Given out every other Thursday, due the next Thursday. This will be selected from problems in the text.

Transcription

1

2 Course Meeting Times Lectures: sessions / week,.5 hours / session Course Objective The aim of the course is to teach the principal techniques and methods of analytic function theory. This is quite different from real analysis and has much more geometric emphasis. It also has significant applications to other fields like analytic number theory. Prerequisite Analysis I (8.00B) or the equivalent. Text Ahlfors, Lars V. Complex Analysis: An Introduction to the Theory of Analytic Functions of One Complex Variable. 3rd ed. New York, NY: McGraw-Hill, 979. ISBN: This is a text with an attractive geometric flavor. I plan to cover most of the material up to p. 3, but will add two lectures devoted to a complete proof of the Prime Number Theorem, which fits very naturally in with the text material on the Riemann zeta function. In most of the lectures I will add some material not in the text, partly giving alternative proofs, and partly working through solutions of some of the more interesting problems. In Lec #, 3, 4, 6, 9,, and the treatment is really quite different from the corresponding material in the text. Lec # and actually contain a complete proof of the Prime Number Theorem. Since the lectures deviate so much from the text, regular attendance is strongly recommended. Homework Assignments Given out every other Thursday, due the next Thursday. This will be selected from problems in the text. Exams Two in-class tests following Lec #4 and 6. Also a final examination. Grading The final grade is based on a cumulative point total. ACTIVITIES PERCENTAGES Homework 0% Two in-class tests (0% each) 40% Final exam 50% General Remarks Working on the homework problems is very important for your understanding and command of the subject. Be sure you allow plenty of time for this. It is ok for you to discuss the problems with another student but you must give your own exposition of the solution. In the process you will often be lead to a better solution. It is very important to do this exposition in a neat fashion. This is the right place for you to develop a clear mathematical writing style. (Tex is of course most welcome). Sloppy homework might not be accepted. There are dozens of books available on the topic of complex variables. I urge you strongly to browse in some of these in the library. Caratheodory's books Theory of Functions Vol. I and II are closely related to our text. Also, I would particularly recommend Pólya and Szegö: Problems and Theorems in Analysis. Several chapters there deal with the subject of complex variables. Rudin's book, Real and Complex Analysis is also a valuable reference. Caratheódory, Constantin. Theory of Functions of a Complex Variable. Rhode Island: AMS Chelsea Pub, 00. ISBN: Caratheódory, Constantin, and F. Steinhardt. Theory of Functions of a Complex Variable. Vol.. New York, NY: Chelsea, 960. Pólya, George, and Gábor Szego. Problems and Theorems in Analysis. Berlin: Springer, 978. ISBN: Rudin, Walter. Real and Complex Analysis. New York, NY: McGraw-Hill, 986. ISBN:

3 Lecture : The algebra of Complex numbers (Text - & 9-0) Remarks on Lecture On p.9-0, it is stated that each circle in C (x a) + (y b) = r () has the form (α 0 α 3 )(x + y ) α x α y + α 0 + α 3 = 0, so the mapping z Z maps circles in the plane to circles on S. Solving the equations α α a =, b =, r a b α 0 + α 3 = α 0 α 3 α 0 α 3 α 0 α 3 for α 0, α, α, α 3 is disagreeable so we instead determine the image of the curve () under the map z Z. Using the formulas (4)-(6) and formula () becomes x 3 =, + z + r a b a + b r + ax + bx + x 3 =. This is a plane which must intersect the sphere so has distance < from 0.

4 The formula (8) can be proved geometrically as follows (Exercise 4): N 0 Z z Let Z S lie on the plane x = 0. The angles at Z are right angles, so by similar triangles: Fig. - d(n, Z) = d(n, z) =. + z Z' N Z Thus d(n, Z) = d(n, z ) + z + z and by symmetry this is z' z d(n, Z ). d(n, z) Fig. - Thus the triangles NZZ and Nzz are similar, so the above ratio is d(z, Z ). z z This proves (8).

5 Finally we show that the spherical representation z Z is conformal. This means that if l and m are two lines in the plane intersecting in z at an angle α, then the corresponding circles C and D through N and Z intersect Z at the same angle α. Consider the tangent plane π to S at the point N. the plane through Z and l intersects π in a line l. Similarly the plane through Z and m intersect π in m. Clearly l and m intersect at N at the same angle α. Since they are tangents to C and D at N, C and D must intersect at the angle α both at N and at Z. 3

6 Lecture : Exponential function & Logarithm for a complex argument (Replacing Text p.0-0) For b >, x R, we defined in 8.00B, b x = sup b t t Q, t x (where b t was easy to define for t Q). Then the formula b x+y = b x b y was hard to prove directly. We shall obtain another expression for b x making proof easy. Then Let x dt L(x) =, x > 0. t L(xy) = L(x) + L(y) and L (x) = > 0. x So L(x) has an inverse E(x) satisfying By 8.00B, so If y = L(x), so x = E(y), we thus have E(L(x)) = x. E (L(x))L (x) =, E (L(x)) = x. E (y) = E(y),

7 It is easy to see E(0) =, so by uniqueness, n x x E(x) = + x and E() = e. n! Theorem b x = E(xL(b)), x R. Proof: Let u = L(x), v = L(y), then and if t = so Since So Now n, m E(u + v) = E(L(x) + L(y)) = E(L(xy)) = xy = E(u)E(v), E(n) = E() n = e n, E(t) m = E(mt) = E(n) = e n. E(t) = e t, t Q, t > 0. E(t)E( t) =, E(t) = e t, t Q. b n = E(nL(b)) and ( ) b m = E L(b) m since both have same m th power. so Now for x R, ( ) ( ) n n ( ) n n b m = b m = E L(b) = E L(b), m m b t = E(tL(b)), t Q. b x = sup (b t ) = sup E(tL(b)) = E(xL(b)) t x, t Q t x, t Q since E(x) is continuous. Q.E.D. Corollary For any b > 0, x, y R, we have b x+y = b x b y.

8 In particular e x = E(x), so we have the amazing formula ( ) x x x n = + x ! n!! n! The formula for e x suggests defining e z for z C by the convergence being obvious. n z z e z = + z ! n! Proposition e z+w = e z e w for all z, w C. Proof: Look at the functions f(t) = e tz+w, g(t) = e tz e w for t R. Differentiating the series for e tz+w and e tz with respect to t, term-by-term, we see that df dg = zf(t), = zg(t) dt dt and f(0) = e w, g(0) = e w. By the uniqueness for these equations, we deduce f g. Thus f() = g(). Q.E.D. Thus Note that if t R, So e it lies on the unit circle. e it e it =, and(e it ) = e it. e it =. Put e it + e it t cos t = = +, e it e it t 3 sin t = = t +. 3! Thus we verify the old geometric meaning e it = cos t+i sin t. Note that the e it (t R) fill up the unit circle. In fact by the intermediate value theorem, {cos t t R} fills up the interval [, ], so e it = cost + i sin t is for a suitable t an arbitrary point on the circle. 3

