Complex Analysis. Preliminaries and Notation For a complex number z = x + iy, we set x = Re z and y = Im z, its real and imaginary components. Any nonzero complex number z = x + iy can be written uniquely in the polar form z = re iθ where r = z = x 2 + y 2 and θ = tan y/x. r is called the norm, modulus, or absolute value of z and θ is called the argument of z, written arg z. Notice that arg z is well defined only up to multiples of 2π. Euler s formulas e iz = cosz + i sinz and e iz = cosz i sinz give cosz = eiz + e iz 2 and sinz = eiz e iz 2i. Complex numbers can sometimes be used an intermediate step to solve problems which are given entirely in terms of real numbers. Example. Let n 2 be an integer. Show that n k= sin kπ n = n 2 n. Solution. Set ω := e 2πi n. Thus ω n/4 = e πi/2 = i and ω n =. sin kπ = ekπi/n e kπi/n n 2i = ωk/2 ω k/2 2i = ωk 2iω k/2 Therefore n k= sin n kπ = n = ω k 2iω = k/2 k= n k= ωk 2 n i n ω n/4 = n n k= ωk 2 n i n ω = +2+...+n /2 n k= ωk 2 n i n i n = n k= ωk 2 n i n ω nn /4 n k= ωk 2 n n Since ω k n = for each k, each factor in the numerator satisfies the equation z+ n =. In other words, they satisfy z n + nz n + n 2 z n 2 +... + nz =. Since none are zero, they satisfy pz := z n + nz n 2 + n 2 z n 3 +... + n =. Therefore they are all the roots of the degree n polynomial pz. In the general, the product of all the roots of a degree d polynomial is d times the constant term. Thus in this case, the numerator is n n.
. Complex Differentiation f : C C. We shall assume that f is defined on a domain D which is open and path connected meaning that any two points in D can be joined by a path within D. Using real and imaginary components we can write z = x+iy and fz = ux, y+ivx, y for real-valued functions u and v, and in this way, when convenient, we can regard f as a function fx, y = ux, y, vx, y : R 2 R 2. For example, when c = a+bi C the function z cz : C C corresponds to the function x, y ax by, ay + bx : R 2 R 2 or equivalently, the function given by x y a b b a The complex derivative of f is defined by x. y f z := lim z z fz fz z z, if the limit exists. The assumption that the limit exists implies that the same value is obtained if the limit is taken as z approaches z along any line. Approaching along the x-axis gives f x + iy = lim x x fx + iy fx + iy x + iy x + iy Similarly approaching along the y-axis gives = x fx, y = x u + i x vx, y. f x + iy = i yu + i y vx, y. 2 Equating real and imaginary parts in and 2 gives Cauchy-Riemann Equations: x u = y v, x v = y u Conversely, suppose f : C C is differentiable when regarded as a function R 2 R 2 and also satisfies the Cauchy-Riemann Equations. The Jacobian matrix for the derivative is given by ux u Df = y, v x which, upon substitution from the Cauchy-Riemann equations, becomes the matrix corresponding to z u x + iv x z. It follows that f is complex differentiable with derivative f z = u x + iv x. Summing up, we have: Theorem. Let f : C C be differentiable when regarded as a function from R 2 R 2. Then f is differentiable as a complex function if and only if it satisfies the Cauchy-Riemann equations. 2 v y
A function which is differentiable at every point in a domain D is called holomorphic on D. In polar form, the CR-equations are as follows. Set x = r cos θ, y = r sin θ. Then and similarly u r = u u v v cos θ + sin θ = cos θ x y y x sin θ = v r θ v r = u r θ. Example. Let D = {z C Re z > }. Recall that arg z is well defined only up to multiples of 2π. Define f : D C by fz = log z + i argz where we choose the value of argz which lies in π, π. u x = logr x = r r x = x r x2 + y = x 2 r r = x r 2 and v x = θ x = θ x = tan y/x x = y/x2 + y/x 2 = y x 2 + y 2 = y r 2 Similarly we can calculate u v and and verify that the Cauchy-Riemmann equations are y y satisfied. Therefore fz is differentiable. According to, f z = u x + i v x and so df dz = x iy r 2 = x + iy = z A matrix of the form x y a b b a x. y corresponds to the linear transformation consisting of the composition of rotation by cos a/a 2 + b 2 and scalar multiplication by a 2 + b 2. Thus the Jacobian matrix for Df corresponds to the composition of rotation by arg f z and multiplication by f z. 3
2. Complex Integration For a differentiable parameterized curve t = xt+iyt : [a, b] C let t C denote the derivative t := x t + iy t. As discussed in MATB42, t gives the tangent vector to the curve t, corresponding to the velocity at time t of a point moving along curve with position t at time t. Let f : B C be continuous where B is a domain containing the curve. That is, [a, b] B. Assume that t is a continuous function of t. Define fz dz by fz dz := b a f t t dt where the right hand side is defined by taking integrals of the real and imaginary components. In other words, b fz dz := Re f t b t dt + i Im f t t dt. a a More generally, for a piecewise differentiable curve, the integral can be defined by adding the integrals on the subintervals on which is differentiable. It is also possible to relax the condition that f be continuous, although we shall not need to consider such cases. As in line integrals in MATB42, the sign of the answer depends upon the orientation of the curve which is determined by the given parameterization. Example. Compute z dz where is the straight line joining to + i/2. Solution. Parameterize by t = t + it/2, t. z dz = t+it/2dt+it/2 = t+it/2+i/2 dt = +i/2 2 t dt = 2 +i/22. Example. Compute C z2 dz where C is the unit circle, oriented counterclockwise. Solution. Parameterize C by Ct = cost + i sint, t 2π. 2π 3 z 2 2d cost + i sint dz = cost + i sint cost + i sint dt = t= 3 =. t= C Example. Compute dz where C is the circle of radius R, oriented counterclockwise. C z Solution. Parameterize C by Ct = R cost + ir sint, t 2π. 2π z dz = 2π R sint + ir cost dt = i dt = 2πi. R cost + ir sint C The calculation could equivalently be expressed by writing Ct = Re it, t 2π, giving 2π 2π z dz = dre it /Re i t = i dt = 2πi. C Note in particular that the answer is independent of the radius of C. 4
