complex analysis

COMPLEX ANALYSIS 1 Douglas N. Arnold 2 References: John B. Conway, Functions of One Complex Variable, Springer-Verlag, 1978. Lars V. Ahlfors, Complex Analysis, McGraw-Hill, 1966. Raghavan Narasimhan, Complex Analysis in One Variable, Birkh¨auser, 1985. CONTENTS I. The Complex Number System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 II. Elementary Properties and Examples of Analytic Fns. . . . . . . . . . . . . . . . 3 Differentiability and analyticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 The Logarithm. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6 Conformality. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6 Cauchy–Riemann Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7 M¨obius transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 III. Complex Integration and Applications to Analytic Fns. . . . . . . . . . . . . 11 Local results and consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Homotopy of paths and Cauchy’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Winding numbers and Cauchy’s Integral Formula. . . . . . . . . . . . . . . . . . . . . . . . . 15 Zero counting; Open Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Morera’s Theorem and Goursat’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 IV. Singularities of Analytic Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .19 Laurent series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .20 Residue integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23 V. Further results on analytic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .26 The theorems of Weierstrass, Hurwitz, and Montel . . . . . . . . . . . . . . . . . . . . . . . 26 Schwarz’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 The Riemann Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Complements on Conformal Mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 VI. Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 The Poisson kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Subharmonic functions and the solution of the Dirichlet Problem . . . . . . . . . 36 The Schwarz Reflection Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 1 These lecture notes were prepared for the instructor’s personal use in teaching a half-semester course on complex analysis at the beginning graduate level at Penn State, in Spring 1 997. They are certainly not meant to replace a good text on the subject, such as those listed on this page. 2 Department of Mathematics, Penn State University, University Park, PA 16802. Web: http://www.math.psu.edu/dna/. 1 2 I. The Complex Number System R is a field. For n > 1, R n is a vectorspace over R, so is an additive group, but doesn’t have a multiplication on it. We can endow R 2 with a multiplication by (a, b)(c, d) = (ac − bd, bc + ad). Under this definition R 2 becomes a field, denoted C. Note that (a/(a 2 + b 2 ), −b/(a 2 + b 2 )) is the multiplicative inverse of (a, b). (Remark: it is not possible to endow R n with a field structure for n > 2.) We denote (0, 1) by i and identify x ∈ R with (x, 0), so R ⊂ C. Thus (a, b) = a + bi, a, b ∈ R. Note that i 2 = −1. C is generated by adjoining i to R and closing under addition and multiplication. It is remarkable that the addition of i lets us not only solve the equation x 2 + 1 = 0, but every polynomial equation. For a and b real and z = a + bi we define Re z = a, Im z = b, ¯z = a − bi, and |z| = (a 2 + b 2 ) 1/2 . Then Re z = (z + ¯z)/2, Im z = (z − ¯z)/(2i), |z| 2 = z¯z, 1 z = ¯z |z| 2 , z ± w = ¯z ± ¯w, zw = ¯z ¯w, z/w = ¯z/ ¯w, |z + w| ≤ |z| + |w|. The map θ → (cos θ, sin θ) defines a 2π-periodic map of the real line onto the unit circle in R 2 . In complex notation this map is θ → cis θ := cos θ + i sin θ. Every nonzero complex number can be written as r cis θ where r > 0 is uniquely determined and θ ∈ R is uniquely determined modulo 2π. The number 0 is equal to r cis θ where r = 0 and θ is arbitrary. The relation z = r cis θ determines the relations z → r which is simply the function r = |z| and z → θ. The latter is denoted θ = arg θ. Note that for z = 0, arg θ is determined