9 Note that z e z takes all values w C except 0. For this note e z = e x e iy, z = x + iy. Choose x with e x = w and then y so that iy w e =, w then e z = w. a If z iϕ iψ z = z e, w = w e, Fig.- w then zw = z w e i(ϕ+ψ) = z w (cos(ϕ + ψ) + i sin (ϕ + ψ)), a Fig. - /n which gives a geometric interpretation of the multiplication. From this we also have the following very useful formula (cosϕ+i sin ϕ) n = e inϕ = cosnϕ+i sin nϕ. Thus n Theorem The roots of z = are, ω, ω,, ω n, where π π ω = cos + i sin. n n Geometric meanings for some useful complex number sets: z a = r circle z a + z b = r, ( a b < r) ellipse z a = z b perpendicular bisector {z z = a + tb, t R} line {z Imz < 0} lower half plane ( ) z a {z Im < 0} general half plane b 4

10 For x real, x e x has an inverse. This is NOT the case for z e z, because z+πi z e = e, thus e z does not have an inverse. Moreover, for w = 0, has infinitely many solutions: So e x = w, e iy w = w z e = w = x = log w, y = arg(w). log w = log w + iarg(w) takes infinitely many values, thus not a function. Define Arg(w) principal argument of w in interval π < Arg(w) < π and define the principal value of logarithm to be Log(w) log w + iarg(w), which is defined in slit plane (removing the negative real axis). We still have log z z = log z + log z in the sense that both sides take the same infinitely many values. We can be more specific: Theorem 3 In slit plane, and n = 0 if Log(z z ) = Log(z ) + Log(z ) + n πi, n = 0 or ± In particular, n = 0 if z > 0. π < Arg(z ) + Arg(z ) < π. Proof: In fact, Arg(z ), Arg(z ) and Arg(z z ) are all in ( π, π), thus π π π < Arg(z ) + Arg(z ) Arg(z z ) < π + π + π, but Arg(z ) + Arg(z ) Arg(z z ) = n πi, 5

11 thus n. If Arg(z ) + Arg(z ) < π, since Arg(z z ) < π, they must agree since difference is a multiple of π. Q.E.D. 6

12 Lecture 3: Analytic Functions; Rational Functions (Text -3) Remarks on Lecture 3 Formula (4) on p.3 was proved under the assumption that R( ) =. On the other hand, if R( ) is finite, then () holds with G 0. Then we use the previous proof on R(β j + ) and we still get the representation (4). ζ For theorem on page 9, we have the following stronger version: Theorem (Stronger version) The smallest convex set which contains all the zeros of P (z) also contains the zeros of P (z). Proof: Let α,, α n be the zeros of P, so Then P (z) = a n (z α ) (z α n ). P (z) = + +. P (z) z α α n If z 0 is a zero of P (z) and z 0 = each α i, then this vanishes for z = z 0 ; conjugating the equation gives z 0 α z 0 α n + + = 0, z 0 α z 0 α n so z 0 = m α + + m n α n, where n m i 0 and m i =. We now only need to prove the following simple result: i=

13 Proposition Given a,, a n C, the set n n { m i a i m i 0, m i = } () i= i= is the intersection C of all convex sets containing all a i (which is called the convex hull of a,, a n ). n Proof: We must show that each point a i m i in () is contained in each convex set i= containing the a i and thus in C. We may assume it has the form p x = m i a i i= where and m i > 0 for i p m j = 0 for j > p. and We prove x C by induction on p. Statement is clear if p =. Put By inductive assumption, a C. But p λ = i= m i p m a = a i. λ i= p x = m i a i = λa + ( λ)a i i= where 0 λ. So x C as stated. Q.E.D.

14 Solution to 4 on p.33 Suppose R(z) is rational and for z =. Then R(z) = R(e iθ ) λ R. Let S(z) be the rational functions obtained by conjugating all the coefficients in R(z), then R(e iθ )S(e iθ ) = R(e iθ )R(e iθ ) =. So R(z)S( ) = on z =. z Clearing denominators we see this relation holds for all z C. R(z)S( ) = z Since a polynomial has only finitely many zeroes, let α,, α p be all the zeroes of R(z) which are not equal to 0 or. Then,, are the poles of S(z) which are not equal to 0 or. So α α p,, ᾱ are the poles of R(z) which are not equal to 0 or because of the definition of S. Then ( ) z α z α p R(z) ᾱ z ᾱ p z has no poles or zeros except possibly 0 and. Hence ᾱ p R(z) = Cz l z α z α p ᾱ z ᾱ p z where C is constant with C =, l is integer. Conversely, such R has R(z) = on z =. 3

15 Lecture 4: Power Series (Text 33-4) Remarks on Lecture 4 Problem 8 on p.4 n We know 0 w converges only for w <. Otherwise the terms do not converge to 0. Now put z = z +, so z z w = =. + z z + So w < is equivalent to or equivalently Rez > 0, Rez >. Problem 9 on p.4 Write Write a n b n if z n =. + z n z n + z n a n c = 0. b n

16 Then if z >, z n, z n + z n and if z <, z n. z n + z n So in both cases we have convergence. If z = e it, we have =, z n + z n cos nt so the terms do not tend to 0, so we have divergence.

17 Lecture 5: Exponentials and Trigonometric Functions (Text 4-47) Remarks on Lecture 5 Since cosz is even, arccosz can just as well defined as arccosz = i log(z + z ). This in fact more appropriate because then the derivative is, z which is better because then the derivative is < 0 for z = 0. Note that in any case cos (arccosz) = z, since z + z and z z are reciprocals.

18

19

20 Lecture 7: Linear Transformations (Text 80-89) Remarks on Lecture 6 Concerning Definition 3 p. 8, formula () shows that the definition does not depend on the choice of z, z, z 3. Exercise on page 88 requires a minor correction. For example w = z is hyperbolic according to definition on page 86, yet when written in the form az + b cz + d with we must take so ad bc =, a = d = i, a + d = 0. The transformation w = z causes other ambiguities. Thus we modify the definition a bit. Definition S is parabolic if either it is the identity or has exactly one fixed point. S is strictly hyperbolic if k > 0 in () but k =. S is elliptic if k = in () p. 86 but S is not identity. Then the statement of Exercise holds with hyperbolic replaced by strictly hyperbolic.

21 (i) the condition for exactly one fixed point for αz + β Sz = γz + δ is (α δ) = 4βγ (wrong sign in text). With the normalization αδ βγ = this amounts to (α + δ) = 4 as desired. (ii) Assume two fixed points are a and b, so which we write as Put and define By linear algebra, and Define Then Now w a z a = k, w b z b αz + β w = Tz =. γz + δ A = α β γ δ (TraceA) Tr (T) =. det A Trace(BAB ) = Trace(A) det (BAB ) = det A. z a z = Sz =, z b w a w = Sw =. w b Tr (T) = Tr (STS ). w = STz = STS z,

22 so w = kz. Then Tr (T) = Tr (STS ) = k + +. k If T is strictly hyperbolic, we have k > 0, k, so which under the assumption amounts to as stated. Conversely, if then k > 0. So the transformation Tr (T) > 4, αδ βγ = (α + δ) > 4 (α + δ) > 4, w = kz maps each line through 0 and into itself. So T maps each circle C into itself with k > 0. Thus T is strictly hyperbolic. (iii) If k =, then and we find since the possibility θ = 0 is excluded. Conversely, if we have Writing iθ w = e z ( ) Tr θ (T) = cos < 4 < α + δ <, αδ βγ = Tr (T) = (α + δ) = k + + < 4. k iθ k = re (r > 0) 3

23 this implies which implies ( ) ( ) r + cos θ + i r sin θ <, r r r = or θ = 0 or θ = π. If r =, then k =, so T is elliptic. Since r +, the possibility θ = 0 is ruled out. r Finally if θ = π, then k = r, so (α + δ) = r +. r But r 0, so since α+δ is real, this implies r =, so k =, and T is thus elliptic. 4

24

25

26

27 Lecture 9: Cauchy-Goursat Theorem (Text 09-5) Remarks on Lecture 9 Property (ii) (page 6) of the winding number follows since a n(γ, a) is a continuous function on C \ γ and the image of each component is a connected set of integers, hence constant.

