges nd let R be its rdius of convergence. Then we hve B(0, R) S B(0, R). The series converges bsolutely on B(0, R) nd if 0 r < R then it converges uniformly on B(0, r). Moreover, we hve 1/R = lim sup n 1/n. n Proof. Let L = lim sup n n 1/n [0, ]. If L = 0 then the sttement should be understood to sy tht the rdius of convergence R is, while if L = we tke R = 0. These two cses re in fct similr but esier thn the cse where L (0, ), so we will only give the detils for the cse where L is finite nd positive. Let s n = sup{ k 1/k : k n} so tht L = lim n s n. If 0 < s < 1/L we cn find n ɛ > 0 such tht (L + ɛ).s = r < 1. Thus by definition, for sufficiently lrge n we hve n 1/n s n < L + ɛ so tht if z s we hve n z n [(L + ɛ) z ] n r n, nd hence by the comprison test, n=0 nz n converges bsolutely nd uniformly on B(0, s). It follows the power series converges everywhere in B(0, 1/L). On the other hnd, if z > 1/L we cn find n ɛ 1 > 0 such tht z (L ɛ 1 ) = r > 1. But then for ll k we hve s k L since (s n ) is decresing, nd hence by the pproximtion property for ech k we cn find n n k k with nk 1/n k > s k ɛ 1 L ɛ nd hence nk z n k > r k. Thus n z n hs subsequence which does not tend to zero, so the series cnnot converge. It follows the rdius of convergence of n=0 nz n is 1/L s climed. The next lemm is reltively stright-forwrd consequence of stndrd lgebr of limits style results: Lemm Let s(z) = k=0 kz k nd t(z) = k=0 b kz k be power series with rdii of convergence R 1 nd R 2 respectively nd let T = min{r 1, R 2 }.

10 44 KEVIN MCGERTY. (1) Let c n = k+l=n kb l, then the power series n=0 c nz n hs rdius of convergence t lest T nd if z < T we hve c n z n = s(z)t(z). n=0 Thus the product of power series is power series. (2) If s(z) nd t(z) re s bove, then k=0 ( k + b k )z k is power series which converges to s(z) + t(z) in B(0, T ), thus the sum of power series is gin power series. Proof. This ws estblished in Prelims Anlysis II. Note tht T is only lower bound for the rdius of convergence in ech cse it is esy to find exmples where the ctul rdius of convergence of the sum or product is strictly lrger thn T. The behviour of power series t its rdius of convergence is in generl rther complicted phenomenon. The following result, which we shll not prove, gives some informtion however. Some of the ides involved in its proof re investigted in Problem Set 4. Theorem (Abel s theorem:) Suppose tht ( n ) is sequence of complex numbers nd n=0 n exists. Then the series n=0 nz n converges for z < 1 nd ( n r n) = n. lim r ( 1,1) r 1 n=0 Proof. Note tht since the series n=0 nz n converges t z = 1 by ssumption, its rdius of convergence is t lest 1, so tht the first sttement holds. For the second see for exmple Exercise 15 of Chpter 1 in the book of Stein nd Shkrchi. Proposition Let s(z) = k 0 kz k be power series, let S be the domin on which it converges, nd let R be its rdius of convergence. Then power series t(z) = k=1 k kz k 1 lso hs rdius of convergence R nd on B(0, R) the power series s is complex differentible with s (z) = t(z). In prticulr, it follows tht power series is infinitely complex differentible within its rdius of convergence. Proof. First note tht the power series k=1 k kz k 1 clerly hs the sme rdius of convergence s k=1 k kz k, nd by Lemm this hs rdius of convergence 22 lim sup k n=0 k k 1/k = lim(k 1/k ) lim sup k 1/k = lim sup k 1/k = R, k k since lim k k 1/k = 1. Thus s(z) = k=0 kz k nd t(z) = k=1 k kz k 1 hve the sme rdius of convergence. To see tht s(z) is complex differentible with derivtive t(z), consider the sequence of polynomils f n in two complex vribles: n 1 f n (z, w) = n ( z i w n 1 i ), (n 1). i=0 Fix ρ < R, then for (z, w) with z, w ρ we hve f n (z, w) = n 1 n z i w n i n 1 n z i w n i n nρ n 1 i=0 22 This uses stndrd property of lim sup which is proved for completeness in Lemm 23.3 in Appendix I. i=0

