MA3111S COMPLEX ANALYSIS I

MA3111S COMPLEX ANALYSIS I 1. The Algebra of Complex Numbers A complex number is an expression of the form a + ib, where a and b are real numbers. a is called the real part of a + ib and b the imaginary part. We write a = Re(a + ib) b = Im(a + ib). A number of the form a + i0 is identified with (and usually written as) the real number a. In this way, the set of real numbers is identified with a subset of the complex numbers. Similarly, 0+ib is usually written as ib, and these are called purely imaginary numbers. The set of complex numbers is denoted by C. Arithmetic of complex numbers is defined as follows: (a + ib) ± (c + id) = (a ± c) + i(b ± d) (a + ib) (c + id) = (ac bd) + i(ad + bc) a + ib c + id ac + bd ad = c 2 + ibc + d2 c 2 + d 2. Since each complex number consists of two real coordinates, we may naturally identify it with a point in the two-dimensional plane. In this context, we call the plane the complex plane. Furthermore, we can give each complex number its polar coordinates, which we ll indicate for now with square brackets [r, θ] (r 0, θ R). Thus [r, θ] = r cos θ + ir sin θ. If z = [r, θ], we call r the modulus of z and denote it by z, and we call θ an argument of z and write arg z = θ. For z 0, the principal argument of 1

2 MA3111S COMPLEX ANALYSIS I z is the unique value of arg z that lies in the interval ( π, π]. The principal argument of z is denoted by Arg z. So, for example a + ib = a 2 + b 2 and tan arg(a + ib) = b a if a 0. Note that if z 0, then arg z is determined only up to multiples of 2π. For z = 0, arg 0 can be any real number. For example, arg(1 + i) = π 4 + 2kπ, k Z, Arg(1 + i) = π 4. In polar coordinates, multiplication and division take on simple forms. [r 1, θ 1 ] [r 2, θ 2 ] = [r 1 r 2, θ 1 + θ 2 ] [r 1, θ 1 ] [r 2, θ 2 ] = [r 1 r 2, θ 1 θ 2 ] if r 2 0. In particular, it follows from the first equation that z 1 z 2 = z 1 z 2. Also, if z = [r, θ] and n is a positive integer, then z n = [r n, nθ]. Definition. If z C and n is a positive integer, the n-th root of z, denoted by z 1/n is defined to be the set of all complex numbers w so that w n = z. Suppose z = [r, θ] 0. If w = [ρ, ξ] and w n = z, then by the above [ρ n, nξ] = [r, θ]. Hence ρ n = r and nξ = θ + 2kπ for some k Z. Thus z 1/n = w = [ n r, θ n + 2kπ ], 0 k n 1. n Here n r denotes the positive n-th root of r. So any nonzero complex number has exactly n n-th roots. The only n-th root of 0 is 0. The complex conjugate of a complex number z = a + ib is defined to be z = a ib. The following are easy to verify: In particular, z 1 ± z 2 = z 1 ± z 2 z 1 z 2 = z 1 z 2 z + z = 2 Re z z z = i2 Im z z 1 = ( z 1 ) z 2 z 2 zz = z 2. z 1 + z 2 2 = (z 1 + z 2 )(z 1 + z 2 ) = (z 1 + z 2 )(z 1 + z 2 ) = z 1 2 + 2 Re(z 1 z 2 ) + z 2 2 z 1 2 + 2 z 1 z 2 + z 2 2 = z 1 2 + 2 z 1 z 2 + z 2 2 = z 1 2 + 2 z 1 z 2 + z 2 2 = ( z 1 + z 2 ) 2. Thus z 1 + z 2 z 1 + z 2.

