Complex Methods (Theorems with Proof)

Part IB — Complex Methods Theorems with proof

Based on lectures by R. E. Hunt Notes taken by Dexter Chua Lent 2016

These notes are not endorsed by the lecturers, and I have modiﬁed them (often signiﬁcantly) after lectures. They are nowhere near accurate representations of what was actually lectured, and in particular, all errors are almost surely mine.

Analytic functions Deﬁnition of an analytic function. Cauchy-Riemann equations. Analytic functions as conformal mappings; examples. Application to the solutions of Laplace’s equation in various domains. Discussion of log z and za. [5]

Contour integration and Cauchy’s Theorem [Proofs of theorems in this section will not be examined in this course.] Contours, contour integrals. Cauchy’s theorem and Cauchy’s integral formula. Taylor and Laurent series. Zeros, poles and essential singularities. [3]

Residue calculus Residue theorem, calculus of residues. Jordan’s lemma. Evaluation of deﬁnite integrals by contour integration. [4]

Fourier and Laplace transforms Laplace transform: deﬁnition and basic properties; inversion theorem (proof not required); convolution theorem. Examples of inversion of Fourier and Laplace transforms by contour integration. Applications to diﬀerential equations. [4]

1 Contents IB Complex Methods (Theorems with proof)

Contents

0 Introduction 3

1 Analytic functions 4 1.1 The complex plane and the Riemann sphere ...... 4 1.2 Complex diﬀerentiation ...... 4 1.3 Harmonic functions ...... 4 1.4 Multi-valued functions ...... 4 1.5 M¨obiusmap ...... 4 1.6 Conformal maps ...... 5 1.7 Solving Laplace’s equation using conformal maps ...... 5

2 Contour integration and Cauchy’s theorem 6 2.1 Contour and integrals ...... 6 2.2 Cauchy’s theorem ...... 7 2.3 Contour deformation ...... 7 2.4 Cauchy’s integral formula ...... 7

3 Laurent series and singularities 9 3.1 Taylor and Laurent series ...... 9 3.2 Zeros ...... 10 3.3 Classiﬁcation of singularities ...... 10 3.4 Residues ...... 10

4 The calculus of residues 11 4.1 The residue theorem ...... 11 4.2 Applications of the residue theorem ...... 11 4.3 Further applications of the residue theorem using rectangular contours ...... 11 4.4 Jordan’s lemma ...... 11

5 Transform theory 13 5.1 Fourier transforms ...... 13 5.2 Laplace transform ...... 13 5.3 Elementary properties of the Laplace transform ...... 13 5.4 The inverse Laplace transform ...... 14 5.5 Solution of diﬀerential equations using the Laplace transform . . 16 5.6 The convolution theorem for Laplace transforms ...... 16

2 0 Introduction IB Complex Methods (Theorems with proof)

0 Introduction

3 1 Analytic functions IB Complex Methods (Theorems with proof)

1 Analytic functions 1.1 The complex plane and the Riemann sphere 1.2 Complex diﬀerentiation Proposition (Cauchy-Riemann equations). If f = u + iv is diﬀerentiable, then

∂u ∂v ∂u ∂v = , = − . ∂x ∂y ∂y ∂x Proposition. Given a complex function f = u + iv, if u and v are real diﬀeren- tiable at a point z and ∂u ∂v ∂u ∂v = , = − , ∂x ∂y ∂y ∂x then f is diﬀerentiable at z.

1.3 Harmonic functions Proposition. The real and imaginary parts of any analytic function are harmonic.

1.4 Multi-valued functions 1.5 M¨obiusmap Proposition. M¨obiusmaps take circlines to circlines. Proof. Any circline can be expressed as a circle of Apollonius,

|z − z1| = λ|z − z2|,

+ where z1, z2 ∈ C and λ ∈ R . This was proved in the ﬁrst example sheet of IA Vectors and Matrices. The case λ = 1 corresponds to a line, while λ 6= 1 corresponds to a circle. Substituting z in terms of w, we get

−dw + b −dw + b − z1 = λ − z2 . cw − a cw − a Rearranging this gives

|(cz1 + d)w − (az1 + b)| = λ|(cz2 + d)w − (az2 + b)|. (∗)

A bit more rearranging gives

az1 + b cz2 + d az2 + b w − = λ w − . cz1 + d cz1 + d cz2 + d This is another circle of Apollonius. Note that the proof fails if either cz1 + d = 0 or cz2 + d = 0, but then (∗) trivially represents a circle.