28 Lecture 0: The Special Cauchy s Formula and Applications (Text 8-6) Remarks on Lecture 0 Exercise 6 on page 08 The values of f(z) lie in the disk w < which is contained in the slit plane where Logw is defined. thus Logf(z) is well-defined and holomorphic in Ω and has derivative f (z). f( z) Thus f (z) dz = 0 γ f(z) by the Primitive theorem. Exercise on page 0 By using the substitution w = ϕ(z) = z we have dw dz =. w + z + ϕ(γ) Since ϕ(γ) = γ (including the orientation). Thus the integral is 0. Also = z + z i z + i γ

29 and so again the total integral is 0. n(γ, i) = n(γ, i), Exercise 3 on page 0 On z = ρ, we can write z = ρe iθ, thus so and Thus dz dθ = ρe iθ i, dz = i dθ, z dz dz = ρdθ = iρ. z dz dz = iρ z =ρ z a z =ρ z(z a)( ρ ā) z dz ā dz = iρ +. ρ a z =ρ z a ρ a z =ρ ρ az If a > ρ, the first term is 0, the other term is dz = πi, ā ρ z =ρ z a so the result is Thus in both cases the result is πρ. ρ a ā πρ. a ρ If a < ρ, then the second is 0 and the other is πρ iρ πi =. ρ a ρ a

30 The Taylor s Theorem (with remainder) proved in pp.5-6 should be stated as follows: Theorem (Taylor s Theorem) If f(z) is analytic in a region Ω containing a, one has f (a) f n (a) f(z) = f(a) + (z a) + + (z a) n + f n (z)(z a) n,! (n )! where f n (z) is analytic in Ω. Moreover, if C is the boundary of a closed disk contained in Ω with center a, then f n (z) has the representation f(ζ) dζ f n (z) = (z inside C). πi C (ζ a) n (ζ z) 3

31 Lecture : Isolated Singularities (Text 6-30) Remarks on Lecture Singularities: Let f(z) be holomorphic in a disk 0 < z a < δ with the center a removed. (i) If exist or if just lim f(z) z a lim f(z)(z a) = 0, z a then a is a removable singularity and f extends to a holomorphic function on the whole disk < δ. z a (ii) If a is said to be a pole. In this case lim f(z) =, z a f(z) = (z a) h f h (z), where h is a positive integer and f h (z) is holomorphic at a and f h (a) = 0. We also have the polar development where ϕ(z) is holomorphic at a. f(z) = B h (z a) h + + B (z a) + ϕ(z), If neither (i) nor (ii) holds, a is said to be an essential singularity.

32 Theorem 9 A holomorphic function comes arbitrarily close to any complex value in every neighborhood of an essential singularity. Simplified Proof: Suppose statement false. Then A C and δ > 0 and ɛ > 0 such that f(z) A > δ for z a < ɛ. Then So has a pole at z = a. Thus lim(z a) (f(z) A) =. z a (z a) (f(z) A) f(z) A = (z a)(z a) h g(z), where h Z + and g(z) is holomorphic at z = a. If h =, f(z) has a removable singularity at z = a. If h >, f(z) A has a pole at z = a and so does f(z). Both possibilities are excluded by assumption, so the proof is complete. Q.E.D. Exercise 4 on p.30. Suppose f is meromorphic in C { }. We shall prove f is a rational function. If is a pole, we work with g = /f, so we may assume is not a pole. It is not an essential singularity, so is a removable singularity. Thus for some R > 0, f(z) is bounded for z R. Since the poles of f(z) are isolated, there are just finitely many poles in the disk z R. (Poles of f(z) are zeroes of /f(z).) At a pole a, use the polar development near a f(z) = B h (z a) h + + B (z a) + ϕ(z). The equation shows that ϕ extends to a meromorphic function on C with one less pole than f(z). We can then do this argument with ϕ(z) and after iteration we obtain n ( ) f(z) = P i + g(z), z a i= i where P i are polynomials and g is holomorphic in C. The formula shows that g is bounded for z > ger and being analytic on z R, it thus must be bounded on C. By Liouville s theorem, it is constant. So f is a rational function.

33 Lecture : The Local Mapping. Schwarz s Lemma and non-euclidean interpretation (Text 30-36) Remarks on Lecture Proof of Theorem, p.3. The last formula on p.3 reads n(γ, z j (a)) = n(γ, z j (b)) () j j provided a and b belong to the same component of the set C Γ. In () we take for γ the circle z(t) = z 0 + ɛe it (0 t π), where ɛ > 0 is so small that z 0 is the only zero of f(z) w 0 = 0 inside γ. (The zeroes of f(z) w 0 = 0 are isolated.) As before let Γ = f(γ) and let Ω 0 denote the component of C Γ containing w 0. Ω 0 Γ z 0 w 0 γ

34 Since γ is a circle, n(γ, z j (a)) = 0 or depending on whether z j (a) is outside γ or inside γ (Since a / Γ, z j (a) / γ). Thus the left hand side of () equals the number of points inside γ where f takes the value a. In () we now take a arbitrary in Ω 0 and take b = w 0. By the choice of γ he right hand side of () equals n. Equation () therefore shows that each value in Ω 0 is taken n times by f inside γ (multiple roots counted according to their multiplicity). In particular, this holds for any disk w w 0 < δ inside Ω 0 with center w 0. Q.E.D. Remark: In dealing with winding numbers n(δ, z), we have to pay attention to the parametrization. Thus if γ(t) = e it (0 t π), f(z) = z n and Γ = f(γ), we have n(γ, 0) =, n(γ, 0) = n although γ and Γ are represented geometrically by the same point set. Non-Euclidean Plane This is the unit disk z < with the following convention: a) Non-Euclidean point = Point in disk; b) Non-Euclidean line = Arc in disk perpendicular to the boundary. This model satisfies all Euclid s axioms except the famous Parallel Axiom: Given a point p outside a line l, there is exactly one line through the point which does not intersect l. This axiom clearly fails in the above model; thereby solving the 000 year old problem of proving the Parallel Axiom on the basis of the other axioms. It cannot be done! We can now introduce distance in the non-euclidean plane D.