modulo 2π (while arg 0 is arbitrary). We can normalize arg by insisting that arg z ∈ (−π, π]. Note that if z 1 = r cis θ 1 and z 2 = r cis θ 2 then z 1 z 2 = r 1 r 2 cis(θ 1 + θ 2 ). The latter formula just encapsulates the formula for the sine and cosine of a sum, and gives arg z 1 z 2 = arg z 1 + arg z 2 . In particular, ir cis θ = r cis(θ + π/2), so multiplication by i is just the operation of rotation by π/2 in the complex plane. Multiplication by an arbitrary complex number r cis θ is just rotation by arg θ followed by (or preceded by) dilation by a factor r. Further, z n = r n cis(nθ). Every nonzero z ∈ C admits n distinct nth roots: the nth roots of r cis θ are n √ r cis[(θ + 2πk)/n], k = 0, 1, . . . , n. Lines and circles in the plane. Circles given by |z − a| = r where a ∈ C is the center and r > 0 is the radius. If 0 = b ∈ C then the line through the origin in the direction b is the set of all points of the form tb, t ∈ R, or all z with Im(z/b) = 0. If t ∈ R and c > 0 then (t + ci)b = tb + cib represents a point in the half plane to the left of b determined by the line tb, i.e., {z : Im(z/b) > 0} is the equation of that half-plane. Similarly, {z : Im[(z − a)/b] > 0} is the translation of that half-plane by a, i.e., the half- plane determined by the line through a parallel to b and in the direction to the left of b. 3 Stereographic projection determines a one-to-one correspondence between the unit sphere in R 3 minus the north-pole, S, and the complex plane via the correspondence z ↔ x 1 + ix 2 1 − x 3 , x 1 = 2 Re z 1 + |z| 2 , x 2 = 2 Im z 1 + |z| 2 , x 3 = |z| 2 − 1 |z| + 1 . If we define C ∞ = C ∪ {∞}, then we have a one-to-one correspondence between S and C ∞ . This allows us to define a metric on C ∞ , which is given by d(z 1 , z 2 ) = 2|z 1 − z 2 |  (1 + |z 1 | 2 )(1 + |z 2 | 2 ) , d(z, ∞) = 2  1 + |z| 2 . II. Elementary Properties and Examples of Analytic Functions For z = 1,  N n=0 z n = (1 − z N+1 )/(1 − z). Therefore the geometric series  ∞ n=0 z n converges (to 1/(1 −z)) if |z| < 1. It clearly diverges, in fact its terms become unbounded, if |z| > 1. Weierstrass M-Test. Let M 0 , M 1 , . . . be positive numbers with  M n < ∞ and suppose that f n : X → C are functions on some set X satisfying sup x∈X |f n (x)| ≤ M n . Then  ∞ n=0 f n (x) is absolutely and uniformly convergent. Theorem. Let a 0 , a 1 , ··· ∈ C be given and define the number R by 1 R = lim sup |a n | 1/n . Then (1) for any a ∈ C the power series  ∞ n=0 a n (z − a) n converges absolutely for all |z −a| < R and it converges absolutely and uniformly on the disk |z −a| ≤ r for all r < R. (2) The sequence a n (z − a) n is unbounded for all |z − a| > R (and hence the series is certainly divergent). Thus we see that the set of points where a power series converges consists of a disk |z −a| < R and possibly a subset of its boundary. R is called the radius of convergence of its series. The case R = ∞ is allowed. Proof of theorem. For any r < R we show absolute uniform convergence on D r = {|z−a| ≤ r}. Choose ˜r ∈ (r, R). Then, 1/˜r > lim sup |a n | 1/n , so |a n | 1/n < 1/˜r for all n sufficiently large. For such n, |a n | < 1/˜r n and so sup z∈D r |a n (z − a) n | < (r/˜r) n . Since  (r/˜r) n < ∞ we get the absolute uniform convergence on D r . If |z − a| = r > R, take ˜r ∈ (R, r). Then there exist n arbitrarily large such that |a n | 1/n ≥ 1/˜r. Then, |a n (z − a) n | ≥ (r/˜r) n , which can be arbitrarily large.  4 Theorem. If a 0 , a 1 , . . . ∈ C and lim |a n /a n+1 | exists as a finite number or infinity, then this limit is the radius of convergence R of  a n (z − a) n . Proof. Without loss of generality we can suppose that a = 0. Suppose that |z| > lim |a n /a n+1 |. Then for all n sufficiently large |a n | < |a n+1 z| and |a n z n | < |a n+1 z n+1 |. Thus the series  a n z n has terms of increasing magnitude, and so cannot be convergent. Thus |z| ≥ R. This shows that lim |a n /a n+1 | ≥ R . Similarly, suppose that z < lim |a n /a n+1 |. Then for all n sufficiently large |a n | > |a n+1 z| and |a n z n | > |a n+1 z n+1 |. Thus the series has terms of decreasing magnitude, and so, by the previous theorem, |z| ≤ R. This shows that lim |a n /a n+1 | ≤ R .  