MA3111S COMPLEX ANALYSIS I 3 2. Sets, Convergence and Continuity If z C and r > 0, let B(z, r) = {w C : w z < r}. This is called the open ball of radius r centered at z. A subset U of C is open if for all z U, there exists r > 0 so that B(z, r) U. A subset S of C is closed if its complement in C is open. A sequence (z n ) n=1 of complex numbers is said to converge to a complex number z 0 if for all ε > 0, there exists N N such that z n z 0 ε for all n N. (z n ) n=1 is Cauchy if for all ε > 0, there exists N N such that z n z m ε for all n, m N. A subset S of C is bounded if there exists M < so that z M for all z S. Proposition 1. If S is a closed set and (z n ) n=1 is a sequence in S that converges to a point z in C, then z S. Proposition 2. A sequence of complex numbers is convergent if and only if it is Cauchy. Proposition 3. (Bolzano-Weierstrass Theorem) Every bounded sequence in C has a convergent subsequence. Proposition 4. (Nested Sets Theorem) Suppose that (K n ) n=1 is a sequence of nonempty closed bounded subsets of C such that K n+1 K n for all n. Then n=1 K n. Sketch of proof. Pick a sequence (z n ) n=1 so that z n K n for all n. Take a convergent subsequence. An infinite series (of complex numbers) is an expression of the form k=0 a k for some a k C. Its sequence of partial sums is the sequence (s n ) n=0, where s n = n k=0 a k. The series converges to a number a if the sequence (s n ) n=0 converges to a. The series converges absolutely if k=0 a k converges. Let U be a subset of C. A point z C is an accumulation point of U if for all ε > 0, the set U B(z, ε) contains a point other than z. If f : U C and w is an accumulation point of U, we say that lim z w f(z) = L if for all ε > 0, there exists δ > 0 so that f(z) L < ε for all z U such that 0 < z w < δ. Proposition 5. Suppose that f, g : U C and w is an accumulation point of U. If lim z w f(z) = L 1 and lim z w g(z) = L 2, then lim (f ± g)(z) = L 1 ± L 2 z w lim (fg)(z) = L 1L 2 z w (f ) L 1 If L 2 0, then lim (z) =. z w g L 2 Let U be a subset of C. A function f : U C is continuous at a point w U if for all ε > 0, there exists δ > 0 so that f(z) f(w) < ε for all z U such that z w < δ.

4 MA3111S COMPLEX ANALYSIS I Proposition 6. Suppose that f, g : U C are both continuous at a point w U. Then f ± g, fg are continuous at w. If g(w) 0, then f/g is also continuous at w. Suppose that f : U C is continuous at w U and g : V C is continuous at v V. If f(u) V and v = f(w), then g f is continuous at w. 3. Differentiation Suppose that a, b R with a < b. We say that : [a, b] C is differentiable at t 0 [a, b] if (t) (t 0 ) (t 0 ) = lim t t0 t t 0 exists. (t 0 ) is called the derivative of at t 0. Suppose that U C contains B(z 0, r) for some z 0 C and r > 0. A function f : U C is said to be differentiable at z 0 if f (z 0 ) = lim z z0 f(z) f(z 0 ) z z 0 exists. f (z 0 ) is called the derivative of f at z 0. f is analytic at z 0 if there exists r > 0 so that f is differentiable at all points in B(z 0, r ). Example. Consider the functions f(z) = z, g(z) = z, h(z) = z 2. Proposition 7. If f is differentiable at z 0, then it is continuous at z 0. Proposition 8. Suppose that f and g are differentiable at z 0. Then f ± g, and fg are differentiable at z 0. If g(z 0 ) 0, then f/g is also differentiable at z 0. (f ± g) (z 0 ) = f (z 0 ) ± g (z 0 ) (fg) (z 0 ) = f (z 0 )g(z 0 ) + f(z 0 )g (z 0 ) (f ) (z0 ) = f (z 0 )g(z 0 ) f(z 0 )g (z 0 ) g g(z 0 ) 2. Proposition 9. (Chain rule) If f is differentiable at z 0, g is differentiable at w 0 and g(w 0 ) = z 0, then f g is differentiable at w 0 and (f g) (w 0 ) = f (z 0 )g (w 0 ). Given a function f : U C, define u, v : U R by u(x, y) = Re f(x + iy) and v(x, y) = Im f(x + iy) respectively. Here we think of U both as a subset of C and also of R 2. u and v are called the real part and the imaginary part of f respectively. Theorem 10. (Cauchy-Riemann equations) If U C and f : U C is differentiable at z 0 = x 0 + iy 0, then the partial derivatives u x, u y, v x, v y all exist at (x 0, y 0 ) and satisfy the Cauchy-Riemann equations u x (x 0, y 0 ) = v y (x 0, y 0 ) u y (x 0, y 0 ) = v x (x 0, y 0 ).