35 u Given z, z D, mark the points u, v of intersection with z =, in the order indicated. We put v z 3 z z d(z, z ) = log (z, z, v, u) = log ( z v : z v ). z u z u The cross ratio is real (page 79) and the geometry shows easily that the cross ratio is. So d(z, z ) 0, d(z, z ) = d(z, z ). It is also easy to show from the formula that d(z, z 3 ) = d(z, z ) + d(z 3, z ). Consider a fractional linear transformation mapping z e iϕ z z z z z z z 0 and z z z. Then by the order of the points v, z, z, u we see that v and u. The invariance of the cross ratio gives ( ) z z d(z, z ) = log 0,,, z z ( ) = log z z + z z z z z z (Exercise 7). Thus ( ) d(z, z + Δz) = log + Δz. z(z + Δz) Δz 3

36 So since we deduce log( + x) lim = x 0 x d(z, z + Δz) lim =. Δz 0 Δz z This suggests defining a non-euclidean length of a curve by γ : z(t) (α t β) dz L(γ) =. γ z Exercises and 6 now give the following geometric interpretation of Schwarz s Lemma: Ex. Formula (30) implies by division, letting z z 0, f (z) f( z) z. Ex 6. Let f : D D be holomorphic and f(γ) the image curve w(t) = f(z(t)) α t β of the curve γ above. Then β w (t) L(f(γ)) = α w(t) dt β f (z(t))z (t) = dt f(z(t)) α β z ( t) α z( t) = L(γ). (by Ex.) Thus as stated. L(f(γ)) L(γ) 4

37 Lecture 3: The General Cauchy Theorem (Replacing Text 37-48) Here we shall give a brief proof of the general form of Cauchy s Theorem. (cf: John D. Dixon, A brief proof of Cauchy s integral theorem, Proc. Amer. Math. Soc. 9, (97) ) Definition A closed curve ζ in an open set Ω is homologous to 0 (written ζ 0) with respect to Ω if n(ζ, a) = 0 for all a / Ω. Definition A region is simply connected if its complement with respect to the extended plane is connected. Remark: If Ω is simply connected and ζ Ω a closed curve, then ζ 0 with respect to Ω. In fact, n(ζ, z) is constant in each component of C ζ, hence constant in C Ω and is 0 for z sufficiently large. Theorem (Cauchy s Theorem) If f is analytic in an open set Ω, then f(z) dz = 0 γ for every closed curve ζ Ω such that ζ 0. In particular, if Ω is simply connected then f(z) dz = 0 for every closed ζ Ω. γ We shall first prove Theorem (Cauchy s Integral Formula) Let f be holomorphic in an open set Ω. Then f(ζ) n(ζ, z)f(z) = dζ () πi γ ζ z where ζ 0 with respect to Ω.

38 Proof: The prove is based on the following three claims. Define g(z, ζ) on Ω Ω by f(ζ) f(z) for z = ζ, ζ z g(z, ζ) = f (z) for z = ζ. Claim : g is continuous on Ω Ω and holomorphic in each variable and g(z, ζ) = g(ζ, z). Clearly g is continuous outside the diagonal in Ω Ω. Let (z 0, z 0 ) be a point on the diagonal and D Ω a disk with center z 0. Let z = ζ in D. Then by Theorem 8 g(z, ζ) g(z 0, z 0 ) = f (ζ) + f (z)(z ζ) f (z 0 ). So the continuous at (z 0, z 0 ) is obvious. For the holomorphy statement, it is clear that for each ζ 0 Ω the function is holomorphic on Ω ζ 0. Since z g(z, ζ 0 ) lim g(z, ζ 0 )(z ζ 0 ) = 0 z ζ 0 the point ζ 0 is a removable singularity (Theorem 7, p.4), so z g(z, ζ 0 ) is indeed holomorphic on Ω. This proves Claim.

39 Let Define function h on C by Ω = {z C (ζ) : n(ζ, z) = 0}. h(z) = g(z, ζ) dζ, πi γ z Ω; () f(ζ) h(z) = dζ, πi γ ζ z z Ω. (3) Since both expression agree on Ω Ω and since Ω Ω = C, this is a valid definition. Claim : h is holomorphic. This is obvious on the open sets Ω and Ω ζ. To show holomorphy at z 0 ζ, consider a disk D Ω with center z 0. Let δ be any closed curve in D. Then ( h(z) dz = g(z, ζ) dζ dz δ πi δ γ ( = g(z, ζ) dz dζ. πi γ δ For each ζ, δ z g(z, ζ) is holomorphic on D (even Ω). So by the Cauchy s theorem for disks, g(z, ζ) dz = 0. Now the Morera s Theorem implies h is holomorphic. Now we can prove: Claim 3: h 0, so () holds. We have z Ω for z sufficiently large. So by (3), lim h(z) = 0. z By Liouville s Theorem, h 0. Q.E.D. 3

40 Proof of Theorem : To derive Cauchy s theorem, let z 0 Ω ζ and put F (z) = (z z 0 )f(z). By (), F (z) f(z) dz = dz πi γ πi γ z z 0 = n(ζ, z 0 )F (z 0 ) = 0. Q.E.D. Note finally that Corollary on p.4 is an immediately consequence of Cauchy s Theorem. 4

41 Lecture 4: The Residue Theorem and Application (Replacing Text 48-54) Let Ω be a region and a Ω. Let f(z) be holomorphic in Ω = Ω a Definition The residue is defined as R = Res z=a f(z) f(z) dz, πi C where C is any circle contained in Ω with center a. C C ` Fig. 4- If C is another circle with center a and C Ω, then Cauchy s Theorem for the annulus shows that Res z=a f(z) is independence of the choice of C. While the definition can be shown to be equivalent to Definition 3 on p.49 in the text, we shall not need this. In place of Theorem 7 (Text p.50) we shall prove the following version: Theorem 7 Let f be analytic except for isolated singularities a j in a region Ω. Let γ be a simple closed curve which has interior contained in Ω and a j / γ (all j). Then f(z) dz = Res z=aj f(z). πi γ where the sum ranges over all a j inside γ. j

42 Proof: Ω γ Fig. 4- By compactness of γ and its interior, the sum above is finite. For simplicity let a, a be the singularities inside γ. a₂ γ Fig. 4-3 a₁ The outside of γ is connected and if we take two disks D, D around a and a and connect their boundaries to γ with bridges as in Fig. 4-3, the piece remaining in the interior of γ is simply connected (the complement is connected). Thus the integral over the boundary of this region is 0. Letting the widths of the bridges tend to 0, the theorem follows. Q.E.D.

43 Calculation of residues.. If lim f(z)(z a) z a exists and is finite, then it equals Res z=a f(z). In fact a is then a pole of f(z), so f(z) = B h (z a) h + + B (z a) + ϕ(z), B h 0. Then f(z) dz = B πi C and since the singular part above equals (z a) h (B h + B h (z a) + + B (z a) h ) the finiteness of the limit implies h.. If f(z) = h g( In fact z) ( z) where g(a) = 0 and h(z) has a simple zero at z = a, then g(a) Res z=a f(z) =. h (a) lim f(a)(z a) = lim g(z) = g(a) z a z a h(z) h(a) h (a). z a 3. If f has a pole of order h, then { } d h Res z=a f(z) =. (h )! dz (z h a)h f(z) In fact f(z) = (z a) h g(z), where g is holomorphic at a. So g (h ) g(z) (a) = (h )! πi (z a) dz = (h )!Res z=af(z). h C Example: (from text p.5.) ( ) e z d f(z) = = Res z=a f(z) = e z = e a. (z a) dz z=a z=a 3