11 METRIC SPACES AND COMPLEX ANALYSIS. 45 It therefore follows from the Weierstrss M-test with 23 tht the series n 0 f n(z, w) converges uniformly (nd bsolutely) on {(z, w) : z, w ρ} to function F (z, w). In prticulr, it follows tht F (z, w) is continuous. But since n k=1 f k(z, z) = n k=1 k kz k 1, it follows tht F (z, z) = t(z). On the other hnd, for z w we hve k 1 i=0 zi w k i = zk w k, so tht z w F (z, w) = k=0 k z k w k z w s(z) s(w) = z w, hence it follows by the continuity of F tht if we fix z with z < ρ then s(z) s(w) lim = F (z, z) = t(z). z w z w Since ρ < R ws rbitrry, we see tht s(z) is differentible on B(0, R) with derivtive t(z). Finlly, since we hve shown tht ny power series is differentible within its rdius of convergence nd its derivtive is gin power series with the sme rdius of convergence, it follows by induction tht ny power series is in fct infinitely differentible within its rdius of convergence. Exmple The previous theorem gives us lrge supply of complex differentible functions. For exmple, z n exp(z) = n!, cos(z) = ( 1) n z2n (2n)!, sin(z) = ( 1) n z2n+1 (2n + 1)!, n=0 n=0 re ll complex differentible on the whole complex plne (since R = in ech cse). Note tht one cn use the bove theorem to show tht cos(z) 2 + sin(z) 2 = 1 for ll z C, but since sin(z) nd cos(z) re not in generl rel, this does not imply tht sin(z) or cos(z) t most 1. (In fct it is esy to check tht they re both unbounded on C). Using wht we hve lredy estblished bout power series it is lso esy to check tht the complex sin function encompses both the rel trigonometric nd rel hyperbolic functions, indeed: n=0 sin( + ib) = sin() cosh(b) + i cos() sinh(b). Exmple Let s(z) = n=1 zn n. Then s(z) hs rdius of convergence 1, nd in B(0, 1) we hve s (z) = n=0 zn = 1/(1 z), thus this power series is complex differentible function which extends the function log(1 z) on the intervl ( 1, 1) to the open disc B(0, 1) C. We will see lter tht we will not be ble to extend the function log to complex differentible function on C\{0} we will only be ble to construct multi-vlued extension. Exmple Recll from Prelims Anlysis tht the binomil theorem generlizes to non-integrl exponents C if we define ( ) k = 1 k!.( 1)... ( k + 1). Indeed we then hve ( ) (1 + z) = z k, k k=0 R. 23 We know n 0 Mn = n nρn 1 converges since ρ < R nd t(z) hs rdius of convergence

12 46 KEVIN MCGERTY. for ll z with z < 1. Indeed it is esy to see from the rtio test tht this series hs rdius of convergence equl to 1, nd then one cn check tht if f(z) denotes the function given by the series inside B(0, 1), then zf (z) = f(z). Note tht, slightly more generlly, we cn work with power series centred t n rbitrry point z 0 C. Such power series re functions given by n expression of the form f(z) = n 0 n (z z 0 ) n. All the results we hve shown bove immeditely extend to these more generl power series, since if g(z) = n 0 n z n, then the function f is obtined from g simply by composing with the trnsltion z z z 0. In prticulr, the chin rule shows tht f (z) = n 1 n n (z z 0 ) n Brnch cuts It is often the cse tht we study holomorphic function on domin D C which does not extend to function on the whole complex plne. Exmple Consider the squre root function f(z) = z 1/2. Unlike the cse of rel numbers, every complex number hs squre root, but just s for the rel numbers, there re two possiblities unless z = 0. Indeed if z = x+iy nd w = u+iv hs w 2 = z we see tht nd so u 2 v 2 = x; 2uv = y, u 2 = x + x 2 + y 2, v 2 = y + x 2 + y where the requirement tht u 2, v 2 re nonnegtive determines the signs. Hence tking squre roots we obtin the two possible solutions for w stifying w 2 = z. (Note it looks like there re four possible sign combintions in the bove, however the requirement tht 2uv = y mens the sign of u determines tht of v.) In polrs