Proposition. For f = u + iv, fz dz = ux, y dx vx, y dy + i Proof. fz dz = = b a b vx, y dx + ux, y dy u xt, yt + iv xt, yt x t + iy t dt u xt, yt x t v xt, yt y t a b + i v xt, yt x t + u xt, yt y t a = ux, y dx vx, y dy + i vx, y dx + ux, y dy Recall MATB42 that a differential form ω is called closed if dω =. Given f = u + iv, let ω = ux, y dx vx, y dy + ivx, y dx + ux, y dy. Then dω = x u y v + i y v+ x u dx dy. Therefore, if f is holomorphic, the Cauchy-Riemann equations imply that dω =. Theorem. Let B be a closed region whose boundary B consists of piecewise differentiable curves. Let f be holomorphic in a domain containing B. Then fz dz =. B Proof. Given f = u + iv, let ω = ux, y dx vx, y dy + ivx, y dx + ux, y dy. Since f is holomorphic, dω =, as above. Therefore using Stokes Theorem MATB42 gives B fz dz = B ω Stokes Thm. dω =. = B Note: The preceding theorem does not require that B be connected. For example, if B is an annular-shaped region lying between an outer curve C 2 and an inner curve C each oriented in the same direction, say counterclockwise then the theorem gives C 2 f dz C f dz = so C 2 f dz = C f dz. As in Stokes Theorem, the signs are determined by picking an orientation on D and using the induced orientations on C 2 and C. If f is holomorphic not only on B but also on the interior of C i.e. the hole in the annular region B then applying the theorem to that region gives the stronger statement that C f dz = and C 2 f dz =. Thus, in particular, if is any simple closed curve which goes once counterclockwise around the origin then the earlier example gives z dz = C z dz = 2πi. The converse to the preceding theorem also holds. Theorem Moreira. Let f : D C be a continuous function such that f dz = for any closed curve D. Then f is holomorphic. Proof. Let f = u + iv, and set ω := ux, y dx vx, y dy + ivx, y dx + ux, y dy. As above, using Stokes Theorem gives that B dw = fz dz = for any bounded B region B D. Since dw integrates to over any bounded region, dw =, which is equivalent to saying that f satisfies the Cauchy-Riemann equations. 5
Theorem Cauchy Integral Formula. Let f be holomorphic on a domain containing a simple closed counterclockwise-oriented curve together with its interior. Then for any z in the interior of, fz = 2πi fw w z dw. Proof. By translation i.e. the change of variable z := z z, it suffices to consider the special case z =. Let D be the interior of. The closure of D is D = D D = D, the union of D with its boundary. Set Fw := { fw f w if w ; f if w =. Since 2πif = f w dw earlier example we need to show that Fw dw =. It is clear that F is continuous throughout D and differentiable in the interior except possible at w =. It is actually also differentiable at w = but this is not so obvious and we do not need it. Since D is closed and bounded i.e. compact in the terminology of MATB43 continuity of F implies MATB43 that F is bounded on D. That is, there exists M R such that Fw M for all w D. Let C be a circle within D centred at, oriented counterclockwise. The preceding theorem implies that Fw dw = Fw dw. C Parameterize C by Ct = R cost + ir sint, t 2π, where R is the radius of C. Then C t = R sint + ir cost = R, so Fw dw = C 2π 2π Fw dw = F Ct 2π C t dt F Ct C t dt MRdt = 2πMR. Since this inequality holds for arbitrarily small R, Fw dw = as desired. The special case when fz = says that if is any simple closed counterclockwiseoriented curve about z, then 2πi z z dz =. more generally, if is a closed curve with z, we define the index, or winding number of about z, denoted I, z, by I, z := 2πi z z dz. It is always an integer, and represents the number of times circles around z, where curves oriented in the clockwise direction are considered to circle a negative number of times. By translation, consider the special case z = in which case we write simply I for I,. If we write z = x + iy, then z = z z 2 = x iy x 2 + y 2 6
so by our earlier Proposition, 2πi z dz = x 2πi x 2 + y dx + y 2 x 2 + y dy + y 2 2π x 2 + y dx + x dy 2 x 2 + y 2 = dlogx 2 + y 2 + y 4πi 2π x 2 + y 2 dx + x x 2 + y 2 dy = + y 2π x 2 + y dx + x 2 x 2 + y dy 2 = y 2π x 2 + y dx + x dy 2 x 2 + y 2 so this agrees with the MATB42 definition of winding number. Another geometrical interpretation of I is as follows. Start at some point w on and let θ = arg w, normalized to lie in [, 2π. As w moves around the curve, argw changes continuously, returning to θ + 2πn when we get back to w. Claim: I = n. Proof. Paramterize by θ = rθe iθ, θ 2πn. Then 2πi z dz = 2πn 2πi r rθe iθ dθ = 2πn θ eiθ+irθeiθ 2πi = 2πi log rθ θ=2πn + θ= 2π dr rθ dθ dθ+ 2π 2πn 2πn dθ dθ = + n = n. We noted earlier that for holomorphic fz, the formula fz dz = does not B require that B be connected. We can similarly generalize Cauchy s Integral Formula so that it applies to cases where B is not a single simple closed curve, but perhaps a disjoint union of curves. Corollary. Let B be a closed region whose boundary B consists of piecewise differentiable curves. Let f be holomorphic in a domain containing B. Then for any z in B, fz = 2πi B fw w z dw. Proof. Pick a simple closed curve in B such that z lies in the interior of and let B = B {interior of } and orient it in the counterclockwise direction. Then gw := fw/w z is holomorphic throughout B so B gw dw =. The boundary of B is the disjoint union of B and, and so, taking orientation into account, we get B fw dw = w z fw w z dw = 2πifz. 7
Notice that Cauchy s Integral Formula implies that for a holomorphic function f the values of f on completely determine the values of f at any point inside. In particular, Corollary. Let f be holomorphic on a domain containing the closed ball B = B R [z ]. Then fz = 2π fz + Re iθ dθ. 2π Proof. fz = 2πi B fw dw = fz + Re iθ w z 2πi B Re iθ ire iθ dθ = 2π fz + Re iθ dθ. 2π In other words, the value of f at the centre of a circle is the average of its values on the circle. Corollary Maximum Modulus Principle. Let f be holomorphic on a domain containing a closed bounded set B. Then the maximum value M of f occurs on B and unless f is constant, fz < M for all z in the interior of B. Proof. A continous function on a closed bounded set attains a maximum MATB43. The average value of a function can equal its maximum value only if the function is constant. Therefore if fz = M for some z in the interior then fz = M in some neighbourhood of z. Since our domains are connected, this implies fz = M for all z in the domain. Therefore it suffices to show Lemma. If fz is constant then fz is constant. Proof. Let f = u + iv and suppose fz = M for all z in the domain. If M = then f = so suppose M >. Differentiating M 2 = fz 2 = u 2 + v 2 gives 2uu x + 2vv x = and 2uu y + 2vv y =. After substituting from the Cauchy-Riemann equations we get uu x + vu y = and uu y vu x =. Therefore u 2 u x = uvu y = v 2 u x and v 2 u y = uvu x = u 2 u y. Thus u 2 + v 2 u x = u 2 + v 2 u y =. Since u 2 + v 2 = M 2, we get u x = u y = so u is constant, and similarly v is constant. Theorem Cauchy Integral Formula for higher derivatives. Let f be holomorphic on a domain containing simple closed counterclockwise-oriented curve together with its interior. Suppose that f is holomorphic throught the interior of. Then f is infinitely differentiable on the interior of with and for any z in the interior of the nth derivative is given by f n z = n! 2πi fw dw. w z n+ Proof. Suppose by induction that the theorem is known for f n, the case n = being the preceding theorem. Thus to prove the theorem it suffices to prove the following lemma which completes the induction. 8