Remark. On the circle of convergence, many different behaviors are possible.  z n diverges for all |z| = 1.  z n /n diverges for z = 1, else converges, but not absolutely (this follows from the fact that the partial sums of  z n are bounded for z = 1 and 1/n ↓ 0).  z n /n 2 converges absolutely on |z| ≤ 1. Sierpinski gave a (complicated) example of a function which diverges at every point of the unit circle except z = 1. As an application, we see that the series ∞  n=0 z n n! converges absolutely for all z ∈ C and that the convergence is uniform on all bounded sets. The sum is, by definition, exp z. Now suppose that  ∞ n=0 a n (z −a) n has radius of convergence R, and consider its formal derivative  ∞ n=1 na n (z−a) n−1 =  ∞ n=0 (n+1)a n+1 (z−a) n . Now clearly  n=0 a n+1 (z−a) n has the same radius of convergence as  ∞ n=0 a n (z − a) n since (z − a) N  n=0 a n+1 (z − a) n = N+1  n=0 a n (z − a) − a 0 , and so the partial sums on the left and right either both diverge for a given z or both converge. This shows (in a roundabout way) that lim sup |a n+1 | 1/n = lim sup |a n | 1/n = 1/R. Now lim(n+1) 1/n = 1 as is easily seen by taking logs. Moreover, it is easy to see that if lim sup b n = b and lim c n = c > 0, then lim sup b n c n = bc. Thus lim sup |(n + 1)a n+1 | 1/n = 1/R. This shows that the formal derivative of a power series has the same radius of convergence as the original power series. Differentiability and analyticity. Definition of differentiability at a point (assumes function is defined in a neighborhood of the point). Most of the consequences of differentiability are quite different in the real and complex case, but the simplest algebraic rules are the same, with the same proofs. First of all, differentiability at a point implies continuity there. If f and g are both differentiable at a point a, then s o are f ± g, f · g, and, if g(a) = 0, f/g, and the usual sum, product, and quotient rules hold. If f is differentiable at a and g is differentiable at f(a), then g ◦ f is differentiable at a and the chain rule holds. Suppose that f is continuous at a, g is continous at f(a), and g(f(z)) = z for all z in a neighborhood of a. Then if g  (f(a)) exists and is non-zero, then f  (a) exists and equals 1/g  (f(a)). 5 Definition. Let f be a complex-valued function defined on an open set G in C. Then f is said to be analytic on G if f  exists and is continuous at every point of G. Remark. We shall prove later that if f is differentiable at every point of an open set in C it is automatically analytic; in fact, it is automatically infinitely differentiable. This is of course vastly different from the real case. If Q is an arbitrary non-empty subset of C we say f is analytic on Q if it is defined and analytic on an open set containing Q. We now show that a power series is differentiable at every point in its disk of convergence and that its derivative is given by the formal derivative obtained by differentiating term- by-term. Since we know that that power series has the same radius of convergence, it follows that a p ower series is analytic and infinitely differentiable in its convergence disk. For simplicity, and without loss of generality we consider a power series centered at zero: f(z) =  n a n z n . Suppose that the radius of convergence is R and that |z 0 | < R . We must show that for any  > 0, the inequality     f(z) − f (z 0 ) z − z 0 − g(z 0 )     ≤  is satisfied for all z sufficiently close to z 0 , where g (z) =  ∞ n=1 na n z n−1 . Let s N (z) =  N n=0 a n z n , R N (z) =  ∞ n=N+1 a n z n . Then     f(z) − f (z 0 ) z − z 0 − g(z 0 )     ≤     s N (z) − s N (z 0 ) z − z 0 − s  N (z 0 )     + |s  N (z 0 ) − g(z 0 )| +     R N (z) − R N (z 0 ) z − z 0     =: T 1 + T 2 + T 3 . Now s  N (z 0 ) is just a partial sum for g(z 0 ), so for N sufficiently large (and all z), T 2 ≤ /3. Also, R N (z) − R n (z 0 ) z − z 0 = ∞  n=N+1 a n z n − z n 0 z − z 0 . Now |z 0 | < r < R for some r, and if we restrict to |z| < r , we have     a n z n − z n 0 z − z 0     = |a n ||z n−1 + z n−2 z 0 + ··· + z n−1 0 | ≤ a n nr n−1 . Since  a n nr n−1 