it looks simpler: if z = re iθ then w = ±r 1/2 e iθ/2. Indeed this expression gives us continuous choice of squre root except t the positive rel xis: for ny z C we my write z uniquely s re iθ where θ [0, 2π), nd then set f(z) = r 1/2 e iθ/2. But now for θ smll nd positive, f(z) = r 1/2 e iθ hs smll positive rgument, but if z = re (2π ɛ)i we find f(z) = r 1/2 e (π ɛ/2)i, thus f(z) in the first cse is just bove the positive rel xis, while in the second cse f(z) is just below the negtive rel xis. Thus the function f is only continuous on C\{z C : I(z) = 0, R(z) > 0}. Using Theorem 12.1 you cn check f is lso holomorphic on this domin. The positive rel xis is clled brnch cut for the multi-vlued function z 1/2. By chosing different intervls for the rgument (such s ( π, π] sy) we cn tke different cuts in the plne nd obtin different brnches of the function z 1/2 defined on their complements. We formlize these concepts s follows:

13 METRIC SPACES AND COMPLEX ANALYSIS. 47 Definition A multi-vlued function or multifunction on subset U C is mp f : U P(C) ssigning to ech point in U subset 24 of the complex numbers. A brnch of f on subset V U is function g : V C such tht g(z) f(z), for ll z V. We will be interested in brnches of multifunctions which re holomorphic. Remrk In order to distinguish between multifunctions nd functions, it is sometimes useful to introduce some nottion: if we wish to consider z z 1/2 s multifunction, then to emphsize tht we men multifunction we will write [z 1/2 ]. Thus [z 1/2 ] = {w C : w 2 = z}. Similrly we write [Log(z)] = {w C : e w = z}. This is not uniform convention in the subject, but is used, for exmple, in the text of Priestley. Thus the squre root z [z 1/2 ] is multifunction, nd we sw bove tht we cn obtin holomorphic brnches of it on cut plne C\R where R = {te iθ : t R 0 }. The point here is tht both the origin nd infinity s brnch points for the multifunction [z 1/2 ]. Definition Suppose tht f : U P(C) is multi-vlued function defined on n open subset U of C. We sy tht z 0 U is not brnch point of f if there is n open disk 25 D U contining z 0 such tht there is holomorphic brnch of f defined on D. We sy z 0 is brnch point otherwise. When C\U is bounded, we sy tht f does not hve brnch point t if there is brnch of f defined on C\B(0, R) U for some R > 0. Otherwise we sy tht is brnch point of f. A brnch cut for multifunction f is curve in the plne on whose complement we cn pick holomorphic brnch of f. Thus brnch cut must contin ll the brnch points. Exmple Another importnt exmple of multi-vlued function which we hve lredy discussed is the complex logrithm: s multifunction we hve Log(z) = {log( z ) + i(θ + 2nπ) : n Z} where z = z e iθ. To obtin brnch of the multifunction we must mke choice of rgument function rg: C R we my define Log(z) = log( z ) + i rg(z), which is continuous function wy from the brnch cut we chose. By convention, the principl brnch of Log is defined by tking rg(z) ( π, π]. Another importnt clss of exmples of multifunctions re the frctionl power multifunctions z [z α ] where α C: These re given by z exp(α.[log(z)]) = {exp(α.w) : w C, e w = z} Note this is includes the squre root multifunction we discussed bove, which cn be defined without the use of exponentil function. Indeed if α = m/n is rtionl, m Z, n Z >0, then [z α ] = {w C : w m = z n }. For α C\Q however we cn only define [z α ] using the exponentil function. Clerly from its definition, nytime we choose brnch L(z) of [Log(z)] we obtin corresponding brnch exp(α.l(z)) of [z α ]. If L(z) is the principl brnch of [Log(z)] then the corresponding brnch of [z α ] is clled the principl brnch of [z α ]. 24 We use the nottion P(X) to denote the power set of X, tht is, the set of ll subsets of X. 25 In fct ny simply-connected domin see our discussion of the homotopy form of Cuchy s theorem.