MA3111S COMPLEX ANALYSIS I 5 In this case, f (z 0 ) = u x (x 0, y 0 ) + iv x (x 0, y 0 ). Conversely, if u x, u y, v x, v y exist on B(z 0, r) for some r > 0, and are continuous and satisfy the Cauchy- Riemann equations u x = v y u y = v x at z 0, then f is differentiable at z 0. Sketch of proof. If f is differentiable at z 0, set z = (x 0 + t) + iy 0, t R, and take the limit of the difference quotient as t 0. Repeat with z = x 0 + i(y 0 + t). Conversely, for z = x + iy, f(z) f(z 0 ) = u(x, y) u(x, y 0 ) + u(x, y 0 ) u(x 0, y 0 ) + i(v(x, y) v(x, y 0 ) + v(x, y 0 ) v(x 0, y 0 )). Apply the Mean Value Theorem to each pair of differences. Proposition 11. If f is analytic on B(z 0, r) and f (z) = 0 for all z B(z 0, r), then f is constant. Sketch of proof. Apply the Mean Value Theorem to the real and imaginary parts of f. Example. Determine all, if any, functions f : C C that are analytic on C and satisfy f (z) = f(z) for all z C. Sketch of proof. If f = u+iv, then u x = u and hence u = g(y)e x. Then v x = u y = g (y)e x and hence v = h(y) g (y)e x. Since v = v x, h(y) = 0. But u x = v y and so we must have g(y)+g (y) = 0. Thus g(y) = C 1 cos y+c 2 sin y. Therefore, f = u + iv = e x [(C 1 cos y + C 2 sin y) + i(c 2 cos y C 1 sin y)]. Definition. Define e x+iy = e x cos y + ie x sin y = [e x, y]. e z is sometimes also written as exp(z). For z C, define cos z = 1 2 (eiz + e iz ) sin z = 1 2i (eiz e iz ). A function f : C C that is analytic on C is said to be entire. e z, cos z and sin z are entire functions, as is any polynomial P (z) = n k=0 a kz k. Define the remaining trigonometric functions in the usual way, at all points where the denominator is nonzero: tan z = sin z cot z = cos z cos z sin z sec z = 1 csc z = 1 cos z sin z. These functions are analytic on their respective domains. If r > 0, denote by Log r the unique real number such that e Log r = r. Definition. If z C, z 0, let log z denote the set of complex numbers w such that e w = z. Log z (the principal logarithm) is the unique complex

6 MA3111S COMPLEX ANALYSIS I number w such that e w = z and π < Im w π. Suppose that a nonzero complex number z is written in polar form as z = re iθ for some r > 0 and θ R. To solve the equation e w = z, write w = u + iv. Then r = e u and e iθ = e iv. Hence u = Log r and v = θ + 2kπ, k Z. Thus log z = Log z + i arg z, Log z = Log z + i Arg z. Let us reiterate that log z has infinitely many values if z 0. Proposition 12. The function f(z) = Log z is analytic on the slit plane {re iθ : r > 0, θ ( π, π)}. f (z) = 1/z for any z in the slit plane. With Log z, what we have done is chosen a particular branch of the logarithm. We can make this precise as follows. If U is a subset of C, we say that f : U C is a branch of log z on U if e f(z) = z for all z U. Proposition 13. For any θ 0 R, there is an analytic branch of log z on the set C \ {re iθ 0 : r 0}. If z, w are complex numbers and z 0, e, define z w to be the set of values e w log z. The principal value of z w is e w Log z. Similar to the case for log z, we can also speak of branches of z w. 4. Line Integrals A path is a continuous function : [a, b] C for some closed bounded interval [a, b] in R. A path : [a, b] C is piecewise smooth if there is a partition P = {a = x 0 < < x n = b} of [a, b] so that is differentiable on [x k 1, x k ] and is continuous on [x k 1, x k ] for all 1 k n. The trace of a path : [a, b] C is the set {} = {(t) : t [a, b]}. Suppose that : [a, b] C is a path. Let : [ b, a] C be the path ( )(t) = ( t). is piecewise smooth if is. If 1 : [a, b] C and 2 : [c, d] C are paths so that 1 (b) = 2 (c), define the path 1 + 2 : [a, b + d c] C to be { 1 (t) if t [a, b] ( 1 + 2 )(t) = 2 (t b + c) if t (b, b + d c] 1 + 2 is piecewise smooth if both 1 and 2 are piecewise smooth. For a function g : [a, b] C, define b a g(t) dt = b a Re g(t) dt + i b a Im g(t) dt if both integrals on the right exist. Given a piecewise smooth path : [a, g] C, the length of is defined to be L() = b a (t) dt.