44 Application: The Argument Principle. Theorem 8 Let f(z) be meromorphic in Ω, γ Ω a simple closed curve with interior inside Ω. Assume γ passes through no zeros nor poles of f. Then f (z) dz = N P, πi γ f(z) where N is the number of zeros, P the number of poles inside γ, all counted with multiplicity. Proof: By theorem 7, the integral is the sum of the residues of f (z)/f(z). At a zero a of order h, we have f(z) = (z a) h f h (z), f h (a) = 0 and f (z) h f h (z) f(z) = z a + f h (z) At a pole b of order k, we have similarly = Residue h, f ( z) k = f( z) z b + f h ( z) f ( z) h = Residue k. Now the result follows from Theorem 7. Q.E.D. Corollary (Rouche s Theorem) Let f and g be holomorphic in a Let γ be a simple closed curve in Ω with interior Ω. Assume region Ω. f(z) g(z) < f(z) on γ. Then f and g have the same number of zeros inside γ, say N f and N g. Proof: (The text does not take into account the case when f and g have common zeros). The inequality implies that f and g are zero-free on γ. Put g(z) ψ(z) =, f(z) then ψ(z) < 4

45 on γ, so the curve Γ = ψ(γ) lies in the disk ζ <. Hence ψ (z) dζ dz = = n(γ, 0) = 0 πi γ ψ(z) Γ ζ (book p.6). Now g (z) N g = dz πi γ g(z) ψ f + ψf = dz πi γ ψf ψ (z) f (z) = dz + dz πi γ ψ(z) πi γ f(z) = N f. This proves the result. Q.E.D. Exercise p.54 We use Rouche s theorem twice, first on γ : z = and then on γ : z =. For γ : z =, take f(z) = z 4, g(z) = z 4 6z + 3. For γ : z =, take f(z) = 6z, g(z) = z 4 6z

46 Lecture 5: Contour Integration and Applications (Text 54-6) Remarks on Lecture 5 In parts 4 and 5 (p ) some clarification of the use of the logarithm are called for. Example 4 p.59 The relation πiα α ( z) α = e z which is crucial for proof deserves explanation. l Fig. 5- We consider the function log θ z = log z + iarg θ z in the region C l θ (the plane with the ray l θ removed) where the angle is fixed by θ < arg θ z < θ + π. In the problem of computing we consider 0 xα R(x) dx log π (z)

47 in the plane C with the negative imaginary axis removed and us the Residue theorem on the contour in Fig As in the text we arrive at the integral ( z α+ R(z ) dz = z α+ + ( z) α+) R(z ) dz. 0 On the right z belongs to (0, ) and ( π π ) π 3π log π (z) = log z + + i, < arg π z <, π 3π log π ( z) = log z + + i Thus for z > 0, = so the last integrals combine to log π (z) + iπ, z > 0. ( z) α+ αiπ ) (α+) log = e π ( z) (α+)(log z +iπ) = e αiπ α+ = e z, α+ ( e z R(z ) dz. 0 For z > 0 we have from the above log π (z) = log z, so α+ ) αiπ α+ z R(z dz = ( e ) x R(x ) dx πi πi 0 απi = e sin πα xα+r(x ) dx. π 0 The left hand side of () is the sum of the residues of z α+ R(z ) = f(z) in the upper half plane. If g(z) R(z ) =, h(z) where g and h are holomorphic, g(a) = 0, and h has a simple zero at a, then () where (α+) (a) g(a) Res z=a f(z) = z, () h (a) α+ (α+) log π z = e (z).

48 Example: Exercise 3(g) p.6 To calculate we use x = t and arrive at 0 x 3 dx, + x 5 dz z 3 + z4 in (). The poles in the upper half plane are π π π i 4 i( + z = e and z = e 4 ). We use () to calculate the residues: ( ) 5 5 ( π ) i Res i π z 3 = z 3 e 4 4 π z=e + z4 i 4(e 4) 3 5 i π log 3 (e 4 π = e ) π 4(e 4 5 π 3(i = e 4) π i 4(e 4) 3 and π i 3, = e 4 i ) 3 ( ) ( ) 5 5 π Res π 3 3 e i3 4 i z=e 3 z = z 4 4 4(ei 3 π + z 4 ) 3 5 π 5 π log 3 π (i( + 4 = e )) 4(ei 3 iπ = e. 4 Thus () gives so π 4 ) 3 π π 5 i iπ πi dx e 3 + e = e 3 sin x 3, 4 4 π x4 5 dx π x 3 =. 0 + x 4 3 3

49 Example 5 p.60 The last four lines on the page are a bit misleading because the specific logarithm has already been chosen. So here is a completion of the proof after the equation We know (Lecture ) that 0 π Log( ie ix sin x) dx = 0. Log(z z ) = Logz + Logz, if π < Argz + Argz < π. (3) Using this for z = sin x we get π π log( sinx) dx + Log( ie ix ) dx = 0. (4) But so since π 0 0 πi ix Log( i) =, Loge = ix (0 < x < π), + x is in ( π, π), (3) implies Now (4) implies the result Log( ie ix πi ) = + ix. 0 π log sin θ dθ = π log. 4

50 Lecture 6: Harmonic Functions (Replacing Text 6-70) While integrals like f(z) dz and M dx+n dy have been defined in the text ϕ ϕ (p.0), differential forms like dx, dy and dz = dx + i dy have not been defined (and the definition is more subtle), we shall develop the theory of harmonic functions (p.6-70) without differential forms. Definition A real-valued function u(z) = u(x, y) in a region Ω is harmonic if it is C and satisfying the equation u u + = 0. x y The Cauchy-Riemann equations for a holomorphic function imply quickly that the real and imaginary parts of a holomorphic function are harmonic. The converse holds if Ω is simply connected: Theorem If Ω is simply connected and u harmonic in Ω, there exists a holomorphic function f(z) such that u(z) = Ref(z). Remark: Note the condition Ω is simply connected can not be removed, for example u(z) = log z is harmonic in the punctured plane C {0}, but it cannot be written as real part of a holomorphic function. Proof: Put Then u u g(z) = i = u + iv. x y u u u v x = x = y = y, u u v y = x y = x.

51 So by the Cauchy-Riemann equation, g is holomorphic. By p.4, since Ω is simply connected, g(z) = f (z) for some holomorphic function f. Writing f(z) = U(x, y) + iv (x, y), we have by the Cauchy-Riemann equation U U g(z) = f (z) = i, x y so Thus u(x, y) = U(x, y) + constant. u(z) = Ref(z) + constant. Q.E.D. Corollary (cf. (34) p.34) If u is harmonic in Ω, then if the disk z z 0 r lies in Ω, π u(z 0 ) = u(z 0 + re iθ ) dα. ϕ 0 More generally, if the annulus r z z 0 r belongs to a region Ω, we have Theorem 0 If u is harmonic in Ω, and {z : r z z 0 r } Ω, then π u(z 0 + re iθ ) dα = α log r +, r r r, () ϕ 0 where α and are constants. Proof: The function z u(z 0 + z) is harmonic, so writing the Laplacian in polar coordinates, = + +. r r r r α Denote the left hand side of () by V (r), then Writing this as the theorem follows. V r V + = 0. r r ( ) V r = 0, r r Q.E.D.