14 48 KEVIN MCGERTY. Exmple Let F (z) be the multi-function [(1 + z) α ] = {exp(α.w) : w C, exp(w) = 1 + z}. Then within the open bll B(0, 1) the power series s(z) = n=0 ( α k) z k yields holomorphic brnch of [(1 + z) α ]. Indeed we hve seen tht (1 + z)s (z) = α.s(z), nd if we tke the principl brnch L(z) of [Log(z)] then on B(0, 1) we hve 26 d dz (L(s(z))) = s (z)/s(z) = α/(1 + z) = d (αl(1 + z)) dz so tht L(s(z)) = α.l(1+z)+c for some constnt c (s B(0, 1) is connected) which by evluting t z = 0 we find is zero. Finlly, it follows tht s(z) = exp(αl(1+z)) so tht s(z) [(1 + z) α ] s required. Exmple A more interesting exmple is the function f(z) = [(z 2 1) 1/2 ]. Using the principl brnch of the squre root function, we obtin brnch f 1 of f on the complement of E = {z C : z 2 1 (, 0]}, which one clcultes is equl to ( 1, 1) ir. If we cross either the segment ( 1, 1) or the imginry xis, this brnch of f is discontinuous. To find nother brnch, note tht we my write f(z) = z 1 z + 1, thus we cn tke the principl brnch of the squre root for ech of these fctors. More explicity, if we write z = 1 + re iθ1 = 1 + se iθ2 where θ 1, θ 2 ( π, π] then we get brnch of f given by f 2 (z) = rs.e i(θ1+θ2)/2. Now the fctors re discontinuous on (, 1] nd (, 1] respectively, however let us exmine the behviour of their product: If z crosses the negtive rel xis t I(z) < 1 then θ 1 nd θ 2 both jumps by 2π, so tht (θ 1 + θ 2 )/2 jumps by 2π, nd hence exp((θ 1 + θ 2 )/2) is in fct continuous. On the other hnd, if we cross the segment ( 1, 1) then only the fctor z 1 switches sign, so our brnch is discontinuous there. Thus our second brnch of f is defined wy from the cut [ 1, 1]. Exmple The brnch points of the complex logrithm re 0 nd infinity: indeed if z 0 0 then we cn find hlf-plne sy H = {z C : I(z) > 0} (where = 1) such tht z 0 H. We cn chose continuous choice of rgument function on H, nd this gives holomorphic brnch of Log defined on H nd hence on the disk B(z 0, r) for r sufficiently smll. The logrithm lso hs brnch point t infinity, since we cnnot chose continous rgument function on C\B(0, R) for ny R > 0. (We will return to this point when discussing the winding number lter in the course.) Note tht if f(z) = [ z 2 1] then the second of our brnches f 2 discussed bove shows tht f does not hve brnch point t infinity, wheres both 1 nd 1 re brnch points s we move in sufficiently smll circle round we cnnot mke continuous choice of brnch. One cn given rigorous proof of this using the brnch f 2 : given ny brnch g of [ z 2 1] defined on B(1, r) for r < 1 one proves tht g = ±f 2 so tht g is not continuous on B(0, r) ( 1, 1). See Problem Sheet 4, question 5, for more detils. Exmple A more sophisticted point of view on brnch points nd cuts uses the theory of Riemnn surfces. As first look t this theory, consider the multifunction f(z) = [ z 2 1] gin. Let Σ = {(z, w) C 2 : w 2 = z 2 1} (this is 26 Any continuous brnch l(z) of [Log(z)] is holomorphic where it is defined nd stisfies exp(l(z)) = z, hence by the chin rule one obtins l (z) = 1/z.