is convergent, we have for N sufficiently large and all |z| < r then T 3 < /3. Now fix a value of N which is sufficiently large by both criteria. Then the differentiability of the polynomial s N shows that T 1 ≤ /3 for all z sufficiently close to z 0 . We thus know that if f(z) =  a n z n , then, within the disk of convergence, f  (z) =  na n z n−1 , and by induction, f  (z) =  n(n−1)a n z n−2 , etc. Thus a 0 = f(0), a 1 = f  (0), a 2 = f  (0)/2, a 3 = f  (0)/3!, etc. This shows that any convergent power series is the s um of its Taylor series in the disk of convergence: f(z) =  f n (a) n! (z − a) n . In particular, exp  = exp. 6 Lemma. A function with vanishing derivative on a region (connected open set) is constant. Indeed, it is constant on disks, since we can restrict to segments and use the real result. Then we can use connectedness to see that the set of points where it takes a give n value is open and closed. Use this to show that exp(z) exp(a−z) ≡ exp a. Define cos z and sin z, get cos 2 + sin 2 ≡ 1, exp z = cos z + i sin z, cis θ = exp(iθ), exp z = exp(Re z) cis(Im z). The Logarithm. If x = 0, the most general solution of exp z = w is z = log |w|+i arg w+ 2πin, n ∈ Z. There is no solution to exp z = 0. Definition. If G is an open set and f : G → C is a continuous function satisfying exp(f(z)) = z, then f is called a branch of the logarithm on G. Examples: C \ R − (principal branch), C \ R + , a spiral strip. By the formula for the derivative of an inverse, a branch of the logarithm is analytic with derivative 1/z. A branch of the logarithm gives a branch of z b := exp(b log z) (understood to be the principal branch if not otherwise noted). Note that e b = exp b. Note that different branches of the logarithm assign values to log z that differ by addition of 2πin. Consequently different branches of z b differ by factors of exp(2πinb). If b is an integer, all values agree. If b is a rational number with denominator d there are d values. Conformality. Consider the angle between two line segments with a common vertex in the complex plane. We wish to consider how this angle is transformed under a complex map. The image will be two smooth curves meeting at a common point, so we have to define the angle between two such curves. Define a path in the complex plane as a continuous map γ : [a, b] → C. If γ  (t) ∈ C exists at some point and is not zero, then it is a vector tangent to the curve at γ(t), and hence its argument is the angle between the (tangent to the) curve and the horizontal. (Note: if γ  (t) = 0, the curve may not have a tangent; e.g., γ(t) = t 2 − it 3 .) Let γ : [a, b] → C and ˜γ : [˜a, ˜ b] → C be two smooth curves, such that γ(a) = ˜γ(˜a) = z, and γ  (a), ˜γ  (˜a) = 0. Then arg ˜γ  (˜a)−arg γ  (a) measures the angle between the curves. Now consider the images under f, e.g., of ρ(t) = f(γ(t)). Then ρ  (a) = f  (z)γ  (a), ˜ρ  (˜a) = f  (z)˜γ  (˜a). Therefore arg ˜ρ  (˜a)−arg ρ  (a) = arg ˜γ  (˜a)+ arg f  (z) −arg γ  (a)−arg f  (z), i.e., the angle is invariant in both magnitude and sign. This says that an analytic mapping is conformal, whenever its derivative is not zero. (Not true if the derivative is zero. E.g., z 2 doubles angles at the origin.) The conformality of an analytic map arises from the fact that in a neighborhood of a point z 0 such a map behaves to first order like multiplication by f  (z 0 ). Since f  (z 0 ) = r exp iθ, near z 0 f simply behave like rotation by θ followed by dilation. By contrast, a smooth map from R 2 → R 2 can behave like an arbitrary linear operator in a neighborhood of a point. At this point, show the computer graphics square1.mps and square2.mps. 7 Cauchy–Riemann Equations. Let f : Ω → C with Ω ⊂ C open. Abuse notation slightly by writing f(x, y) as an alternative for f(x + iy). If f  (z) exists for some z = x + iy ∈ Ω, then f  (z) = lim h∈R h→0 f(z + h) − f (z) h = lim h∈R h→0 f(x + h, y) − f(x, y) h = ∂f ∂x , and f  (z) = lim h∈R h→0 f(z + ih) − f (z) ih = lim h∈R h→0 −i f(x, y + h) − f(x, y) h = −i ∂f ∂y . Thus complex-differentiability of f at z implies not only that the partial derivatives of f exist there, but also that they