MA3111S COMPLEX ANALYSIS I 7 Suppose that : [a, b] C is a piecewise smooth path. If f is a complex function defined on {}, let b f(z) dz = f((t)) (t) dt if the integral on the right exists. a Proposition 14. (1) If f is continuous on {} and is a piecewise smooth path, then f(z) dz exists. (2) If f(z) dz exists, then f(z) dz exists and f(z) dz = f(z) dz. (3) If 1 f(z) dz and 2 f(z) dz exist, then 1 + 2 f(z) dz exist and f(z) dz = f(z) dz + f(z) dz. 1 + 2 1 2 A path such that {} U is said to be a path in U. Proposition 15. (Fundamental Theorem of Calculus) Let U be an open set and let : [a, b] C be a piecewise smooth path in U. Suppose that F : U C is differentiable on U and F = f. If f(z) dz exists, then f(z) dz = F ((b)) F ((a)). Example. Let : [ π, π] C be the path (t) = e it. Consider the integrals zn dz for any n Z. Proposition 16. (M-L estimate) Let f be a complex function on {} for some piecewise smooth path. Suppose that f(z) M for all z {} and that f(z) dz exists. Then f(z) dz ML(). Corollary 17. Suppose that (f k ) k=0 is a sequence of continuous functions defined on {} for some piecewise smooth path. If (f k ) k=0 converges uniformly on {} to a function f, then f k (z) dz = f(z) dz. lim k 5. Cauchy Theory on a Disc One of the main results of this section, and indeed of the theory of complex functions, is that f(z) dz = 0 for any analytic function on a suitable set U and any piecewise smooth path in U that begins and ends at the

8 MA3111S COMPLEX ANALYSIS I same point. But we first begin with triangles. Definition. If a and b are distinct complex numbers, the segment [a, b] is the path : [0, 1] C given by (t) = tb + (1 t)a. If a, b, c are distinct points, then the triangle (a, b, c) is the set {ra+sb+tc : 0 r, s, t 1, r+s+t = 1}. The boundary of (a, b, c) is the path (a, b, c) = [a, b] + [b, c] + [c, a]. Theorem 18. (Cauchy s Theorem for triangles) Suppose that f is analytic on an open set. If (a, b, c) is a triangle contained in U, then f(z) dz = 0. (a,b,c) Sketch of proof. Suppose that (a,b,c) f(z) dz = ε > 0. Divide (a, b, c) into 4 triangles using the midpoints of the 3 sides. One of these triangles, which we label by 1 satisfies f(z) dz ε 1 4. Now divide 1 into 4 pieces, etc. In this way, we obtain a nested sequence of triangles ( n ) n=1 so that n f(z) dz ε 4 and L( n n ) = L( (a, b, c))/2 n. There exists z 0 that belongs to all n. Now, for any δ > 0, if n is sufficiently large, f(z) dz = n f(z) f(z 0 ) f (z 0 )(z z 0 ) dz n δ z z 0 dz n This is a contradiction if δ is small enough. δl( (a, b, c))2 4 n. A subset U of C is said to be convex if whenever a, b U, then the segment [a, b] lies in U. Theorem 19. (Existence of antiderivative) Suppose that f is continuous on an open convex set U. Assume that for every triangle (a, b, c) contained in U, (a,b,c) f(z) dz exists and is equal to 0. Then there exists F : U C such that F (z) = f(z) for all z U. Sketch of proof. Fix z 0 U and define F (z) = [z 0,z] f(w) dw for all z U. A path : [a, b] C is called a closed path if (a) = (b). Theorem 20. (Cauchy s Theorem) Suppose that f is analytic on an open convex set U. Then f(z) dz = 0 for any piecewise smooth closed path in U.