15 METRIC SPACES AND COMPLEX ANALYSIS. 49 n exmple of Riemnn surfce). Then we hve two mps from Σ to C, projecting long the first nd second fctor: p 1 (z, w) = z nd p 2 (z, w) = w. Now if g(z) is brnch of f, it gives us mp G: C Σ where G(z) = (z, g(z)). If we tke f 2 (z) = z 1 z + 1 (using the principl brnch of the squre root function in ech cse, then let Σ + {(z, f 2 (z)) : z / [ 1, 1]} nd Σ = {(z, f 2 (z)) : z / [ 1, 1]}, then Σ + Σ covers ll of Σ prt from the pirs (z, w) where z [ 1, 1]. For such z we hve w = ±i 1 z 2, nd Σ is obtined by gluing together the two copies Σ + nd Σ of the cut plne C\[ 1, 1] long the cut locus [ 1, 1]. However, we must exmine the discontinuity of g in order to see how gluing works: the upper side of the cut in Σ + is glued to the lower side of the cut in Σ nd similrly the lower side of the cut in Σ + is glued to the upper side of Σ. Notice tht on Σ we hve the (single-vlued) function p 2 (z, w) = w, nd ny mp q : U Σ from n open subset U of C to Σ such tht p 1 q(z) = z gives brnch of f(z) = z 2 1 given by p 2 q. Such function is clled section of p 1. Thus the multi-vlued function on C becomes single-vlued function on Σ, nd brnch of the multifunction corresponds to section of the mp p 1 : Σ C. In generl, given multi-vlued function f one cn construct Riemnn surfce Σ by gluing together copies of the cut complex plne to obtin surfce on which our multifunction becomes single-vlued function. 14. Pths nd Integrtion Pths will ply crucil role in our development of the theory of complex differentible functions. In this section we review the notion of pth nd define the integrl of continuous function long pth Pths. Recll tht pth in the complex plne is continuous function : [, b] C. A pth is sid to be closed if () = (b). If is pth, we will write for its imge, tht is = {z C : z = (t), some t [, b]}. Although for some purposes it suffices to ssume tht is continuous, in order to mke sense of the integrl long pth we will require our pths to be (t lest piecewise) differentible. We thus need to define wht we men for pth to be differentible: Definition We will sy tht pth : [, b] C is differentible if its rel nd imginry prts re differentible s rel-vlued functions. Equivlently, is differentible t t 0 [, b] if (t) (t 0 ) lim t t 0 t t 0 exists, nd denote this limit s (t). (If t = or b then we interpret the bove s one-sided limit.) We sy tht pth is C 1 if it is differentible nd its derivtive (t) is continuous. We will sy pth is piecewise C 1 if it is continuous on [, b] nd the intervl [, b] cn be divided into subintervls on ech of which is C 1. Tht is, there is finite sequence = 0 < 1 <... < m = b such tht [i, i+1] is C 1. Thus in prticulr, the left-hnd nd right-hnd derivtives of t i (1 i m 1) my not be equl.