52 The Poisson Formula Let u be harmonic on z. Then u = Re(f) where f is holomorphic on z. Consider z + a S(z) = + āz, ( a < ) which maps the unit disk onto itself. Then f S is holomorphic and u S is harmonic (the real part of f S). Use the corollary on it with z 0 = 0, then u(a) = u(s(0)) = π u(s(e iϕ )) dϕ. ϕ 0 But so Hence or This gives Poisson s Formula ((63) in text) iθ S(e iϕ ) = eiϕ + a = e, + āe iϕ iθ iϕ e a e =. ae iθ ie iϕ dϕ = ieiθ a ie iθ, dα ( ae iθ ) dϕ ie iθ a ie iθ ae iθ = dα ( ae iθ ) i e iθ a () a = iθ e a. u(a) = π iθ dϕ a u(e ) dα = u(z) dα. ϕ 0 dα ϕ z a z = 3

53 Schwarz Theorem Theorem (Schwarz Theorem) Let U be a real piecewise continuous function on z = and define the Poisson integral u(z) = P U (z) by u(a) = ϕ 0 π a a e U(e iϕ ) dϕ, a <. (3) iϕ Then u is harmonic, and if U is continuous at e iϕ 0. lim u(z) = U(e iϕ 0 ) z e iϕ 0 Proof: We may assume ϕ 0 = 0. Since ( z e iϕ + z ) = Re, z e iϕ e iϕ z u is the real part of a holomorphic function, hence harmonic. Because of () formula (3) can be written u(s(0)) = π U(S(e iϕ )) dϕ. ϕ 0 Taking a = tanh t we obtain as t ( ) π e iϕ + tanh t u(tanh t) = U dϕ ϕ 0 tanh te iϕ + π U() dϕ ϕ 0 = U(). Q.E.D. 4

54 Exercise 5, p.7 Since log + z is harmonic in z < we have by the mean-value theorem π log + re iθ dα = log = 0 (4) ϕ for r <. We shall now show that log + re iθ π is bounded by an integrable function g(α). So by the dominated convergence theorem we can let r under the integral sign, giving the desired result π π log + e iθ dα = 0. (5) Since the integrand log + e iθ changes sign on the circle, we split the circle π π 4π into the two arcs (, ) and ( π, ), where we have and e iθ + e iθ respectively. In the first interval we have cos α so 3 iθ iθ α ϕ + re + e = cos, α, and r. (6) 3 In the second interval we put α = ϕ +ϕ and we see from the geometry, since ϕ π, 3 that α ϕ 4ϕ + re iθ = re iϕ cos ϕ = cos, 3 α 3. (7) α iθ Since log cos is integrable, the estimates (6) and (7) show that log + re is bounded by an integrable function g(α), so (5) is established. 5

55 Lecture 7: Mittag-Leffer s Theorem (Text 87-90) Theorem (Mittag-Leffer s Theorem) Let {b ν } be a sequence in C such that lim b ν =, ν and P ν (ζ) polynomials without constant term. Then there exist functions f meromorphic in C with poles at just the points b ν and corresponding singular parts ( ) P ν. z b ν The most general f(z) of this kind can be written [ ( ) ] f(z) = g(z) + P ν p ν (z) () z b ν ν where g is holomorphic in z and the p ν are polynomials. ( ) Proof: We may assume all b ν = 0. Consider the Taylor series for P ν z b ν n around z = 0. It is analytic for z < b ν. Let p ν (z) be the partial sum up to z ν (n ν to be determined later). Consider the finite Taylor series of ( ) ϕ(z) = P ν z b ν in a disk D with center 0. By (9) on p.6, ϕ(ζ) ϕ n (z) = dζ. πi C ζ n (ζ z)

56 where b ν Taking C as the circle with center 0 and radius and n = n ν + we deduce b ν M ν b ν ϕ nν+(z) π for z, π ( b ν ) nν+ b ν 4 Thus by Theorem 8 on p.5, Then 4 ( ) M ν = max P ν. z C z b ν ( ) ( ) n z ν+ b ν P ν p ν (z) M ν for z. () z b ν b ν 4 We now select n ν large enough so that nν M ν ν. ( ) n z ν+ b ν M ν ν for z. b ν 4 We claim now that the sum () converges uniformly in each disk z R (except at the poles) and thus represents a meromorphic function h(z). To see this we split the sum in (): h(z) = bν 4 R ( ( ) ) ( ( ) ) P ν p ν (z) + P ν p ν (z). (3) z b ν z b ν bν 4 >R Because of (), the second sum is holomorphic for z R since R first sum is finite and has ( ) P ν z b ν as the singular part at the pole b ν. b ν 4. The This proves the existence. If f is any other meromorphic function with these properties, then f(z) h(z) is holomorphic. Q.E.D.

57 Exercise 3 on p.78 Here we need some preparation on series of the form n= a n v n and use on a n = ( ) n, v n = ( + n) s, s = σ + it. We have if A n = a a n, then N N A 0 v 0 + (A n A n )v n A n (v n v n+ ) = A N v N. n= n=0 Lemma If (A n ) is bounded, v n 0, and then L a n v n converges. n=0 n= v n v n+ <, This is obvious from the identity above. In our example, v n = ( + n) s =, ( + n) σ so v n 0 even uniformly on compact subsets of Res > 0. For v n v n+ we have n+ v n v n+ = = s x s dx, (n + ) s (n + ) s n+ so Thus v n v n+ s. (n + ) σ+ ( ) n n s n= converges, and actually uniformly on compact sets in the region σ > 0 because this L is the case with v n 0 and v n v n+. 3

58 Lecture 8: Infinite Products (Text 9-00) Remarks on Lecture 8 Problem on p.97: Suppose that a n (all different, a condition missing in text) and A n arbitrary complex numbers. Show that there exists an entire function f(z) which satisfies f(a n ) = A n. Proof: (A simpler alternative to the hint in text). Let g(z) be an analytic function with simple zeros at the a n. By the Mittag-Leffler theorem, there exists a meromorphic function h on C with poles exactly at the points a n with the corresponding singular part A n /b n z a n, g(z) = (z a n )k(z), k(a n ) = b n 0. Then f(z) = g(z)h(z) has the desired property. Remarks on the formula for π cotπz (line 8 p.97) Q.E.D. Since the product formula for sin πz has infinitely many factors taking the logarithmic derivative requires justification. Generally, write I IN f(z) = f n (z) = lim f n (z) = lim g N (z) N the convergence being uniform on compacts. By Theorem, f (z) = lim g N (z) N so f (z) g N (z) = lim. f(z) N g N (z) N

59 Here g N(z)/g N (z) is given by the rule for differentiating a product. This remark justifies to proof of (7) as well. In the text the Gamma function is defined by means of the product formula (9) in.4 and the integral formula (4) derived by an interesting residue calculus due to Lindelöf. Here we go a shorter way and derive the product formula from the definition in terms of the integral formula. The Gamma function can be defined by Γ(z) = t z e t dt Rez > 0. 0 Writing n f t z n (z) = e t dt 0 f n is holomorphic and z Rez Γ(z) f n (z) t e t dt t e t dt n which 0 uniformly in each half plane Rez > δ (δ > 0). Thus Γ(z) is holomorphic in Rez > 0. Here are some of its properties (i) Γ(z + ) = zγ(z). This follows by integration by parts. (ii) Γ(z) extends to a meromorphic function on C with simple poles at z = 0,,,.... The function Γ(z + ) H(z) = z is meromorphic in Rez > with a pole at z = 0. Since lim zh(z) 0 z 0 the pole is simple. The residue is Γ() =. Also H(z) = Γ(z) for Rez > 0. Thus Γ(z) is meromorphic in Rez > with simple pole at z = 0. Statement (ii) follows by repetition. (iii) For x > 0, y > 0 Γ(x)Γ(y) t x ( t) y dt =. 0 Γ(x + y) r Γ(z)Γ(w) 0 Γ/+w Extend to Re z > 0, Rew > 0 t z ( t) w dt = Proof: t x 0 0 Γ(x)Γ(y) = e t dt s y e s ds n