MA3111S COMPLEX ANALYSIS I 9 Corollary 21. (Cauchy s Theorem for an annulus) Let z 0 and w be complex numbers and R 1, R 2 be positive real numbers so that z 0 w + R 1 < R 2. Let 1, 2 : [ π, π] C be the paths 1 (t) = w+r 1 e it, 2 (t) = z 0 +R 2 e it. If f is analytic on an open set U containing {z C : z w 0 R 1, z z 0 R 2 }, then f(z) dz = 1 f(z) dz. 2 Sketch of proof. If n is large enough, then the shaded wedge is contained in an open convex set on which f is analytic. Take the sum of the integrals along the boundary paths (traversed in the counterclockwise direction) of the n wedges formed in this manner. Theorem 22. (Cauchy Integral Formula) Suppose that f is analytic on B(z 0, r) and 0 < R < r. Let : [ π, π] C be the path (t) = z 0 + Re it. Then, for any w B(z 0, R), f(w) = 1 2πi f(z) z w dz. Sketch of proof. For small s > 0, consider the path s : [ π, π] C, s (t) = w + se it. Then 1 f(z) 2πi z w dz = 1 f(z) 2πi s z w dz. But for small s, is small. 1 f(z) f(w) f (w) dz 2πi s z w Proposition 23. Suppose that f is continuous on {}, where : [ π, π] C is a piecewise smooth path. For each n N, define F n : C \ {} C by F n (w) = f(z) (z w) dz. Then F n n is analytic on C \ {} and F n(w) = nf n+1 (w) for all w C \ {}.

10 MA3111S COMPLEX ANALYSIS I Sketch of proof. Observe that f(z) (z u) n f(z) n (z w) n = (u w)f(z) 1 (z u) n k (z w) k+1. k=1 Consider F n (u) F n (w) nf n+1 (w) u w in integral form and use the M-L estimate. Corollary 24. If f is analytic on an open set U, then f has derivatives of any order on U. Theorem 25. (Cauchy Integral Formula for Derivatives) Suppose that f is analytic on B(z 0, r) and 0 < R < r. Let : [ π, π] C be the path (t) = z 0 + Re it. Then, for any w B(z 0, R) and any n N, f (n) (w) = n! 2πi f(z) dz. (z w) n+1 The next result is a kind of converse to Cauchy s Theorem. Theorem 26. (Morera s Theorem) Suppose that f is a continuous function on an open set U and (a,b,c) f(z) dz = 0 for any triangle (a, b, c) contained in U. Then f is analytic on U. Theorem 27. (Liouville s Theorem) A bounded entire function is constant. Sketch of proof. Apply the M-L estimate to the Cauchy formula for f to show that f = 0. Theorem 28. (Fundamental Theorem of Algebra) Every non-constant complex polynomial has a zero in C. Sketch of proof. If P is a complex polynomial with no zeros, then 1/P is a constant entire function. 6. Harmonic Functions Let U be an open subset of R 2. A function u : U R is said to be harmonic on U if all second order partial derivatives of u exist and are continuous on U, and (1) u xx + u yy = 0 on U. Equation (1) is called Laplace s equation. It is a fundamental equation in mathematical physics. Here we discuss some aspects of harmonic functions and their connections with analytic functions, motivated by the Cauchy-Riemann equations and Cauchy Theory on a disk. Proposition 29. If f is analytic on an open set U, then the real part Re f and the imaginary part Im f of f are harmonic on U. (We think of U both as a subset of C and of R 2.