16 50 KEVIN MCGERTY. Remrk Note tht C 1 pth my not hve well-defined tngent t every point: if : [, b] C is pth nd (t) 0, then the line {(t)+s (t) : s R} is tngent to, however if (t) = 0, the imge of my hve no tngent line there. Indeed consider the exmple of : [ 1, 1] C given by { t 2 1 t 0 (t) = it 2 0 t 1. Since (0) = 0 the pth is C 1, even though it is cler there is no tngent line to the imge of t 0. If s: [, b] [c, d] is differentible mp, then we hve the following version of the chin rule, which is proved in exctly the sme wy s the rel-vlued cse. It will be crucil in our definition of the integrl of functions f : C C long pths. Lemm Let : [c, d] C nd s: [, b] [c, d] nd suppose tht s is differentible t t 0 nd is differentible t s 0 = s(t 0 ). Then s is differentible t t 0 with derivtive ( s) (t 0 ) = s (t 0 ). (s(t 0 )). Proof. Let ɛ: [c, d] C be given by ɛ(s 0 ) = 0 nd (x) = (s 0 ) + (s 0 )(x s 0 ) + (x s 0 )ɛ(x), (so tht this eqution holds for ll x [c, d]), then ɛ(x) 0 s x s 0 by the definition of (s 0 ), i.e. ɛ is continuous t t 0. Substituting x = s(t) into this we see tht for ll t t 0 we hve (s(t)) (s 0 ) t t 0 = s(t) s(t 0) t t 0 ( (s(t)) + ɛ(s(t)) ). Now s(t) is continuous t t 0 since it is differentible there hence ɛ(s(t)) 0 s t t 0, thus tking the limit s t t 0 we see tht s required. ( s) (t 0 ) = s (t 0 )( (s 0 ) + 0) = s (t 0 ) (s(t 0 )), Definition If φ: [, b] [c, d] is continuously differentible with φ() = c nd φ(b) = d, nd : [c, d] C is C 1 -pth, then setting = φ, by Lemm 14.3 we see tht : [, b] C is gin C 1 -pth with the sme imge s nd we sy tht is reprmetriztion of. Definition We will sy two prmetrized pths 1 : [, b] C nd 2 : [c, d] C re equivlent if there is continuously differentible bijective function s: [, b] [c, d] such tht s (t) > 0 for ll t [, b] nd 1 = 2 s. It is stright-forwrd to check tht equivlence is indeed n equivlence reltion on prmetrized pths, nd we will cll the equivlence clsses oriented curves in the complex plne. We denote the equivlence clss of by []. The condition tht s (t) > 0 ensures tht the pth is trversed in the sme direction for ech of the prmetriztions 1 nd 2. Moreover 1 is piecewise C 1 if nd only if 2 is. Recll tht we sw before (in generl metric spce) tht ny pth : [, b] C hs n opposite pth nd tht two pths 1 : [, b] C nd 2 : [c, d] C with 1 (b) = 2 (c) cn be conctented to give pth 1 2. If, 1, 2 re piecewise C 1 then so re nd 1 2. (Indeed piecewise C 1 pth is precisely finite conctention of C 1 pths).

17 METRIC SPACES AND COMPLEX ANALYSIS. 51 Remrk Note tht if : [, b] C is piecewise C 1, then by choosing reprmetriztion by function ψ : [, b] [, b] which is strictly incresing nd hs vnishing derivtive t the points where fils to be C 1, we cn replce by = ψ to obtin C 1 pth with the sme imge. For this reson, some texts insist tht C 1 pths hve everywhere non-vnishing derivtive. In this course we will not insist on this. Indeed sometimes it is convenient to consider constnt pth, tht is pth : [, b] C such tht (t) = z 0 for ll t [, b] (nd hence (t) = 0 for ll t [, b]). Exmple The most bsic exmple of closed curve is circle: If z 0 C nd r > 0 then the pth z(t) = z 0 + re 2πit (for t [0, 1]) is the simple closed pth with positive orienttion encircling z 0 with rdius r. The pth z(t) = z 0 + re 2πit is the simple closed pth encircling z 0 with rdius r nd negtive orienttion. Another useful pth is line segment: if, b C then the pth [,b] : [0, 1] C given by t + t(b ) = (1 t) + tb trverses the line segment from to b. We denote the corresponding oriented curve by [, b] (which is consistent with the nottion for n intervl in the rel line). One of the simplest clsses of closed pths re tringles: given three points, b, c, we define the tringle, or tringulr pth, ssocited to them, to be the conctention of the ssocited line segments, tht is T,b,c =,b b,c c, Integrtion long pth. To define the integrl of