60 Put s = tv. Since integrands are positive, integrals can be interchanged. We get Γ(x)Γ(y) = t x e t dt t y v y e tv dv 0 0 u = v y dv t x+y e (v+)t dt t = v x+y = v y dv u e u ( + v) x y du 0 0 y v = Γ(x + y) dv = Γ(x + y) sx ( s) y ds () 0 ( + v) x+y 0 the last expression coming from v = s ( s). This proves (iii), and it extends to Re z > 0, Re w > 0. π (iv) Γ(z)Γ( z) = sinπz From () we obtain x v Γ(x)Γ( x) = dv 0 + v which evaluates to π/ sin(πx) by the method of Exercise 3(g) p. 6, done in Lecture 5. This proves (iv) by meromorphic continuation. Since the poles of Γ(z) are canceled by zeros of sin πz, Γ( z) is never 0. By x (iii) we have for 0 < h < z = x + iy Γ(z h)γ(h) = ( t) z h t h dt Γ(z) 0 = + [( t) z h ]t h dt. h 0 In the integral we use the dominated convergence theorem to justify letting h 0 under the integral sign. In the interval [, ] there is no problem bounding the x integrand uniformly for h <. On the interval [0, ] we have (with α = z h ) h ( t) α (( t) α )t t and by l Hospital s rule this has limit α = z h z + so again the integrand is bounded. Thus we let h 0 and obtain Γ(z h)γ(h) = + [( t) z ]t dt + o() as h 0. Γ(z) h 0 The left hand side is using Taylor for h Γ(z h) and Laurent for h Γ(h) both at h = 0 (Γ(z) hγ (z) + ) + A + Bh Γ(z) h 3

61 where { } is the Laurent series for Γ(h) with center h = 0. Equating the constant terms on left hand side and right hand side we get Γ (z) = ( ( t)) z t dt A x > 0. Γ(z) 0 L Writing t = 0 ( t) n the expression is L [( t) n ( t) n+z ] dt A, 0 0 ( t) and since expression in [ ] equals n ( ( t) z ) t which is bounded by ( t) n Kt, t r L with integral K/(n + )(n + ) we can exchange and n by the dominated convergence theorem. Thus our expression equals L [( t) n ( t) n+z ] dt A so 0 0 L L = A = + n + n + z z n + n + z 0 A = L L + n n + z n + n + z A L = + A z n n + z L Γ (z) = + = A. Γ(z) z n n + z Having justified taking logarithmic derivative of an infinite product this gives I ( z ) z Cz = ze + e n C = const. Γ(z) n Putting z = we have so so I C = e + e n n ( ) C (+ + = e lim (N + )e N ) N 4

62 Lecture 9: Normal Families (Replacing Text 9-7) Theorem Let Ω C be a region, F a family of holomorphic functions on Ω such that for each compact E Ω, F is uniformly bounded on E. Then F has a subsequence converging uniformly on each compact subset of Ω. First we prove that on each compact subset E Ω, the family F is equicontinuous. This means, given π > 0 there exists a δ > 0 such that for all f F, f(z ) f(z ) < π if z z < δ, z, z E. () The distance function x d(x, C Ω) is continuous and has a minimum > 0 on the compact set E. Let d > 0 be such that (D denoting disk) F = x E D(x, d) has closure F Ω. Let z, z E satisfy and let ζ denote the circle z z < d ζ : z z = d. Then ζ F and z and z are both inside ζ. Also ζ z = d, ζ z d for ζ ζ. By Cauchy s formula for f F, z z f(ζ) f(z ) f(z ) = dζ, πi γ (ζ z )(ζ z ) so if M(F ) is the maximum of f on F Hence () follows. M(F ) f(z ) f(z ) z z. d To conclude the proof of Theorem choose any sequence (z j ) which is dense in Ω. Let f m be any sequence in F. The sequence f m (z ) is bounded so f m has

63 a subsequence f m, converging at z. Form this take a subsequence f m, which converges at z. Continuing we see that the subsequence f m,m converges at each z j. By the first part of the proof, F is equicontinuous on the compact set F. Given π > 0 there exists a δ < d such that () holds for all z, z F, f F. If z E the disk D(z, δ) contains some z j so D(z j, δ) contains z. By the compactness of E, p E D(z i, δ) i= for some z,, z p. Thus given z E there exists a z i = z i (z) such that z z i (z) < δ. Then z i (z) F. Thus by () for F, There exists N > 0 such that Given z E we have with z i = z i (z) f(z) f(z i (z)) < π. f F. () f r,r (z i ) f s,s (z i ) < π i p, r, s > N. (3) f r,r (z) f s,s (z) f r,r (z) f r,r (z i ) + f r,r (z i ) f s,s (z i ) + f s,s (z i ) f s,s (z) 3π by () and (3). The proves the stated uniform convergence on E. Remark: In the text, p. 3, it is erroneously assumed (and used) that ζ k E. This error occurs in many other texts.

64 Lecture 0: The Riemann Mapping Theorem (Text 9-3) Remark: The first step is to show that there exists an analytic and univalent function f, mapping Ω into the unit disk D satisfying f(z 0 ) = 0 and f (z 0 ) > 0. In order to expand f(ω) we take f (z 0 ) as large as possible. This indeed accomplishes f(ω) = D. The proof concludes with the contradiction G (z 0 ) > B = f (z 0 ). After the proof it is indicated in the text that this contradiction f (z 0 ) < G (z 0 ) is a consequence of Schwarz s lemma by writing f(z) = H(W ), W = G(z). However, H is then only defined on G(Ω), so until we know that G(Ω) = D Schwarz s lemma (p.35) does not apply. Thus the phrase a consequence of Schwarz s lemma should properly read consistent with Schwarz s lemma, which probably was the author s intention. Koebe s proof from 95 is outlined in Pólya and Szegö, Vol. II,. This is based on considering the distance of the origin to the boundary of the image region f(ω). The square root map is then used to increase this distance. By iteration this leads to an explicit sequence converging to the desired limit map.

65 Lecture and : The Prime Number Theorem (New lecture, not in Text) The location of prime numbers is a central question in number theory. Around 808, Legendre offered experimental evidence that the number π(x) of primes < x behaves like x/ log x for large x. Tchebychev proved (848) the partial result that the ratio of π(x) to x/ log x for large x lies between 7/8 and 9/8. In 896 Hadamard and de la Valle Poussin independently proved the Prime Number Theorem that the limit of this ratio is exactly. Many distinguished mathematicians (particularly Norbert Wiener) have contributed to a simplification of the proof and now (by an important device by D.J. Newman and an exposition by D. Zagier) a very short and easy proof is available. These lectures follows Zagier s account of Newman s short proof on the prime number theorem. cf: () D.J.Newman, Simple Analytic Proof of the Prime Number Theorem, Amer. Math. Monthly 87 (980), () D.Zagier, Newman s short proof of the Prime Number Theorem. Amer. Math. Monthly 04 (997), The prime number theorem states that the number π(x) of primes which are x less than x is asymptotically like : log x π(x) x/ log x as x. Through Euler s product formula (I) below (text p.3) and especially through Riemann s work, π(x) is intimately connected to the Riemann zeta function ζ(s) =, n s n= which by the convergence of the series in Res > is holomorphic there.