4 1 Analytic functions IB Complex Methods (Theorems with proof)

Proposition. Given six points α, β, γ, α0, β0, γ0 ∈ C∗, we can ﬁnd a M¨obius map which sends α 7→ α0, β 7→ β0 and γ → γ0.

Proof. Deﬁne the M¨obiusmap β − γ z − α f (z) = . 1 β − α z − γ By direct inspection, this sends α → 0, β → 1 and γ → ∞. Again, we let

β0 − γ0 z − α0 f (z) = . 2 β0 − α0 z − γ0

0 0 0 −1 This clearly sends α → 0, β → 1 and γ → ∞. Then f2 ◦ f1 is the required mapping. It is a M¨obiusmap since M¨obiusmaps form a group.

1.6 Conformal maps Proposition. A conformal map preserves the angles between intersecting curves.

Proof. Suppose z1(t) is a curve in C, parameterised by t ∈ R, which passes 0 through a point z0 when t = t1. Suppose that its tangent there, z1(t1), has a 0 well-deﬁned direction, i.e. is non-zero, and the curve makes an angle φ = arg z1(t1) to the x-axis at z0. Consider the image of the curve, Z1(t) = f(z1(t)). Its tangent direction at t = t1 is 0 0 0 0 0 Z1(t1) = z1(t1)f (z1(t1)) = z1(t0)f (z0), and therefore makes an angle with the x-axis of

0 0 0 0 arg(Z1(t1)) = arg(z1(t1)f (z0)) = φ + arg f (z0),

0 0 noting that arg f (z0) exists since f is conformal, and hence f (z0) 6= 0. 0 In other words, the tangent direction has been rotated by arg f (z0), and this is independent of the curve we started with. Now if z2(t) is another curve passing through z0. Then its tangent direction 0 will also be rotated by arg f (z0). The result then follows.

1.7 Solving Laplace’s equation using conformal maps

5 2 Contour integration and Cauchy’sIB Complex theorem Methods (Theorems with proof)

2 Contour integration and Cauchy’s theorem 2.1 Contour and integrals Proposition.

(i) We write γ1 + γ2 for the path obtained by joining γ1 and γ2. We have Z Z Z f(z) dz = f(z) dz + f(z) dz γ1+γ2 γ1 γ2 Compare this with the equivalent result on the real line:

Z c Z b Z c f(x) dx = f(x) dx + f(x) dx. a a b

(ii) Recall −γ is the path obtained from reversing γ. Then we have Z Z f(z) dz = − f(z) dz. −γ γ Compare this with the real result

Z b Z a f(x) dx = − f(x) dx. a b

(iii) If γ is a contour from a to b in C, then Z f 0(z) dz = f(b) − f(a). γ This looks innocuous. This is just the fundamental theorem of calculus. However, there is some subtlety. This requires f to be differentiable at every point on γ. In particular, it must not cross a branch cut. For example, 1 our previous example had log z as the antiderivative of z . However, this does not imply the integrals along different paths are the same, since we need to pick different branches of log for different paths, and things become messy. (iv) Integration by substitution and by parts work exactly as for integrals on the real line.

(v) If γ has length L and |f(z)| is bounded by M on γ, then Z

f(z) dz ≤ LM. γ This is since Z Z Z

f(z) dz ≤ |f(z)|| dz| ≤ M |dz| = ML. γ γ γ We will be using this result a lot later on.

6 2 Contour integration and Cauchy’sIB Complex theorem Methods (Theorems with proof)

2.2 Cauchy’s theorem Theorem (Cauchy’s theorem). If f(z) is analytic in a simply-connected domain D, then for every simple closed contour γ in D, we have I f(z) dz = 0. γ

Proof. (non-examinable) The proof of this remarkable theorem is simple (with a catch), and follows from the Cauchy-Riemann equations and Green’s theorem. Recall that Green’s theorem says

I ZZ ∂Q ∂P (P dx + Q dy) = − dx dy. ∂S S ∂x ∂y Let u, v be the real and imaginary parts of f. Then I I f(z) dz = (u + iv)(dx + i dy) γ γ I I = (u dx − v dy) + i (v dx + u dy) γ γ ZZ ∂v ∂u ZZ ∂u ∂v = − − dx dy + i − dx dy S ∂x ∂y S ∂x ∂y But both integrands vanish by the Cauchy-Riemann equations, since f is diﬀer- entiable throughout S. So the result follows.

2.3 Contour deformation

Proposition. Suppose that γ1 and γ2 are contours from a to b, and that f is analytic on the contours and between the contours. Then Z Z f(z) dz = f(z) dz. γ1 γ2

Proof. Suppose ﬁrst that γ1 and γ2 do not cross. Then γ1 − γ2 is a simple closed contour. So I f(z) dz = 0 γ1−γ2 by Cauchy’s theorem. Then the result follows. If γ1 and γ2 do cross, then dissect them at each crossing point, and apply the previous result to each section.