complex-vlued function long pth, we first need to be ble to integrte functions F : [, b] C on closed intervl [, b] tking vlues in C. Lst yer in Anlysis III the Riemnn integrl ws defined for function on closed intervl [, b] tking vlues in R, but it is esy to extend this to functions tking vlues in C: Indeed we my write F (t) = G(t) + ih(t) where G, H re functions on [, b] tking rel vlues. Then we sy tht F is Riemnn integrble if both G nd H re, nd we define: b F (t)dt = b G(t)dt + i b H(t)dt Note tht if F is continuous, then its rel nd imginry prts re lso continuous, nd so in prticulr Riemnn integrble 27. The clss of Riemnn integrble (rel or complex vlued) functions on closed intervl is however slightly lrger thn the clss of continuous functions, nd this will be useful to us t certin points. In prticulr, we hve the following: Lemm Let [, b] be closed intervl nd S [, b] finite set. If f is bounded continuous function (tking rel or complex vlues) on [, b]\s then it is Riemnn integrble on [, b]. Proof. The cse of complex-vlued functions follows from the rel cse by tking rel nd imginry prts. For the cse of function f : [, b]\s R, let = x 0 < x 1 < x 2 <... < x k = b be ny prtition of [, b] which includes the elements of S. Then on ech open intervl (x i, x i+1 ) the function f is bounded nd continuous, nd hence integrble. We my therefore set b f(t)dt = x1 f(t)dt + x2 x 1 f(t)dt +... xk x k 1 f(t)dt + b x k f(t)dt. 27 It is cler this definition extends to give notion of the integrl of function f : [, b] R n we sy f is integrble if ech of its components is, nd then define the integrl to be the vector given by the integrls of ech component function.

18 52 KEVIN MCGERTY. The stndrd dditivity properties of the integrl then show tht b f(t)dt is independent of ny choices. Remrk Note tht normlly when one speks of function f being integrble on n intervl [, b] one ssumes tht f is defined on ll of [, b]. However, if we chnge the vlue of Riemnn integrble function f t finite set of points, then the resulting function is still Riemnn integrble nd its integrl is the sme. Thus if one prefers the function f in the previous lemm to be defined on ll of [, b] one cn define f to tke ny vlues t ll on the finite set S. It is esy to check tht the Riemnn integrl of complex-vlued functions is complex liner. We lso note version of the tringle inequlity for complex-vlued functions: Lemm Suppose tht F : [, b] C is complex-vlued function. Then we hve b F (t)dt b F (t) dt. Proof. First note tht if F (t) = u(t) + iv(t) then F (t) = u 2 + v 2 so tht if F is integrble F (t) is lso 28. We my write b F (t)dt = reiθ, where r [0, ) nd θ [0, 2π). Now tking the components of F in the direction of e iθ nd e i(θ+π/s) = ie iθ, we my write F (t) = u(t)e iθ + iv(t)e iθ. Then by our choice of θ we hve b F (t)dt = eiθ b u(t)dt, nd so b F (t)dt = b u(t)dt b u(t) dt b F (t) dt, where in the first inequlity we used the tringle inequlity for the Riemnn integrl of rel-vlued functions. We re now redy to define the integrl of function f : C C long piecewise- C 1 curve. Definition If : [, b] C is piecewise-c 1 pth nd f : C C, then we define the integrl of f long to be b f(z)dz = f((t)) (t)dt. In order for this integrl to exist in the sense we hve defined, we hve seen tht it suffices for the functions f((t)) nd (t) to be bounded nd continuous t ll but finitely mny t. Our definition of piecewise C 1 -pth ensures tht (t) is bounded nd continuous wy from finitely mny points (the boundedness follows from the existence of the left nd right hnd limits t points of discontinuity of (t)). For most of our pplictions, the function f will be continuous on the whole imge of, but it will occsionlly be useful to weken this to llow f((t)) finitely mny (bounded) discontinuities. 28 The simplest wy to see this is to use tht fct tht if φ is continuous nd f is Riemnn integrble, then φ f is Riemnn integrble.

