Chapter 1 Complex integration 1.1 Complex number quiz 1 1. Simplify 3+4i . 1 2. Simplify | 3+4i |. 3. Find the cube roots of 1. 4. Here are some identities for complex conjugate. Which ones need correc- tion? z + w =z ¯+w ¯, z − w =z ¯− w¯, zw =z ¯w¯, z/w =z/ ¯ w¯. Make suitable corrections, perhaps changing an equality to an inequality. 5. Here are some identities for absolute value. Which ones need correction? |z + w| = |z| + |w|, |z − w| = |z| − |w|, |zw| = |z||w|, |z/w| = |z|/|w|. Make suitable corrections, perhaps changing an equality to an inequality. 6. Define log(z) so that −π < = log(z) ≤ π. Discuss the identities elog(z) = z and log(ew) = w. 7. Define zw = ew log z. Find ii. 8. What is the power series of log(1 + z) about z = 0? What is the radius of convergence of this power series? 9. What is the power series of cos(z) about z = 0? What is its radius of convergence? 10. Fix w. How many solutions are there of cos(z) = w with −π < <z ≤ π. 1 2 CHAPTER 1. COMPLEX INTEGRATION 1.2 Complex functions 1.2.1 Closed and exact forms In the following a region will refer to an open subset of the plane. A differential form p dx + q dy is said to be closed in a region R if throughout the region ∂q ∂p = . (1.1) ∂x ∂y It is said to be exact in a region R if there is a function h defined on the region with dh = p dx + q dy. (1.2) Theorem. An exact form is closed. The converse is not true. Consider, for instance, the plane minus the origin. The form (−y dx + x dy)/(x2 + y2) is not exact in this region. It is, however, exact in the plane minus the negative axis. In this region −y dx + x dy = dθ, (1.3) x2 + y2 where −π/2 < θ < π/2. Green’s theorem. If S is a bounded region with oriented boundary ∂S, then Z ZZ ∂q ∂p p dx + q dy = ( − ) dx dy. (1.4) ∂S S ∂x ∂y Consider a region R and an oriented curve C in R. Then C ∼ 0 (C is homologous to 0) in R means that there is a bounded region S such that S and its oriented boundary ∂S are contained in R such that ∂S = C. Corollary. If p dx + q dy is closed in some region R, and if C ∼ 0 in R, then Z p dx + q dy = 0. (1.5) C If C is an oriented curve, then −C is the oriented curve with the opposite orientation. The sum C1 + C2 of two oriented curves is obtained by following one curve and then the other. The difference C1 − C2 is defined by following one curve and then the other in the reverse direction. Consider a region R and two oriented curves C1 and C2 in R. Then C1 ∼ C2 (C1 is homologous to C2) in R means that C1 − C2 ∼ 0 in R. Corollary. If p dx + q dy is closed in some region R, and if C1 ∼ C2 in R, then Z Z p dx + q dy = p dx + q dy. (1.6) C1 C2 1.2. COMPLEX FUNCTIONS 3 1.2.2 Cauchy-Riemann equations Write z = x + iy. Define partial differential operators ∂ ∂ 1 ∂ = + (1.7) ∂z ∂x i ∂y and ∂ ∂ 1 ∂ = − (1.8) ∂z¯ ∂x i ∂y The justification for this definition is the following. Every polynomial in x, y may be written as a polynomial in z, z¯, and conversely. Then for each term in such a polynomial ∂ zmz¯n = mzm−1z¯n (1.9) ∂z and ∂ zmz¯n = zm nz¯n−1. (1.10) ∂z¯ Let w = u + iv be a function f(z) of z = x + iy. Suppose that this satisfies the system of partial differential equations ∂w = 0. (1.11) ∂z¯ In this case we say that f(z) is an analytic function of z in this region. Explicitly ∂(u + iv) ∂(u + iv) − = 0. (1.12) ∂x ∂iy This gives the Cauchy-Riemann equations ∂u ∂v = (1.13) ∂x ∂y and ∂v ∂u = − . (1.14) ∂x ∂y 1.2.3 The Cauchy integral theorem Consider an analytic function w = f(z) and the differential form w dz = f(z) dz = (u + iv)(dx + idy) = (u dx − v dy) + i(v dx + u dy). (1.15) According to the Cauchy-Riemann equations, this is a closed form. Theorem (Cauchy integral theorem) If f(z) is analytic in a region R, and if C ∼ 0 in R, then Z f(z) dz = 0. (1.16) C 4 CHAPTER 1. COMPLEX INTEGRATION Example: Consider the differential form zm dz for integer m 6= 1. When m ≥ 0 this is defined in the entire complex plane; when m < 0 it is defined in the punctured plane (the plane with 0 removed). It is exact, since 1 zm dz = dzm+1. (1.17) m + 1 On the other hand, the differential form dz/z is closed but not exact in the punctured plane. 