MA3111S COMPLEX ANALYSIS I 11 Theorem 30. Suppose that u is harmonic on B(z 0, r) for some z 0 = x 0 +iy 0 (identified with the point (x 0, y 0 ) in R 2 ). There is an analytic function f on B(z 0, r) such that u = Re f. If g is any analytic function on B(z 0, r) such that u = Re g, then f g is a purely imaginary constant. Sketch of proof. Solve for the imaginary part of f from the Cauchy-Riemann equations. Theorem 31. (Mean Value Property) Suppose that u is harmonic on B(z 0, r) and 0 < R < r. Then u(z 0 ) = 1 2π 2π 0 u(z 0 + Re it ) dt. Sketch of proof. Cauchy Integral Formula. Theorem 32. (Poisson Integral Formula) Suppose that u is harmonic on B(0, 1) and continuous on B(0, 1). Then for any re is B(0, 1), u(re is ) = 1 2π 2π 0 u(e it )(1 r 2 ) 1 2r cos(t s) + r 2 dt. Sketch of proof. Find an analytic function f on B(0, 1) such that Re f = u and expand it as a power series. 7. Power Series Definition. A sequence of complex functions (f n ) n=0 defined on a set U is said to (1) converge uniformly on U to a function f : U C if for all ε > 0, there exists N N such that f n (x) f(x) ε for all n N and all x U; (2) converge compactly on U to a function f : U C if the sequence converges uniformly to f on K for every compact (= closed and bounded) subset K of U. Given a series of complex functions k=0 f k, define the partial sums to be s n = n k=0 f k. We say that the series k=0 f k converges uniformly, respectively, compactly, on a set U if and only if the sequence (s n ) n=0 converges uniformly, respectively, compactly on U. We also say that k=0 f k converges absolutely on U if k=0 f k(z) converges absolutely for all z U. Theorem 33. (Weierstrass Double Series Theorem) Suppose that (f n ) n=0 is a sequence of analytic functions on an open set U that converges compactly on U to a function f. Then f is analytic on U and (f n (k) ) n=0 converges compactly on U to f (k) for each k N. Sketch of proof. Analyticity of f follows from Morera s Theorem. Compact convergence of (f n (k) ) n=0 on U to f (k) follows from the Cauchy Integral Formula for Derivatives. A power series is an expression of the form k=0 a k(z z 0 ) k for some a k, z 0 C. z 0 is called the center of the power series and the number

12 MA3111S COMPLEX ANALYSIS I ρ = 1/ lim sup k a k 1/k is called the radius of convergence of the power series. For the next result, we take B(z 0, ρ) = C if ρ =. Proposition 34. The power series k=0 a k(z z 0 ) k converges absolutely and compactly on B(z 0, ρ) to a function f : B(z 0, ρ) C and diverges at any z with z z 0 > ρ. The function f is analytic on B(z 0, ρ) and f (z) = ka k (z z 0 ) k 1 for all z B(z 0, ρ). k=1 Corollary 35. Suppose that k=0 a k(z z 0 ) k has radius of convergence ρ > 0 and denote the sum by f(z) for all z B(z 0, ρ). Then a k = f (k) (z 0 )/k! for all k 0. Theorem 36. (Taylor Series Expansion on a Disc) If f is analytic on B(z 0, r), then the power series f (k) (z 0 ) k=0 k! (z z 0 ) k has radius of convergence at least r and converges to f(z) on B(z 0, r). Theorem 37. Suppose that f is analytic on an open set U and f(z 0 ) = 0 for some z 0 U. Either (1) There exists r > 0 so that f(z) = 0 for all z B(z 0, r); or (2) There exist a unique m N and a unique analytic function g on U such that g(z 0 ) 0 and f(z) = (z z 0 ) m g(z) for all z U. Definition. If case (2) of the last theorem occurs, we say that f has an isolated zero at z 0 with order m. If f is analytic at z 0 and f(z 0 ) 0, we say that f has a zero of order 0 at z 0. Sketch of proof. Look at the Taylor series expansion of f on a small disc centered at z 0. Definition. An open set U in C is said to be connected if given any pair of points z and w in U, there is a path in U : [a, b] U such that (a) = z and (b) = w. Remark. Properly speaking, the property just defined is called path connectedness. In general topology, there is a concept of connectedness that applies to topological spaces. For open subsets of C, it can be shown that what we call connected is equivalent to the usual definition. Theorem 38. (Weierstrass Identity Theorem) If f is analytic on an open connected set U and the set Z = {z U : f(z) = 0} has an accumulation point in U, then f(z) = 0 for all z U. Sketch of proof. Let W = {z U : z is an accumulation point of Z}. Then both W and U \ W are open sets. For any w Z, let be a path in U that begins at z 0 and ends at w. Consider the function h so that h(t) = 1 if (t) W and h(t) = 0 if (t) U \ W. Apply the Intermediate Value Theorem to conclude that h is constant.