Lemma. Let f be holomorphic on a domain containing a simple closed counterclockwiseoriented curve. For z in the complement of define functions F n z by F n z := n! fw 2πi w z dw. Then F n+ n is differentiable on the complement of with derivative equal to F n+. For future reference, notice that the Lemma is stronger than needed in the proof of the theorem in that the definition of the functions F n and the proof that they are differentiable do not require f to be defined throughout the interior of and the conclusion also applies to points outside. Proof. Let z lie in the complement of. F n z F n z z z n +! 2πi fw dw w z n+2 = n! z z 2πi n +! fw 2πi w z = n! 2πi fw n! dw w z n+ 2πi n+2 dw fw dw w z n+ fw w z n+ fw w z n+ n + fw w z n+ For any a and b, applying xk+ y k+ = k x y j= xj y k j with x = /a and y = /b gives b a a n+ b n+ n + b n+2 = b a a n b a j b n j n + b n+2 = ab = = = = j= n j= j= a j b n + n j b n+2 n a j+ b n j+ n + n a j+ b n j+ b n+2 j= b n j+ b n j+ = b a n j= n j= n j= i= Setting a = w z and b = w z gives fw w z n+ fw w z n+ n fw w z n+ 9 a j+ b j+ b n+2 a j b a i b j i i= j a i+ b n i+2 dw = z z gz, w dw dw
where Thus gz = n j j= i= F n z F n z n +! z z 2πi fw w z i+ w z n i+2. fw w z n+2 dw z z gz, w dw Since gz, w and t are continuous on the closed bounded set they are bounded MATB43, and so gz, w dw is bounded, showing lim z z z z gz, w dw =. Hence F n z n +! fw dw =. 2πi w z n+2 Corollary Liouville. If f is holomorphic throughout C and f is bounded then f is constant. Proof. Suppose fz < M for all z. Then given z C, letting applying the Cauchy Integral formula for the first derivative to the circle of radius R about z gives f z = fz +Re iθ 2πi ire iθ dθ B Re iθ 2π M 2 2π R R dθ = M/R. Since R is arbitrary, this implies 2 f z for all z and so f is constant. 3. Power Series Expansions Recall the following from MATB43. A sequence of functions f n z is said to converge uniformly to fz if ǫ > N independent of z such that n N fz f n z < ǫ for all z in the domain. 2 If f n continuous on D and f n converges uniformly to f then f is continuous on D and lim n D f nz dz = fz dz. D 3 To any power series fz = n= c nz n there is an associated radius of convergence R [, ] such that fz converges absolutely for z < R and diverges for z > R. Within its radius of convergence, fz is differentiable and integrable on bounded regions with f z = n= c nnz n and anti-derivative c n n= n+ zn+ and the radius of convergence of the differentiated and integrated series are R. 4 A continuous function fz on a closed bounded domain has both a minimum and a maximum. In particular fz is bounded. For any curve our assumption that t is continous implies by 4 that it is bounded. It follows that: Proposition. If f n z converges uniformly to fz on a domain including the image of then f n t t converges uniformly to f t t and so applying gives lim n f nz dz = fz dz.
For differentiable functions of a real variable, even if f n x converges uniformly to fx in the vicinity of a point a, it need not be true that fx is differentiable at a. However for functions of a complex variable, we can use the fact that corresponding statement for integration the preceding proposition to deduce the result for differentiation. Theorem. Suppose f n z is a sequence of functions which converge uniformly to fz on D. If f n z is holomorphic on D for all n, then fz is holomorphic on D. Proof. For evey closed curve in D, fz = lim f n z = lim =. n n Therefore fz is holomorphic in D by Moreira s theorem, From 3 we know that a function which is representable by a power series is differentiable within its radius of convergence. For complex differentiable functions we show that the converse is true. a x = a x n a n+ within its radius of con- Observe that x/a = x n a n= a = n= vergence x < a. Suppose that fz is differentiable on a domain D and let z belong to D. Since D is open, the closed ball B r [z ] := {z C z z r} is contained in D for sufficiently small r. Let C be the circle C := B r [z ] := {z C z z = r}, oriented counterclockwise. Cauchy says fz = dw for all z in the open ball C w z B r z := {z C z z < r}. For simplicity we will use translation to consider expansions about z =. Then for z B r we have z < w for any w C and so fz = C fw w z dw = C fw fwz n w n+ dw = z n n= n= C fw w n+ dw = n= z n fn. n! Thus within C, the Taylor series of fz converges to fz. Since r was arbitrary subject to the condition that B r z be contained in D the Taylor series expansion is valid for any z whose distance to z is less than the distance of z to the boundary of D. In particular, the radius of convergence of the Taylor series of fz about z is at least as large as the distance from z to the boundary of D. It is also clear that the power series expansion of fz about any point is unique. That is, if n= a nz z n is any series which converges to fz in some neighbourhood of z then by successively differentiating and evaluating at z = z we find that a n = f n z /n! so the series is the Taylor series of fz. A function with the property that at every point in its domain, the Taylor series of the function converges to the function within some sufficiently small radius of the point is called analytic. Therefore for complex functions, differentiable and analytic are equivalent. This is in contract to functions of a real varable where if fx is differentiable: a f not need be differentiable more than once so the Taylor series of f might not be defined. b Even if f is infinitely differentiable, the Taylor series of f need not converge to f.