19 METRIC SPACES AND COMPLEX ANALYSIS. 53 Lemm If : [, b] C be piecewise C 1 pth nd : [c, d] C is n equivlent pth, then for ny continuous function f : C C we hve f(z)dz = f(z)dz. In prticulr, the integrl only depends on the oriented curve []. Proof. Since is equivlent to there is continuously differentible function s: [c, d] [, b] with s(c) =, s(d) = b nd s (t) > 0 for ll t [c, d]. Suppose first tht is C 1. Then by the chin rule we hve d f(z)dz = f((s(t)))( s) (t)dt = c d c b = = f((s(t)) (s(t))s (t)dt f((s)) (s)ds f(z)dz. where in the second lst equlity we used the chnge of vribles formul. If = x 0 < x 1 <... < x n = b is decomposition of [, b] into subintervls such tht is C 1 on [x i, x i+1 ] for 1 i n 1 then since s is continuous incresing bijection, we hve corresponding decomposition of [c, d] given by the points s 1 (x 0 ) <... < s 1 (x n ), nd we hve d f(z)dz = f((s(t)) (s(t))s (t)dt c n 1 = s 1 (x i+1) i=0 s 1 (x i) n 1 xi+1 = = i=0 b x i f((s(t)) (s(t))s (t)dt f((x)) (x)dx f((x)) (x)dx = f(z)dz. where the third equlity follows from the cse of C 1 pths estblished bove. Definition If : [, b] C is C 1 pth then we define the length of to be l() = b (t) dt. Using the chin rule s we did to show tht the integrls of function f : C C long equivlent pths re equl, one cn check tht the length of prmetrized pth is lso constnt on equivlence clsses of pths, so in fct the bove defines length function for oriented curves. The definition extends in the obvious wy to give notion of length for piecewise C 1 -pths. More generlly, one cn define the

20 54 KEVIN MCGERTY. integrl with respect to rc-length of function f : U C such tht U to be b f(z) dz = f((t)) (t) dt. This integrl is invrint with respect to C 1 reprmetriztions s: [c, d] [, b] if we require s (t) 0 for ll t [c, d] (the condition s (t) > 0 is not necessry becuse of this integrl tkes the modulus of (t)). In prticulr l() = l( ). The integrtion of functions long piecewise smooth pths hs mny of the properties tht the integrl of rel-vlued functions long n intervl possess. We record some of the most stndrd of these: Proposition Let f, g : U C be continuous functions on n open subset U C nd, η : [, b] C be piecewise-c 1 pths whose imges lie in U. Then we hve the following: (1) (Linerity): For α, β C, (αf(z) + βg(z))dz = α (2) If denotes the opposite pth to then f(z)dz = f(z)dz. f(z)dz + β g(z)dz. (3) (Additivity): If η is the conctention of the pths, η in U, we hve f(z)dz = f(z)dz + f(z)dz. η (4) (Estimtion Lemm.) We hve f(z)dz sup f(z).l(). z Proof. Since f, g re continous, nd, η re piecewise C 1, ll the integrls in the sttement re well-defined: the functions f((t)) (t), f(η(t))η (t), g((t)) (t) nd g(η(t))η (t) re ll Riemnn integrble. It is esy to see tht one cn reduce these clims to the cse where is smooth. The first clim is immedite from the linerity of the Riemnn integrl, while the second clim follows from the definitions nd the fct tht ( ) (t) = ( + b t). The third follows immeditely for the corresponding dditivity property of Riemnn integrble functions. For the fourth prt, first note tht ([, b]) is compct in C since it is the imge of the compct set [, b] under continuous mp. It follows tht the function f is bounded on this set so tht sup z ([,b]) f(z) exists. Thus we hve f(z)dz b = f((t)) (t)dt b η f((t)) (t) dt b sup f(z) (t) dt z = sup f(z).l(). z

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