satisfy the Cauchy–Riemann equation ∂f ∂x = −i ∂f ∂y . If f = u + iv, then this equation is equivalent to the system ∂u ∂x = ∂v ∂y , ∂v ∂x = − ∂u ∂y . Another convenient notation is to introduce ∂f ∂z := 1 2  ∂f ∂x − i ∂f ∂y  , ∂f ∂¯z := 1 2  ∂f ∂x + i ∂f ∂y  . (These are motivated by the equations x = (z + ¯z)/2, y = (z − ¯z)/(2i), which, if z and ¯z were independent variables, would give ∂x/∂z = 1/2, ∂y/∂z = −i/2, etc.) In terms of these, the Cauchy–Riemann equations are exactly equivalent to ∂f ∂¯z = 0, which is also equivalent to ∂f ∂z = ∂f ∂x . Thus, if f is analytic on Ω, then the partial derivaties of f exist and are continuous on Ω, and the Cauchy-Riemann equations are satisfied there. The converse is true as well: Theorem. If the partial derivatives of f exist and are continuous on Ω, and the Cauchy- Riemann equations are satisfied there, then f is analytic on Ω. 8 Proof. Let z = x + iy ∈ Ω. We must show that f  (z) exists. Let r be small enough that the disk of radius r around z belongs to Ω and choose h = s + it with 0 < |s + it| < r. Then, by the mean value theorem, f(z + h) − f (z) h = f(x + s, y + t) − f(x, y) s + it = f(x + s, y + t) − f(x, y + t) s s s + it + f(x, y + t) − f(x, y) t t s + it = ∂f ∂x (x + s ∗ , y + t) s s + it + ∂f ∂y (x, y + t ∗ ) t s + it , where |s ∗ | < s and |t ∗ | < t. Note that ∂f ∂x (x + s ∗ , y + t) − ∂f ∂x (x, y) → 0 as h → 0, and similarly for the second partial derivative. Moreover |s/(s + it)| stays bounded (by 1). Thus f(z + h) − f (z) h = ∂f ∂x (x, y) s s + it + ∂f ∂y (x, y) t s + it + R(h), where lim h→0 R(h) = 0. Now if the Cauchy–Riemann equations hold, then the the first two terms on the right hand side sum to ∂f/∂x(x, y), which is independent of h, so the limit exists and is equal to ∂f/∂x(x, y).  Remark. The theorem can be weakened to say that if f is continuous on Ω and the partial derivatives exist and satisfy the Cauchy–Riemann equations there (without assuming that the partial derivatives are continuous), then the complex derivative of f exists on Ω (which is equivalent to f being analytic on Ω. This is the Looman–Menchoff Theorem. See Narasimhan, Complex Analysis in One Variable, for a proof. We do need at least continuity, since otherwise we could take f to be the characteristic function of the coordinate axes. Note that ∂ ∂z ∂ ∂¯z = ∂ ∂¯z ∂ ∂z = 1 4 ∆. This shows that any analytic function is harmonic (equivalently, its real and imaginary parts are harmonic). It also shows that the conjugate of an analytic function, while not analytic, is harmonic. If u(x, y) = Re f(x + iy), then u is harmonic. Conversely, if u is harmonic on a region Ω, does there exist a harmonic conjugate, i.e., a function v on Ω such that f = u + iv is analytic? Clearly, if there exists v it is determined up to addition of a real constant (or f is determined up to addition of an imaginary constant). The book gives an elementary construction of the harmonic conjugate in a disk and in the whole plane. We sketch the idea of a more general proof. We require that the domain satisfy the condition that if γ is any piecewise simple smooth closed curve in Ω, then γ is the b oundary of a subset G of 9 Ω. In other words, Ω has no holes (is simply-connected). Our proof follows directly from the following result, which holds on simply-connected domains: let f : Ω → R 2 be a C 1 vectorfield. Then there exists v : Ω → R such that f = ∇v if and only if ∂f 1 /∂y = ∂f 2 /∂x. The “only if” part is obvious. To prove the “if” part, fix a point (x 0 , y 0 ) in Ω and for any (x, y) ∈ Ω, let γ (x,y) be a piecewise smooth path in Ω from (x 0 , y 0 ) to (x, y), and let τ be the unit tangent to the path pointing from (x 0 , y 0 ) towards (x, y). Define v(x, y) =  γ (x,y) f · τ ds. It is essential that this quantity doesn’t depend on the choice of path, or, equivalently, that  γ f · τ ds = 0 for all piecewise smooth simple closed paths γ. But  γ f · τ ds =  γ (f 2 , −f 1 ) · n ds =  ∇ · (f 2 , −f 1 ) dx dy = 0, by the divergence theorem. It is easy to check that ∇v = f. For a non-simply connected region, there may exist no harmonic conugate. E.g., if u = log |z| on C \ {0}, then u + i arg z is analytic on C \ {z ≤ 0}, but cannot be extended to an analytic function on C \ {0}. M¨obius transformations. Definition. Let a, b, c, d be complex numbers with ad = bc. Then the mapping S(z) = az + b cz + d , z ∈ C, z = −d/c, is a M¨obius transformation. Remarks. 