Since the n= zn f n n! converges within C, its radius of convergence is at least r. Thus within C we can define a function gz by the convergent series gz := n= z n+ fn n +! and it will have the property with it is differentiable within C with g z = fz. In summary, unlike functions of a real variable, complex differentiable functions have the following properties: Theorem. fz is analytic at every point in its domain 2 f is locally the derivative of some complex function. That is, at every point of the domain of f there a function gz defined in some neighbourhood of that point such that g z = fz. 3 If f n z is a sequence of differentiable functions which converge uniformly to fz on D then fz is differentiable on D. Notice that 2 does not say that there is a single function gz defined throughout all of the domain of fz such that g z = fz. Laurent Series Consider a function fz which is holomorphic throughout an annular region A := {z C R < z z < R 2 } where R < R 2. By translation, assume z =. Given p A, choose concentric circles C and C 2 about such that R < radius of C < p < radius of C 2 < R 2 and orient them in the counterclockwise direction. By the Corollary to Cauchy s Integral Formula, fz = fw 2πi C 2 w z dw fw 2πi C w z dw. Set f z := fw 2πi C w z dw and f 2z := fw 2πi C 2 w z dw so that fz = f 2z f z. According to the lemma in the proof of Cauchy s Integral Formula, the functions f and f 2 are defined and analytic on the complement of C and C 2 respectively. Hence f 2 is analytic throughout the interior of C 2, and given in this region by the power series expansion f 2 z = n= a nz n where a n = f n 2 /n! = fw 2πi w dw, which holds, in particular, n+ at the point p. For the analogous statement concerning the function f make the change of variable ζ := /z. That is, set f ζ := f /ζ = fw 2πi C w /ζ dw. Letting C be the circle with counterclockwise orientation whose radius is the reciprocal of the radius of C, the change of variable w := /ω gives f ζ = f ω 2πi /w /ζ C ω 2 dω = ζ f ω 2πi C ω ω ζ dω = ζ gω 2πi C ω ζ dω 2
where gω = f ω ω. As with f 2, we use the fact that hζ := gω 2πi C ω ζ the interior of C to obtain the expansion hζ = n= c nζ n where c n = 2πi Therefore f ζ = ζhζ = n= c nζ n+ = n= b nζ n where b n = c n = gω 2πi C ω dω = n 2πi = fww n+ w 2πi C 2 dw = 2πi ω C f ω n+ dω C fww n dw. dω is analytic in gω C ω dω. n+ The expansion f ζ = n= b nζ n is valid in the interior of C which contains /p. Therefore f z = f ζ = n= b nζ n = n= b n/z n holds at p. Using the Corollary to Cauchy s theorem, if we choose a curve about p which lies within the annulus and give it the counterclockwise orientation, we can write our coefficents as a n = fw dw and b 2πi w n+ n = 2πi fwwn dw. To summarize, Theorem. Let fz be holomorphic in an annular region R < z z < R 2 where R < R 2. Then fz = a n z z n + n= n= b n z z n where a n = fw 2πi w z dw and b n+ n = 2πi fww z n dw for any curve within the annulus such that p lies in the interior of. This is called the Laurent expansion of fz in the region. If fw is holomorphic within the entire ball of radius R 2 about z, then the b n s are all and the Laurent expansion of fz reduces to its Taylor expansion. 3
4. Analytic Continuation Theorem. Suppose that fz are gz are analytic within a domain D. For any z D, if there exists a sequence of points z n z with z n z, such fz n = gz n for all n then fz = gz for all z in D. Note: Recall that our definition of domain was an open path connected set. Proof. By substraction, it suffices to consider the special case where gz =. Let d be the distance from z to the boundary of D. Let fz = k= a kz z k be the Taylor series expansion of fz, which, as noted earlier, converges to fz throughout B := B d z. Since fz is differentiable at z it is continuous so fz = lim n fz n =. Therefore substituting into the power series gives a =. Write fz = z z hz where { fz/z z if z z hz = ; f z if z =. The series k= a k+z z n converges to hz for z z, so by continuity it also converges to gz = f z when z = z. Since z n z, hz n = fz n /z n z = so substituting into the power series gives a =. Proceeding, we inductively conclude that a k = for all k and therefore fz = throughout B. Now let p be an arbitrary point in D. Let X = {z D fz = gz}. Since we assumed that D is path connected, there exists a path : [, ] D joining z to p. Set t := sup{t [, ] t X}. Since the definition of supremum implies that there is a sequence of points in X converging to t, the preceding shows that X contains an open ball about t. This can happen only if t =, and so p lies in X. An analytic function g which extends an analytic f to a larger domain is called an analytic continuation of f. That is, if fz and gz are analytic functions with D := domain of f D := domain of g and fz = gz for all z D then g is called an analytic continuation of f to D. The preceding theorem implies that any two analytic continuations g z, g 2 z of fz to the same domain D are equal. However this does not imply that analytic continuations of fz to different domains must agree on the intersection of those domains. In other words it might be possible that fz has analytic continuations g z defined on D D and g 2 z defined on D 2 D for which there exists p D D 2 for which g p g 2 p. This can happen only if there is no path in the intersection D D 2 joining p to a point in D. Indeed, if D D 2 contains a path joining p to a point q in D, we can form an open connected subset D of D D 2 containing. Then D is a domain and since by hypothesis g z and g 2 z agree in a neighbourhood of q where both are given by fz applying the preceding to the domain D shows that g z and g 2 z agree on D and in particular at p. Example. Let D = C {x, x } and D 2 = C {, y y }. Recall that arg z is well defined only up to multiples of 2π. Define g : D C by g z = log z +i argz where we choose the value of argz which lies in π, π. Define g 2 : D 2 C by g 2 z = log z + i argz where we choose the value of argz which lies in π/2, 3π/2. We showed earlier that g and g 2 are differentiable with g z = g 2 z = /z. Since z = z e i arg z by definition, e gz = e log z e i arg z = z. Similarly e g2z = z. 4