2.4 Cauchy’s integral formula Theorem (Cauchy’s integral formula). Suppose that f(z) is analytic in a domain D and that z0 ∈ D. Then 1 I f(z) f(z0) = dz 2πi γ z − z0

for any simple closed contour γ in D encircling z0 anticlockwise.

7 2 Contour integration and Cauchy’sIB Complex theorem Methods (Theorems with proof)

Proof. (non-examinable) We let γε be a circle of radius ε about z0, within γ.

z0 γε

f(z) Since is analytic except when z = z0, we know z−z0 I f(z) I f(z) dz = dz. γ z − z0 γε z − z0

iθ We now evaluate the right integral directly. Substituting z = z0 + εe , we get

I Z 2π iθ f(z) f(z0 + εe ) iθ dz = iθ iεe dθ γε z − z0 0 εe Z 2 = i π(f(z0) + O(ε)) dθ 0 → 2πif(z0)

as we take the limit ε → 0. The result then follows.

Theorem (Liouville’s theorem*). Any bounded entire function is a constant. Proof. (non-examinable) Suppose that |f(z)| ≤ M for all z, and consider a circle of radius r centered at an arbitrary point z0 ∈ C. Then I 0 1 f(z) f (z0) = 2 dz. 2πi |z−z0|=r (z − z0) Hence we know 1 1 M ≤ · 2πr · → 0 2πi 2π r2 0 as r → ∞. So f (z0) = 0 for all z0 ∈ C. So f is constant.

8 3 Laurent series and singularitiesIB Complex Methods (Theorems with proof)

3 Laurent series and singularities 3.1 Taylor and Laurent series

Proposition (Laurent series). If f is analytic in an annulus R1 < |z − z0| < R2, then it has a Laurent series ∞ X n f(z) = an(z − z0) . n=−∞ This is convergent within the annulus. Moreover, the convergence is uniform within compact subsets of the annulus.

Proof. (non-examinable) We wlog z0 = 0. Given a z in the annulus, we pick r1, r2 such that R1 < r1 < |z| < r2 < R2,

and we let γ1 and γ2 be the contours |z| = r1, |z| = r2 traversed anticlockwise respectively. We choose γ to be the contour shown in the diagram below.

z r1 r2 z0 γ

We now apply Cauchy’s integral formula (after a change of notation): 1 I f(ζ) f(z) = dζ. 2πi γ ζ − z We let the distance between the cross-cuts tend to zero. Then we get 1 I f(ζ) 1 I f(ζ) f(z) = dz − dζ, 2πi γ2 ζ − z 2πi γ1 ζ − z We have to subtract the second integral because it is traversed in the opposite direction. We do the integrals one by one. We have I f(ζ) 1 I f(ζ) = − dζ ζ − z z ζ γ1 γ1 1 − z 1 Taking the Taylor series of ζ , which is valid since |ζ| = r1 < |z| on γ1, we 1− z obtain ∞ m 1 I X ζ = − f(ζ) dζ z z γ1 m=0 ∞ X I = − z−m−1 f(ζ)ζm dζ. m=0 γ1

9 3 Laurent series and singularitiesIB Complex Methods (Theorems with proof)

This is valid since H P = P H by uniform convergence. So we can deduce

−1 1 I f(ζ) X − dζ = a zn, 2πi ζ − z n γ1 n=−∞ where I 1 −n−1 an = f(ζ)ζ dζ 2πi γ1 for n < 0. Similarly, we obtain

∞ 1 I f(ζ) X dζ = a zn, 2πi ζ − z n γ2 n=0

for the same deﬁnition of an, except n ≥ 0, by expanding

∞ 1 1 1 X zn = = . ζ − z ζ 1 − z ζn+1 ζ n=0

This is again valid since |ζ| = r2 > |z| on γ2. Putting these result together, we obtain the Laurent series. The general result then follows by translating the origin by z0. We will not prove uniform convergence — go to IB Complex Analysis.