1.2.4 Polar representation The exponential function is defined by X∞ zn exp(z) = ez = . (1.18) n! n=0 It is easy to check that ex+iy = exeiy = ex(cos(y) + i sin(y)). (1.19) Sometimes it is useful to represent a complex number in the polar represen- tation z = x + iy = r(cos(θ) + i sin(θ)). (1.20) This can also be written z = reiθ. (1.21) From this we derive dz = dx + i dy = dr eiθ + rieiθ dθ. (1.22) This may also be written dz dr = + i dθ. (1.23) z r Notice that this does not say that dz/z is exact in the punctured plane. The reason is that the angle θ is not defined in this region. However dz/z is exact in a cut plane, that is, a plane that excludes some line running from the origin to infinity. Let C(0) be a circle of radius r centered at 0. We conclude that Z Z 2π f(z) dz = f(z)z i dθ. (1.24) C(0) 0 In particular, Z Z 1 2π dz = i dθ = 2πi. (1.25) C(0) z 0 By a change of variable, we conclude that for a circle C(z) of radius r centered at z we have Z 1 dξ = 2πi. (1.26) C(z) ξ − z 1.3. COMPLEX INTEGRATION AND RESIDUE CALCULUS 5 1.2.5 Branch cuts Remember that dz dr = + i dθ (1.27) z r is exact in a cut plane. Therefore dz = d log(z) (1.28) z in a cut plane, log(z) = log(r) + iθ (1.29) Two convenient choices are 0 < θ < 2π (cut along the positive axis and −π < θ < π (cut along the negative axis). In the same way one can define such functions as √ 1 z = exp( log(z)). (1.30) 2 Again one must make a convention about the cut. 1.3 Complex integration and residue calculus 1.3.1 The Cauchy integral formula Theorem. (Cauchy integral formula) Let f(ξ) be analytic in a region R. Let C ∼ 0 in R, so that C = ∂S, where S is a bounded region contained in R. Let z be a point in S. Then Z 1 f(ξ) f(z) = dξ. (1.31) 2πi C ξ − z 0 Proof: Let Cδ(z) be a small circle about z. Let R be the region R with the 0 point z removed. Then C ∼ Cδ(z) in R . It follows that Z Z 1 f(ξ) 1 f(ξ) dξ = dξ. (1.32) 2πi C ξ − z 2πi Cδ (z) ξ − z It follows that Z Z 1 f(ξ) 1 f(ξ) − f(z) dξ − f(z) = dξ. (1.33) 2πi C ξ − z 2πi Cδ (z) ξ − z Consider an arbitrary ² > 0. The function f(ξ) is continuous at ξ = z. Therefore there is a δ so small that for ξ on Cδ(z) the absolute value |f(ξ)−f(z)| ≤ ². Then the integral on the right hand side has integral with absolute value bounded by Z 1 2π ² δdθ = ². (1.34) 2π 0 δ Therefore the left hand side has absolute value bounded by ². Since ² is arbitrary, the left hand side is zero. 6 CHAPTER 1. COMPLEX INTEGRATION 1.3.2 The residue calculus Say that f(z) has an isolated singularity at z0. Let Cδ(z0) be a circle about z0 that contains no other singularity. Then the residue of f(z) at z0 is the integral Z 1 res(z0) = f(z) dz. (1.35) 2πi Cδ (z0) Theorem. (Residue Theorem) Say that C ∼ 0 in R, so that C = ∂S with the bounded region S contained in R. Suppose that f(z) is analytic in R except for isolated singularities z1, . , zk in S. Then Z Xk f(z) dz = 2πi res(zj). (1.36) C j=1 Proof: Let R0 be R with the singularities omitted. Consider small circles 0 C1,...,Ck around these singularities such that C ∼ C1 + ··· + Ck in R . Apply the Cauchy integral theorem to C − C1 − ... − Ck. If f(z) = g(z)/(z − z0) with g(z) analytic near z0 and g(z0) 6= 0, then f(z) is said to have a pole of order 1 at z0. Theorem. If f(z) = g(z)/(z − z0) has a pole of order 1, then its residue at that pole is res(z0) = g(z0) = lim (z − z0)f(z).