Therefore each of g and g 2 might deserve the name logz. They are called branches of the logarithm function. If we let D = open st quadrant = {z C Re z > and Imz > } and define f : D C by fz = log z + i argz where we choose the value of argz which lies in, π/2, then each of g and g 2 are analytic continuations of fz. However although the points in the 3rd quadrant are in D D 2, the values of g z and g 2 z on these points differ by 2π. Proposition. Let R be the radius of convergence of fz = n= a nz p n and suppose < R <. Then there must be at least one point q with q p = R such that f cannot be analytically continued to any domain containing q. Proof. Let B = B R p be the open ball of radius R about p, and let B be the boundary circle of B. Suppose that for each point q B, there exists an analytic continuation g q z of fz to a domain D q containing q. Let ˆD = q B D q. For any q, q 2 on the boundary, applying the preceding uniqueness theorem to D q D q2 shows that g q z and g q2 z agree on D q D q2. Therefore the functions {g q z} q B piece together to produce a well defined function on gz on ˆD. The function g z is differentiable since in the neighbourhood of each point of ˆD it equals some differentiable function g q z. Since ˆD is open and B is closed, the distance d of B to the boundary of ˆD is positive, which means that the distance R+d from p to the boundary of ˆD is greater than R. But according to our earlier discussion Section 3, the radius of convergence of the Taylor series of gz about p which is the same as fz is at least as large as the distance of p to the boundary of ˆD, contradicting the definition of R. Let fz be analytic throughout D. At any point p in D we can expand fz into a power series fz = a n z p n whose radius of convergence R is at least as large as the distance d from p to the boundary of D where R and d depend upon p. If there is a point p in D at which d < R, then we can create an analytic continuation gz of fz to D B R p by defining { fz if z D; gz := a n z p n if z B R p.. By applying this procedure to more points in the extended domain one might be able to extend the domain still further. Recall MATB42: a two curves,, from p to q in a subset X of R n are called homotopic in X, written, if one can be continuously deformed into the other within the subset X. More precisely, if there exists a continuous function H : [, ] [, ] X such that H, t = t, H, t = t and Hs, = p for all s and Hs, = q for all q. In other words, the family of curves defined by s t := Hs, t interpolates continuously from to. b a connected subset X is called simply connected if any two curves in X with the same endpoints are homotopic in X. For subsets X of R 2, this is equivalent to saying that X is simply connected iff for every closed curve lying in X, the interior of also lies in X. 5
Theorem Monodromy. Let fz be analytic on a domain D. Let X be a simply connected domain containing D. Suppose that for every curve X there is an analytic continuation of fz to some domain containing. Then there exists a unique analytic continuation of fz to X. Proof. If the analytic continuation exists, it is unique by an earlier theorem. Pick a point p D. Given q X, choose a path joining p to q. By hypothesis, there exists an analytic continuation g z to some domain D containing. We wish to show that g q is independent of the choices involved. It is clear from the uniqueness theorem that the same value of g q would be obtained if we choose a different domain containing, but what happens if we choose a different path from p to q. Suppose that is another path joining p to q. Since X is simply connected, there exists a homotopy H : [, ] [, ] X from to and set s t := Hs, t. Let d be the distance of ImH to the boundary of X. d >, since Im H is compact closed and bounded and X is open. Suppose h is some point in H and suppose k h z is any analytic continuation of fz to some domain containing h. The straight line joining h to any point whose distance to h is less than d lies in X, and therefore by hypothesis there is an analytic continuation of fz to some domain containing that line. By the preceding proposition, this means that the radius of convergence of the Taylor series of k h z about h must be at least d. Let S ={s [, ] an analytic continuation of fz to some domain containing s } and let ŝ = sup S. We show that ŝ =. Choose s S such that ŝ s < d. Let g s z be an analytic continuation of fz to a domain D s containing s. According to the preceding discussion, for each h s, the radius of convergence of the Taylor series of g s z about h is at least d, so fz has an analytic continuation g h z to D s B d h. According to the uniqueness theorem, these functions for various h s agree whenever their domains overlap, so they piece together to produce a well defined analytic continuation ĝz on ˆD := D s h s B d h. Since s + d > ŝ, unless ŝ =, ˆD contains s for some s > ŝ, contradicting the definition of ŝ. Therefore ŝ = and ĝz is an analytic continuation of fz to a domain containing all of H and in particular contains both and. Therefore the value g q obtained using the path equals the value g q obtained using the path. The conclusion is that there is a well defined extension of fz to a function gz defined on X where for any q X, gq is defined by chosing any path from p to q, and setting gq := g q where g z is any analytic continuation of fz to a domain containing z. The resulting function gz is differentiable at each point q since it equals some differentiable function in the neighbourhood of q. Theorem Schwarz Reflection Principle. Let D be a domain which is symmetrical about the x-axis. i.e. z D if and only if z D. Let I = D x axis. Let U = {z D Imz > } the upper half of D and let Û = U I. Let f : Û C be a continuous function such that the restriction f U is holomorphic and the restriction f I is real-valued. Then f has a holomorphic extension to D. Proof. Extend f to D Û by setting fz := f z for z D Û. Since f is continuous on Û, by symmetry f D has a continuous extension to I and since fp = fp for Û p I, the symmetry in the definition of f implies that this extension agrees with fi. 6