3.2 Zeros 3.3 Classiﬁcation of singularities 3.4 Residues Proposition. At a simple pole, the residue is given by

res f(z) = lim (z − z0)f(z). z=z0 z→z0 Proof. We can simply expand the right hand side to obtain a−1 lim (z − z0) + a0 + a1(z − z0) + ··· = lim (a−1 + a0(z − z0) + ··· ) z→z0 z − z0 z→z0

= a−1, as required. Proposition. At a pole of order N, the residue is given by

N−1 1 d N lim N−1 (z − z0) f(z). z→z0 (N − 1)! dz Theorem. γ be an anticlockwise simple closed contour, and let f be analytic within γ except for an isolated singularity z0. Then I f(z) dz = 2πia−1 = 2πi res f(z). γ z=z0

10 4 The calculus of residues IB Complex Methods (Theorems with proof)

4 The calculus of residues 4.1 The residue theorem Theorem (Residue theorem). Suppose f is analytic in a simply-connected region except at a ﬁnite number of isolated singularities z1, ··· , zn, and that a simple closed contour γ encircles the singularities anticlockwise. Then

n I X f(z) dz = 2πi res f(z). z=zk γ k=1 Proof. Consider the following curve γˆ, consisting of small clockwise circles γ1, ··· , γn around each singularity; cross cuts, which cancel in the limit as they approach each other, in pairs; and the large outer curve (which is the same as γ in the limit).

z2 γˆ

H Note that γˆ encircles no singularities. So γˆ f(z) dz = 0 by Cauchy’s theorem. So in the limit when the cross cuts cancel, we have

n I X I I f(z) dz + f(z) dz = f(z) dz = 0. γ k=1 γk γˆ But from what we did in the previous section, we know I f(z) dz = −2πi res f(z), z=zk γk

since γk encircles only one singularity, and we get a negative sign since γk is a clockwise contour. Then the result follows.

4.2 Applications of the residue theorem 4.3 Further applications of the residue theorem using rectangular contours 4.4 Jordan’s lemma Lemma (Jordan’s lemma). Suppose that f is an analytic function, except for a ﬁnite number of singularities, and that f(z) → 0 as |z| → ∞.

11 4 The calculus of residues IB Complex Methods (Theorems with proof)

γR × ×

−R R ×

0 γR

Then for any real constant λ > 0, we have Z f(z)eiλz dz → 0 γR

as R → ∞, where γR is the semicircle of radius R in the upper half-plane. 0 For λ < 0, the same conclusion holds for the semicircular γR in the lower half-plane.

π Proof. The proof relies on the fact that for θ ∈ 0, 2 , we have 2θ sin θ ≥ . π So we get Z Z π iλz iθ iλReiθ iθ f(z)e dz = f(Re )e iRe dθ γR 0 π Z iθ iθ iλRe ≤ R |f(Re )| e dθ 0 Z π/2 ≤ 2R sup |f(z)| e−λR sin θ dθ z∈γR 0 Z π/2 ≤ 2R sup |f(z)| e−2λRθ/π dθ z∈γR 0 π = (1 − e−λR) sup |f(z)| λ z∈γR → 0,

0 as required. Same for γR when λ < 0.

12 5 Transform theory IB Complex Methods (Theorems with proof)

5 Transform theory 5.1 Fourier transforms 5.2 Laplace transform 5.3 Elementary properties of the Laplace transform Proposition. (i) Linearity: L(αf + βg) = αL(f) + βL(g).

(ii) Translation: −pt0 ˆ L(f(t − t0)H(t − t0)) = e f(p).

(iii) Scaling: 1 p L(f(λt)) = fˆ , λ λ where we require λ > 0 so that f(λt) vanishes for t < 0. (iv) Shifting: p0t ˆ L(e f(t)) = f(p − p0).

(v) Transform of a derivative:

L(f 0(t)) = pfˆ(p) − f(0).

Repeating the process,

L(f 00(t)) = pL(f 0(t)) − f 0(0) = p2fˆ(p) − pf(0) − f 0(0),

and so on. This is the key fact for solving ODEs using Laplace transforms. (vi) Derivative of a transform:

fˆ0(p) = L(−tf(t)).

Of course, the point of this is not that we know what the derivative of fˆ is. It is we know how to ﬁnd the Laplace transform of tf(t)! For example, this lets us ﬁnd the derivative of t2 with ease. In general, fˆ(n)(p) = L((−t)nf(t)).

(vii) Asymptotic limits ( f(0) as p → ∞ pfˆ(p) → , f(∞) as p → 0

where the second case requires f to have a limit at ∞.

Proof.