1) Note that if ad = bc the same expression would yield a constant. 2) The coefficents aren’t unique, since we can multiply them all by any nonzero complex constant. To each MT we associate the nonsingular matrix  a b c d  of its coefficients, which is determined up to a non-zero multiple. 3) The linear (but non-constant) polynomials are MTs, namely the ones for which c = 0. 4) If z = −d/c then S(z) ∈ C and S  (z) = (ad − bc)/(cz + d) 2 = 0. 5) We can define S(−d/c) = ∞ and S(∞) = a/c, so S can be viewed of as a map from C ∞ into itself (which is, as we will now see, 1-1 and onto). If S and T are MTs, then so is S ◦ T , its coefficient matrix being the product of the coefficient matrices of S and T . Any MTs admits an inverse, namely the transform with the inverse coefficient matrix. With the understanding that S(∞) = a/c and S(−d/c) = ∞ (or, if c = 0, then S(∞) = ∞, the MT maps C ∞ 1-1 onto itself. This is evident from the 10 existence of its inverse S −1 (z) = (dz −b)/(−cz + a). Moreover, the composition of M¨obius transformations is again one, so they form a group under composition. If S is a non-linear MT we can write it as az + b z + d = a + b − ad z + d . This shows that any MT can be written as a comp osition of translations, rotations and dilations (multiplication by a complex number), and inversion (reciprocals). Since the geometric action of translation, dilation, and rotation are quite evident, let’s consider inversion. It takes the unit disk D to C ∞ \ ¯ D. We now show that inversion maps lines and circles to other lines and circles. Consider the equation |z − p| = k|z − q| where k > 0, p, q ∈ C. This is (x − a) 2 + (y − b) 2 = k 2 [(x − c) 2 + (y − d) 2 ] which is clearly the equation of a circle if k = 1 and the equation of a line if k = 1. In either case we have a circle in the Riemann sphere, which we call C 1 . Substituting 1/z for z and doing some simple manipulations gives the equation |z − 1/p| = (k|q|/|p|)|z −1/q|, which is another such circle, C 2 . Thus we have 1/z ∈ C 1 ⇐⇒ z ∈ C 2 , showing that inversion maps circles in the sphere onto other circles in the sphere. Since the same property is evident for translation, dilation, and rotation, we conclude that this property holds for all MT. Some further analysis, which will be omitted, establishes two further properties. 1) A MT is orientiation preserving in the sense that, if we traverse a circle in the order of three distinct points on it, z 1 , z 2 , z 3 , the region to the left of the circle will map to the region to the left of the image circle, with respect to the image orientation. 2) A MT preserves the property of two points being symmetric with respect to a circle, i.e., lying on the same ray from the center, and such that the geometric mean of their distances from the center equals the radius. Let z 2 , z 3 , z 4 be distinct points in C. Then it is easy to see that there is a unique MT taking these points to 1, 0, and ∞, respectively, namely Sz = z − z 3 z − z 4 z 2 − z 4 z 2 − z 3 . We can also handle, as a special case, the possibility that one of the z i is ∞. We infer from this and the invertibility of the MTs that given two ordered triples of distinct points in C ∞ , there is a unique MT taking the first onto the second. If we have any two disks or halfplanes, or one of each, we can map one to the other by a MT, and can arrange that the image of any three points on the boundary of one go to three points on the boundary of the other. The book makes heavy use of the the notation (z 1 , z 2 , z 3 , z 4 ) (cross ratio) for the image of z 1 under the transformation taking z 2 , z 3 , z 4 to 1, 0, and ∞, respectively, so (z 1 , z 2 , z 3 , z 4 ) = z 1 − z 3 z 1 − z 4 z 2 − z 4 z 2 − z 3 . From the definition, (T z 1 , T z 2 , T z 3 , T z 4 ) = (z 1 , z 2 , z 3 , z 4 ) for any MT T . Example: z → (1 − z)/(1 + z) takes the right half plane onto the unit disk. [...]