Thus f is continous on D. By Moreira s theorem, it suffices to show that fz = for any closed curve D. For a curve which is entirely contained in U, this is clear, since fz is holomorphic on U and for a curve contained entirely in the refection D Û of U it is also clear, by symmetry. Therefore consider a curve which intersects I. We may write as a union of curves each of which lies entirely in the upper half plane or the lower half plane so this reduces the problem to showing that fz = for a curve which lies entirely in one the two half-planes. By slightly perturbing the portion of running along the x-axis, such a curve can be approximately as closely as desired by a curve which lies entirely within U or D Û. Since, as noted above, fz dz =, and we can choose the so as to make the difference fz dz fz dz arbitrarily small, it follows that fz dz =. 5. Residues Proposition. Let be a simple closed counterclockwise-oriented curve about a point p. Then { 2πi if n = ; w pn dw = otherwise. Proof. If n, z p n is holomorphic everywhere so the integral is by Cauchy s theorem. According to Cauchy s Integral Formula for higher derivatives, 2πi dw = w p m m! gm p, where gz =, and applying this with m := n gives the result when n <. Let fz be homorphic in an annular region < z p R about some point p. Let B := B R [p] be the closed ball of radius R about p and let B = B {p} be the punctured ball obtained by removing the point p. Let fz = n= a nz p n + b n n= be the z p n Laurent expansion of fz in B. Then applying the proposition gives fz dz = 2πi B 2πi n= c n B z p n dz = c. The coefficient c in the Laurent expansion of fz about p is called the Residue of f and p, written Res p fz. Notice that if is any simple closed counterclockwise-oriented curve about a point p and fz is holomorphic in a domain containing together with all of its interior except possibly p then for any circle C about p contained in, applying Cauchy s Theorem to the region between and C gives 2πi fz = 2πi fz, so the answer is again given C by Res p fz. Consider now a simple closed counterclockwise-oriented curve and a function fz which is holomorphic on a domain containing together will all of its interior except possibly a finite number of points p, p 2,..., p k. Carve the interior of into subregions B, B 2,..., B k where B j contains p j but none of the other p s. Then, assuming all the 7
curves are given the counterclockwise orientation, fz dz = k j= B j fz dz, since the integrals over the extra curves in the RHS introduced by the division into subregions each appear twice with opposite directions and cancel out. Therefore fz dz = 2πi 2πi k j= B j fz dz = k Res pj fz. Suppose fz is a function which is holomorphic in a punctured neighbourhood of a point p. That is, there is a ball B = B R p about p such that fz is holomorphic on the punctured ball B {p}. Then p is called an isolated singularity of f. Let fz = n= a nz p n + n= b n z p n be the Laurent expansion of fz in the neighbourhood of an isolated singular p. If b n = for all n then p is called a removable singularity of f. If there exists an integer N such that b n = for all n > N, then p is called a pole of f of order k, where k is the largest integer for which b k. A pole of order is called a simple pole. If p is not a pole of f i.e. b n for infinitely many n, then p is called an essential singularity of f. If f is holomorphic on a region A everywhere except for a finite number of islated singularties none of which are essential singularities, then f is called meromorphic on A. Note that the uniqueness theorem in the analytic continuation section implies that the zeros of a nonconstant meromorphic function are isolated. Our previous result can be stated as Residue Theorem. Let be a simple closed curve counterclockwise-oriented curve and let fz be meromorphic on a domain containing with no zeros or poles on. Then fz dz = 2πi where p, p 2,..., p k are the poles of f within A. j= k Res pj fz Proposition. Let p be an isolated singularity of fz. Then p is a removable singularity if and only if any of the following conditions hold in which case all must hold: fz is bounded in a deleted neighbourhood of p. 2 lim z p fz exists 3 lim z p z pfz = Proof. If p is a removable singularity then in some deleted neighbourhood of p, fz = n= a nz p n. In this case we can extend fz to an analytic function in the neighbourhood of p by setting fp := a. It follows that all three conditions are satisfied. Conversely, if either conditions or 2 holds then obviously so does 3, so it suffices to show that condition 3 implies that the singularity is removable. Let C r denote the circle of radius r around p. b k = 2πi C r fww p k dw. Given ǫ >, condition 3 says that there exists r such that z p fz < ǫ whenever z p r. In particular, if z lies on C r then r fz < ǫ. Choosing a smaller r if necessary, we may assume r <. Therefore b k 2π j= C r r k ǫ/r = 2π 2πrrk ǫ/r = r k ǫ ǫ for every ǫ > and so b k =. 8
Computation of residues can be complicated, but there are some tricks which handle many common situations. If fz has a removable singularity at p then lim z p fz = a exists and the domain of f can be extended to include p by setting fp := a. Conversely, it is clear from the Laurent expansion that if lim z p fz exists then the singularity must removable with the limit equalling a. If the singularity of f at p is a pole of order k then z p k fz has a removable singularity at p. In particular, if k = then lim z p z pfz exists and equals the residue Res p fz and conversely if lim z p z pfz exists then k = and Res p fz = lim z p z pfz. As a consequence we have Proposition. Let fz = gz/hz where gz and hz are holomorphic at p with hp = and h p. Then Res p fz = gp/h p. Proof. h p = lim z p hz hp z p = lim z p hz z p and so lim z p z p hz = h p. Thus limz pfz = lim gzz p z p z p hz = gp h p. As above, the existence of the limit shows Res p fz = lim z p z pfz = gp/h p. Corollary. a If f has a zero of order k at p, then Res p f z/fz = k. b If f has a pole of order k at p, then Res p f z/fz = k. Proof. Use the convention that a pole of order k can also be called a zero of order k. With this convention, let m be the order of the zero of f at p, where m may be either positive or negative. In a punctured neighbourhood of p, set gz := f z/z p m and hz := fz/z p m. Since fz has a zero of order m at p, f z has a zero of order m at p, and so both gz and hz have removable singularites at p and thus have extensions to holomorphic functions at p. Therefore the proposition applies to give Res p f z/fz = Res p gz/hz = gp/h p f z/z p m = lim z p f z/z p m m fz/z p m f z = lim z p f z m fz/z p z pf z/fz = lim z p z pf z/fz m m = m m = m using that lim z p = z pf z fz is the order of the zero at p. 9