66 The prime number theorem is approached by use of the functions log p Φ(s) =, V(x) = log p. p prime p s p x prime Simple properties of Φ will be used to show ζ(s) = 0 and Φ(s) holomorphic s for Res. Deeper properties result from writing Φ(s) as an integral on which Cauchy s theorem for contour integration can be used. This will result in the relation V(x) x from which the prime number theorem follows easily. I = ( p s n ) for Res >. ζ(s) Proof: For each n ( p s n ) = p ms n. Putting this into the finite product m=0 N N ( p s ) we obtain ( p s n ) = n s k k=. Now let N. II ζ(s) extends to a holomorphic function in Res > 0. s Proof: In fact for Res >, ζ(s) = s n s n= n+ = n= n dx x s ( n s x s ) dx But x dy s s = s max, n s x s y s+ x n y y s+ R s n n e + so the sum above converges uniformly in each half-plane Res δ (δ > 0). III V(x) = O(x) (Sharper form proved later). Proof: Since the p in the interval n < p n divides (n)! but not n! we have n n = ( + ) n = k=0 ( ) n k ( ) n n p = e V(n) V(n), n< p n

67 Thus V(n) V(n) n log. () x If x is arbitrary, select n with n < n +, then V(x) V(n + ) V(n + ) + (n + ) log (by ()) ( x ) = V + + (x + ) log ( x ) ( x ) = V + log + + (x + ) log. Thus if C > log, ( x ) V(x) V Cx for x x 0 = x 0 (C). () Consider the points x x x x x x 0 r+ r r- Use () for the points right of x 0, ( x V ) x ) V ( r. ( x ) x V C, V ( x ) x C. r+ r Summing, we get ( x ) V(x) V(x 0 ) V(x) V r+ x Cx + + C, r so V(x) C(x) + O(). 3

68 IV ζ(s) = 0 and Φ(s) is holomorphic for Res. s Proof: If Res >, part I shows that ζ(s) = 0 and ζ (s) log p = log p = Φ(s) +. ζ(s) p s p s (p s ) p p (3) The last sum converges for Res >, so by II, Φ(s) extends meromorphically to Res > with poles only at s = and at the zeros of ζ(s). Note that ζ(s) = 0 = ζ(s ) = 0. Let α R. If s 0 = + iα is a zero of ζ(s) of order µ 0, then So ζ (s) µ = + function holomorphic near s 0. ζ(s) s s 0 We exploit the positivity of each term in for ǫ > 0. It implies so lim ǫφ( + ǫ + iα) = µ. ǫ 0 l Φ( + ǫ) = og p p +ǫ p log p ( i ) p + α p +ǫ + p iα 0, p Φ( + ǫ + iα) + Φ( + ǫ iα) + Φ( + ǫ) 0. (4) By II, s = is a simple pole of ζ(s) with residue +, so lim ǫφ( + ǫ) =. ǫց0 Thus (4) implies so µ + 0, µ. 4

69 This is not good enough, so we try Putting log p ( ) p + p p p +ǫ + iα iα 4 0. lim ǫφ( + ǫ ± iα) = ν, ǫց0 where ν 0 is the order of ± iα as a zero of ζ(s), the same computation gives 6 8µ ν 0, which implies µ = 0 since µ, ν 0. Now II and (3) imply Φ(s) holomorphic s for Res. V(x) x V dx is convergent. x Proof: The function V(x) is increasing with jumps log p at the points x = p. Thus Φ(s) = log p p s p V(x) = s dx x s+ p In fact, writing as i+ ( ) i p i this integral becomes i= V(p i) s p s i p s i+ which by V(p i+ ) V(p i ) = log p i+ reduces to Φ(s). Using the substitution x = e t we obtain Φ(s) = s e st V(e t ) dt Res >. Consider now the functions 0 f(t) = V(e t )e t, Φ(z + ) g(z) =. z + z f(t) is bounded by III and we have e T V(x) x dx = x 5 0 T f(t) dt. (5)

70 Also, by IV, Φ(z + ) = where h is holomorphic in Rez 0, so + h(z), z is holomorphic in Rez 0. For Rez > 0 we have g(z) = Φ(z + ) h(z) g(z) = = z + z z + = 0 Now we need the following theorem: 0 e zt (f(t) + ) e zt f(t) dt. 0 e zt dt Theorem (Analytic Theorem) Let f(t) (t 0) be bounded and locally integrable and assume the function g(z) = e zt f(t) dt Re(z) > 0 0 extends to a holomorphic function on Re(z) 0, then T lim f(t) dt exists and equals g(0). T 0 This will imply Part V by (5). Proof of Analytic Theorem will be given later. VI V(x) x. Proof: Assume that for some λ > we have V(x) λx for arbitrary large x. Since V(x) is increasing we have for such x λx V(t) t λx λx t λ λ s dt dt = ds = δ(λ) > 0. t x t s x On the other hand, V implies that to each ǫ > 0, K such that K V(x) x dx < ǫ for K, K > K. K x 6

71 Thus the λ cannot exist. Similarly if for some λ <, V(x) λx for arbitrary large x, then for t x, V(t) V(x) λx, so x V(t) t x λx t λ s = ds = δ(λ) < 0. λx t λx t λ s Again this is impossible for the same reason. Thus both V(x) β = lim sup > x x and V(x) α = lim inf < x x are impossible. Thus they must agree, i.e. V(x) x. Proof of Prime Number Theorem: so We have Secondly if 0 < ǫ <, V(x) = log p log x = π(x) log x, p x p x π(x) log x V(x) lim inf lim inf =. x x x x V(x) x ǫ p x ( ǫ) log p x ǫ p x log x = ( ǫ) log x ( π(x) + O(x ǫ ) ) thus for each ǫ. Thus π(x) log x V(x) lim sup lim sup x x ǫ x x π(x) log x lim =. x x Q.E.D. 7

72 Proof of Analytic Theorem: (Newman). Put T g T (z) = e zt f(t) dt, 0 which is holomorphic in C. We only need to show lim g T (0) = g(0). T z = R Fix R and then take δ > 0 small enough so that g(z) is holomorphic on C and its interior. By Cauchy s formula δ 0 C g(0) g T (0) = ( ) z dz (g(z) g T (z)) e zt +. (6) πi R z C On semicircle C + : C (Rez > 0) integrand is bounded by B, where R B = sup f(t). t 0 In fact for Rez > 0, g(z) g T (z) = = T B f(t)e zt dt T Be RezT Rez e zt dt and ( ) e zt z + R z = e ezt Re (z = Re iθ ). R z R 8

73 B So the contribution to the integral (6) over C + is bounded by, namely R Be RezT T z B e Rez Re πr =. Rez R π R Next consider the integral over C_ = 0. Look at g(z) and g T (z) separately. For g T (z) which is entire, this contour can be replaced by C'_ = 0. B Again the integral is bounded by because R T g T (z) = f(t)e zt dt 0 = T B e zt dt 0 T B e zt dt Be RezT Rez 9