...11 III Complex Integration and Applications to Analytic Functions If γ : [a, b] → C is piecewise smooth, and f is a continuous complex- valued function defined on the image of γ, define b f (z) dz = γ f (γ(t))γ (t) dt a This is simply the line integral of f along γ It can... an interesting example, take G to be the complement of the spiral r = θ.) We can at long last show that a function which is complex differentiable in a region (but without assuming it is continuously differentiable) is analytic Goursat’s Theorem Let G be a region and f : G → C complex differentiable at each point of G Then f is analytic Proof We will show that the integral of f vanishes over every triangular... path in G is homotopic to 0 As an immediate consequence of the theorem above, we have Cauchy’s theorem (previously proved only for G a disk) Cauchy’s Theorem If G is is a simply-connected region in the complex plane then f (z) dz = 0 γ for every function f analytic in G and every piecewise smooth closed curve γ in G Winding numbers and Cauchy’s Integral Formula Intuitive meaning of winding number: if... a path around the boundary of the disk Then f (w) dw = 0 γ Proof Fix z in the open disk and apply Cauchy’s formula to the function w → f (w)(w − z) We can now show that an analytic (i.e., continuously complex differentiable) function has a power series expansion around each point in its domain (And we already know the converse is true.) Theorem Let f be analytic in B(a; R) Then f (z) = some an ∈ C an... the series n=−∞ an (z − a) ∞ ∞ −n n a−n (z − a) and an (z − a) converge (and their sum is the value of the n=1 n=0 doubly infinite summation) Lemma Let γ be a piecewise smooth curve and g a continuous complex- valued function on γ Define g(w) dw, z ∈ γ / f (z) = γ z−w The f is analytic on C \ γ Proof Fix z ∈ γ Then / g(w) ζ−w − g(w) z−w =− ζ−z g(w) g(w) →− (ζ − w)(z − w) (z − w)2 as ζ → z with the convergence... if n is large, is bounded by 2 on the horizontal segments It follows easily that the integral goes to 0 with n, so ∞ 1 π2 = (n − a)2 sin2 πa n=−∞ To state the argument principle we define: Definition A complex function on an open set G is called meromorphic if it is analytic on G except for a set of poles Note that the pole set of a meromorphic function is discrete (but may be infinite) A meromorphic... in the unit disk (take f = z exp(λ − z), φ = −1) V Further results on analytic functions The theorems of Weierstrass, Hurwitz, and Montel Consider the space of continuous functions on a region in the complex plane, endowed with the topology of uniform convergence on compact subsets (In fact this is a metric space topology: we can define a metric such that a sequence of continuous functions converge... root g(z), which we may determine uniquely by insisting that g(0) = b Defining r = φb ◦ g, we have r(0) = 0, and q(r(z)) ≡ z Thus if 0 = z ∈ G, r(z) = 0 and |z| = |q(r(z))| < |r(z)| Two regions in the complex plane are called conformally equivalent if they are analytically isomorphic, that is, if there exists a one-to-one analytic mapping of the first domain onto the second (whose inverse is then automatically... to airfoil design An important class of regions are the polygons The Schwarz–Christoffel formula is an explicit formula for the conformal map of a given polygon onto the disk The formula involves a few complex parameters which are determined by the vertices of the polygon Given the vertices, one cannot usually compute the parameters analytically, but they are not difficult to compute numerically (with . COMPLEX ANALYSIS 1 Douglas N. Arnold 2 References: John B. Conway, Functions of One Complex Variable, Springer-Verlag, 1978. Lars V. Ahlfors, Complex Analysis, McGraw-Hill,. Ahlfors, Complex Analysis, McGraw-Hill, 1966. Raghavan Narasimhan, Complex Analysis in One Variable, Birkh¨auser, 1985. CONTENTS I. The Complex Number System . . . . . . . . . . . . . . . . . . . are continuous), then the complex derivative of f exists on Ω (which is equivalent to f being analytic on Ω. This is the Looman–Menchoff Theorem. See Narasimhan, Complex Analysis in One Variable,