Corollary Argument Principle. Let be a simple closed counterclockwise-oriented curve and let fz be meromorphic on a domain containing with no zeros or poles on. Then f z dz = total number of zeros of fz within 2πi fz where the zeros are counted with multiplicity and poles are consider to be zeros of negative multiplicity. Since making the change of variable w = fz gives f w dw = dz the argument principle can be rewritten in terms of the index If of the curve f about as follows: Corollary Argument Principle. Let be a simple closed counterclockwise-oriented curve and let fz be meromorphic on a domain containing with no zeros or poles on. Then If = total number of zeros of fz within where the zeros are counted with multiplicity and poles are consider to be zeros of negative multiplicity. Example. How many zeros does the function fz = z 6 + 6z + have in the first quadrant? Solution. Let C be the portion of a large circle z = R which lies in the first quadrant. Let be the curve consisting of the line segment L := [, R] on the x-axis followed by the curve C from to R, followed by the line segment L 2 := [ir, ] on the y-axis from ir to. The restriction of fz to L is fx = x 6 + 6x + which is a positive real number for all x and thus arg fz remains constant at along L. For large R, the value of fz is approximately the same as the value of z 6 and in particular, the change in the argument of fz along C is the same as the change in the argument of z 6 on C. Over a complete circle, argz 6 changes from to 2π so on the quarter circle C it changes from to 3π. The restriction of fz to L 2 is fiy = y 6 + + 6iy, from which we see that, along L 2, fz begins in the second quadrant, slightly above the x-axis, argfz slightly less than 3π, moves into the first quadrant as we pass through y = 6 finishing at, on the x-axis. Thus along L 2, argz decreases from approximately 3π to 2π. Putting it all together, arg fz stays at along L, increases from to slighly less than 3π along C and then decreases by around π, going from from approximately 3π to 2π along L 2. Hence, for large R, the index If is and so fz has zero in the first quadrant. f z fz Example. How many zeros does the function fz = z 3 + 5z 2 + 8z + 6 have in the first quadrant? Solution. Again let C be the portion of a large circle z = R which lies in the first quadrant and let be the curve consisting of the line segment L := [, R] on the x-axis followed by the curve C from to R, followed by the line segment L 2 := [ir, ] on the y-axis from ir to. 2
The restriction of fz to L is fx = x 3 + 5x 2 + 8x + 6 which is a positive real number for all x and thus arg fz remains constant at along L. For large R, the value of fz is approximately the same as the value of z 3 and in particular, the change in the argument of fz along C is the same as the change in the argument of z 3 on C. Over a complete circle, argz 6 changes from to 6π so on the quarter circle C it changes from to 3π/2. The restriction of fz to L 2 is fiy = uy + ivy where uy = y 2 + 6 and vy = y 3 + 8y. Along L 2, arg fz = tan v u. uy > for y < 6/5 and uy < for y > 6/5. vy > for < y < 8 and vy < for y > 8. Therefore as we traverse L 2, arg fz begins in the third quadrant at approximately 3π/2, moves into the second quadrant as we pass y = 6/5, and then into the first quadrant as we pass y = 8, finishing at. Hence, for large R, the index If is and so fz has no zeros in the first quadrant. In some cases, determination of the number of zeros can sometimes be done more easily by comparison with a function whose number of zeros in known, according to the following corollary of the Argument Principle. Corollary Rouché s Theorem. Let be a simple closed counterclockwise-oriented curve. Let fz and gz be holomorphic on a domain containing and its interior, with no zeros or poles on. Suppose that fz gz < fz for all z on. Then total number of zeros of fz within = total number of zeros of gz within where the zeros are counted with multiplicity. Proof. Set hz = gz/fz. The hypothesis implies that hz < for all z on. Therefore the curve h is contained within the open disk of radius about and in particular does not contain in its interior. Thus Ih =. Applying the Argument Principle gives = h z hz dz = fzg z gzf z fz 2 fz gz dz = g z gz dz f z fz dz so another application of the Argument Principle gives that the result Example. How many zeros does the function kz = e z 4z 7 have inside the unit circle? Solution. Let fz = 4z 7. Then kz fz = e z. If z lies on the unit circle then kz fz = e z = e Rez e. However for z on the unit circle fz = 4z 7 = 4 z 7 = 4 > e. Thus fz kz < fz for all z on the unit circle. It follows that the number of zeros within the unit circle of kz is the same as that of fz, which is 7. Theorem. Let p be a zero of order k of a nonconstant holomorphic function fz. Then there exist an open neighbourhood U of p such that fu is an open neighbourhood of and every point in fu aside from has precisely k-preimages under f. Proof. Since fz is nonconstant, neither fz nor f z is identically zero. Therefore we can choose a closed ball B in the domain of fz containing p but no other zeros of f and 2