13 5 Transform theory IB Complex Methods (Theorems with proof)

(v) We have Z ∞ Z ∞ 0 −pt −pt ∞ −pt ˆ f (t)e dt = [f(t)e ]0 + p f(t)e dt = pf(p) − f(0). 0 0 (vi) We have Z ∞ fˆ(p) = f(t)e−pt dt. 0 Differentiating with respect to p, we have Z ∞ fˆ0(p) = − tf(t)e−pt dt. 0 (vii) Using (v), we know Z ∞ pfˆ(p) = f(0) + f 0(t)e−pt dt. 0 As p → ∞, we know e−pt → 0 for all t. So pfˆ(p) → f(0). This proof looks dodgy, but is actually valid since f 0 grows no more than exponentially fast. Similarly, as p → 0, then e−pt → 1. So Z ∞ pfˆ(p) → f(0) + f 0(t) dt = f(∞). 0 5.4 The inverse Laplace transform Proposition. The inverse Laplace transform is given by 1 Z c+i∞ f(t) = fˆ(p)ept dp, 2πi c−i∞ where c is a real constant such that the Bromwich inversion contour γ given by Re p = c lies to the right of all the singularities of fˆ(p). Proof. Since f has a Laplace transform, it grows no more than exponentially. So we can find a c ∈ R such that g(t) = f(t)e−ct decays at infinity (and is zero for t < 0, of course). So g has a Fourier transform, and Z ∞ g˜(ω) = f(t)e−cte−iωt dt = fˆ(c + iω). −∞ Then we use the Fourier inversion formula to obtain 1 Z ∞ g(t) = fˆ(c + iω)eiωt dω. 2π −∞ So we make the substitution p = c + iω, and thus obtain 1 Z c+i∞ f(t)e−ct = fˆ(p)e(p−c)t dp. 2πi c−i∞ Multiplying both sides by ect, we get the result we were looking for (the requirement that c lies to the right of all singularities is to fix the “constant of integration” so that f(t) = 0 for all t < 0, as we will soon see).

14 5 Transform theory IB Complex Methods (Theorems with proof)

Proposition. In the case that fˆ(p) has only a ﬁnite number of isolated singu- ˆ larities pk for k = 1, ··· , n, and f(p) → 0 as |p| → ∞, then

n X f(t) = res (fˆ(p)ept) p=pk k=1 for t > 0, and vanishes for t < 0.

0 Proof. We ﬁrst do the case where t < 0, consider the contour γ0 + γR as shown, which encloses no singularities.

c + iR

0 γR × γ0 ×

c − iR

Now if fˆ(p) = o(|p|−1) as |p| → ∞, then

Z fˆ(p)ept dp ≤ πRect sup |fˆ(p)| → 0 as R → ∞. 0 0 γR p∈γR

Here we used the fact |ept| ≤ ect, which arises from Re(pt) ≤ ct, noting that t < 0. If fˆ decays less rapidly at inﬁnity, but still tends to zero there, the same result holds, but we need to use a slight modiﬁcation of Jordan’s lemma. So in either case, the integral Z fˆ(p)ept dp → 0 as R → ∞. 0 γR

Thus, we know R → R . Hence, by Cauchy’s theorem, we know f(t) = 0 for γ0 γ t < 0. This is in agreement with the requirement that functions with Laplace transform vanish for t < 0. Here we see why γ must lie to the right of all singularities. If not, then the contour would encircle some singularities, and then the integral would no longer be zero. When t > 0, we close the contour to the left.

15 5 Transform theory IB Complex Methods (Theorems with proof)

c + iR

γR γ × 0 ×

c − iR

This time, our γ does enclose some singularities. Since there are only ﬁnitely many singularities, we enclose all singularities for suﬃciently large R. Once again, we get R → 0 as R → ∞. Thus, by the residue theorem, we know γR

n Z Z X fˆ(p)ept dp = lim fˆ(p)ept dp = 2πi res (fˆ(p)ept). R→∞ p=pk γ γ0 k=1 So from the Bromwich inversion formula,

n X f(t) = res (fˆ(p)ept), p=pk k=1 as required.

5.5 Solution of diﬀerential equations using the Laplace transform 5.6 The convolution theorem for Laplace transforms Theorem (Convolution theorem). The Laplace transform of a convolution is given by L(f ∗ g)(p) = fˆ(p)ˆg(p). Proof.

Z ∞ Z t L(f ∗ g)(p) = f(t − t0)g(t0) dt0 e−pt dt 0 0 We change the order of integration in the (t, t0) plane, and adjust the limits accordingly (see picture below)

Z ∞ Z ∞ = f(t − t0)g(t0)e−pt dt dt0 0 t0

16 5 Transform theory IB Complex Methods (Theorems with proof)

We substitute u = t − t0 to get

∞ ∞ Z Z 0 = f(u)g(t0)e−pue−pt du dt0 0 0 ∞ ∞ Z Z 0 = f(u)e−pu du g(t0)e−pt dt0 0 0 = fˆ(p)ˆg(p).

Note the limits of integration are correct since they both represent the region below t0