no zeros of f except possibly p. Let M be the minimum value of the restriction of fz to B. M > by choice of B. Set U := f B M Interior of B. For q B M define gq by gq := f z 2πi B fz q dz. The zeros of the function h q z := fz q count, with multiplicity, the number of preimages in B of q under f. By the argument principle # of preimages of q = h q z dz = gq. 2πi B h q z The function g takes on only integer values, so by continuity gq = g = k. If u p lies in U, then by choice of B, h fu u = f u and so u is a zero of h fu of multiplicity. Since the multiplicity of each preimage of any point q is, each such point must have k pre-images in B. Note also that since every point in B M has preimages in U, so that fu = B M and is, in particular, an open neighbourhood of. Corollary Complex Inverse Function Theorem. Let fz be holomorphic at p with f p. Then there exists an open neighbourhood U of p such that V := fu is an open neighbourhood of fp and the restriction f : U V is a bijection. The inverse function f : V U is also holomorphic with derivative given by f z. Proof. Apply the preceding theorem to gz := fz fp. Since g p = f p, the multiplicity of p as a zero of g is. Therefore k =, and choosing U as in the theorem, every point in gu including has precisely one preimage in U. Equivalently, f : U V is a bijection. Let g : V U be the inverse to f. If is any closed curve in V, then making the change of variable z = fw gives gz dz = g wf w dw = since it is the integral of the holomorphic function wf w over the closed curve g. Therefore g is holomorphic by Moreira s theorem, and its derivative is determined by applying the chain rule to z = f gz. Residues can be used to evaluate integrals MATC34. We will now examine how this process can sometimes be usefully reversed. The function sinπz has zeros precisely at the integers. We can sometimes make use of this to compute n= fn in cases where fn is the restriction to the integers of some meromorphic function. Suppose fz is a meromorphic function. Then fz is meromorphic with poles sinπz at the integers in addition to the poles of fz. The poles at the integers which are not singularities of fz are simple poles, while those at singularities of fz have higher order. If fn is not a singularity of f, according to our earlier formula, fz Res n = sinπz fn d sinπz /dz z=n = fn π cosπn = n fn π This is sometimes useful in evaluating alternating series, but for series of positive terms it is more useful to replace sin πz in the preceding discussion by tanπz and consider the 22
function π fz tanπz = π cotπzfz. Since d tanπz /dz z=n = π sec 2 πn =, repeating the preceding calculation gives Res n π cotπzfz = fn. To apply the method we will need to assume that zfz is bounded as z. That is, there exists R and M such zfz M for all z R. Consider a large square N centred at the origin and having side length 2N + for a large integer N. Then π cotπzfz dz = all residues inside N of π cotπzfz N N = {fn n is not a singularity of f} n= N + residues of π cotπzfz at singularities of fz inside N We will show that our boundedness assumptions on fz imply that lim π cotπzfz dz = N N so that we might get a formula for n= fn. Lemma. Let fz be a meromorphic function such that zfz is bounded as z. Then lim cotπzfz dz = N N where N is the square of side length 2N + centred at the origin. Proof. Euler s formulas e iz = cosz + i sinz and e iz = cosz i sinz give cosz = e iz +e iz 2 and sinz = eiz e iz 2i. Therefore cotz = eiz +e iz e iz e = e2iz + iz e 2iz. On the vertical sides of N, z = ±2N + /2 + iy so cotπz = e ±2πiN e ±iπ e 2πy + e ±2πiN e ±iπ e 2πy = e 2πy + e 2πy = e 2πy + e 2πy = e 2πy + e 2πy On the horizontal sides of N, z = x ± i2n + /2 so cotπz = e 2πix e ±π2n+ + e 2πix e ±π2n+ = e 2πix + e ±π2n+ e 2πix e ±π2n+ The right hand side is a continuous periodic function of x so it is bounded. A continuous function on a closed bounded set is bounded MATB43 and in the case of a periodic function we may restrict attention to one period, which is a bounded set. In fact, it 23
represents the ratio of the distances of some point on the unit circle to the points w and to w for some fixed w and is bounded, for example, by 2. Thus 2 is an upper bound for cotπz on N. Our assumption on fz is that there exists R and M such that zfz M for all z R. Let w = /z. Set gw := zfz for < w /R or equivalently z /ger. Since gw is bounded on < w < /R, its singularity at is removable so it extends to a holomorphic function on w < /R. Let gw = a + a w + a 2 w 2 + a 3 w 3... be the Taylor series of gw about. SubLemma. z 2 fz a z is bounded as z. gw a Proof. Since lim w w = g exists, gw a w has a removable singularity at so it represents an analytic, thus continuous, function on w /R. Therefore MATB43 it has a bound on any closed set in its domain. Let K be a bound for gw a on w /2R. If z 2R then w /2R and so for z 2R. z 2 fz a z Proof of Lemma cont.. If 2N + > 2R, = zzfz a = N gw a w K cotπz fz a dz N z z 2 fz a /z cotπz N z 2 dz N 2K 42N + 2K dz = 8K 2N + 2 2N + 2 2N + Therefore cotπz lim cotπzfz dz = lim a N N N N z w 2K z 2 dz cotπz dz = a Residues of inside N z We noted earlier if that fz does not have a zero at n then Res n cotπzfz = fn/π. cotπz Applying this in the case fz = /z shows Res n = /πn. Therefore, in the above z sum, for n, the residues at n and n cancel out leaving lim cotπzfz dz = a Res N N since the pole of cotπz z at has order 2. cotπz = z Substituting this into equation * and taking the limit as N gives 24
Theorem. Let fz be a meromorphic function such that zfz is bounded as z. Then N {fn n is not a singularity of f} n= N = residues of π cotπzfz at singularities of fz N Note: lim N n= N fn is not exactly the same as n= fn since the former might exist in cases where the latter does not converge. For example, if fz is some function for which fn = /n for n then, because of the cancellation, the limit on the left exists and equals f, but the series on the right does not converge. Example. Let fz = /z 2. Then zfz = /z is bounded as z. The only singularity N of fz is at. Therefore lim N n= n + N 2 n= n = Res 2 π cotπz z. In this case 2 we know that the two series on the left converge individually and are equal by symmetry. Thus we get cotπz n = π 2 2 Res = π πz 2 z 2 2 Res /2 +... z 2 πz πz 3 /6 +... n= = π 2 Res = π π = π2 2 3 6 π 2 z 2 /2 +... + π 2 z 2 /6 +... πz 3 = π 2 Res π 2 z 2 /3... πz 3 Example. Suppose p is not an integer and let fz = /z p. Then zfz = z/z p is bounded as z. The only singularity of fz is at p. Therefore lim N N n= N n p = Res p π cotπz = π cotπp. z p In other words, for every point in the domain of cotπz we have the identity π cotπz = lim N N n= N = z + lim N = z + lim N = z + n= N n z = lim n= N n= 2z z 2 n 2 N z + n + N n= N n= z n + z + n z 2 n 2 N z n z n 25 = z + lim N N = z + lim N n= N n= z + n + z n 2z z 2 n 2