MA3111S COMPLEX ANALYSIS I 13 8. Singularities If 0 r 1 < r 2, the annulus with center z 0, inner radius r 1 and outer radius r 2 is the set Ann(z 0, r 1, r 2 ) = {z C : r 1 < z z 0 < r 2 }. If 0 = r 1 < r <, the annulus Ann(z 0, r 1, r) is also called a punctured disc B (z 0, r). An annulus and a punctured disc Definition. If a complex function f is defined and analytic on a punctured disc B (z 0, r) but not defined or not analytic at z 0, then we say that f has an isolated singularity at z 0. Theorem 39. (Riemann Removable Singularity Theorem) Suppose that f has an isolated singularity at z 0. If there exists R > 0 so that f is bounded on B (z 0, R), then there is an analytic function g on B(z 0, R) such that g(z) = f(z) for all z B (z 0, R). In the above situation, if we extend the function f by defining f(z 0 ) to be g(z 0 ), then the extended function is analytic at z 0. For this reason, we call the isolated singularity z 0 of f a removable singularity. Sketch of proof. Consider the function g(z) = (z z 0 )f(z) defined on a punctured disc centered at z 0 and set g(z 0 ) = 0. Use Morera s Theorem to show that g is analytic at z 0. Consider the Taylor series expansion of g at z 0 and divide it by z z 0. Definition. An isolated singularity z 0 of f is called a pole if lim z z0 f(z) =, i.e., for any real number M <, there exists r > 0 such that f(z) M for all z B (z 0, r). Theorem 40. Suppose that f has a pole at z 0. There is a unique m N so that there is an analytic function g on B(z 0, r) for some r > 0 with g(z 0 ) 0 and f(z) = g(z)/(z z 0 ) m for all z B (z 0, r). Definition. The number m in the last theorem is called the order of the pole z 0 of f.

14 MA3111S COMPLEX ANALYSIS I Sketch of proof. The function 1/f has a removable singularity at z 0. Then 1/f(z) = (z z 0 ) m h(z), where m is the order of the zero of 1/f at z 0. An isolated singularity of f that is neither removable nor a pole is called an essential singularity. Theorem 41. (Casorati-Weierstrass Theorem) Suppose that f has an essential singularity at z 0. For any r > 0, any w C and any ε > 0, there exists z B (z 0, r) such that f(z) w < ε. Sketch of proof. Suppose there exists r > 0, w C and ε > 0 so that f is defined on B (z 0, r) and f(z) w ε for all z B (z 0, r). Then h(z) = 1/(f(z) w) has a removable singularity at z 0. If h has a zero of order 0 at z 0, then f has a removable singularity at z 0. If h has a zero of order m > 0 at z 0, then f has a pole at z 0. Lemma 42. Suppose that 0 r 1 < R 1 < z z 0 < R 2 < r 2 and that f is analytic on the annulus Ann(z 0, r 1, r 2 ). Let j : [ π, π] C be the path j (t) = z 0 + R j e it, j = 1, 2. Then f(z) = 1 2πi for all z Ann(z 0, R 1, R 2 ). 2 f(w) w z dw 1 2πi 1 f(w) w z dw Sketch of proof. Define g : Ann(z 0, r 1, r 2 ) C by g(w) = { f(w) f(z) w z if w z f (z) if w = z. g is analytic on Ann(z 0, r 1, r 2 ) by Morera s Theorem. Apply Cauchy s Theorem for an annulus to g to obtain 1 f(w) 2πi 2 w z dw 1 f(w) 2πi 1 w z dw 1 1 = f(z)[ 2πi 2 w z dw 1 1 2πi 1 w z dw].

MA3111S COMPLEX ANALYSIS I 15 By Cauchy s Theorem for an annulus, the first integral on the right has value 1. By Cauchy s Theorem, the second integral on the right is 0. Theorem 43. (Laurent Series Expansion) Suppose that 0 r 1 < R < r 2 and that f is analytic on the annulus Ann(z 0, r 1, r 2 ). Let : [ π, π] C be the path (t) = z 0 + Re it. For each z Ann(z 0, r 1, r 2 ), (2) f(z) = a k (z z 0 ) k b k + (z z 0 ) k, where k=0 k=1 a k = 1 f(w) 2πi (w z 0 ) k+1 dw b k = 1 f(w)(w z 0 ) k 1 dw. 2πi Moreover, the first series in (2) converges absolutely and compactly on B(z 0, r 2 ) and the second series converges absolutely and compactly on Ann(z 0, r 1, ). Sketch of proof. Let the notation be as in the last theorem, where we take R 1 < z z 0 < R 2. For the integral f(w) 2 w z dw, expand 1 w z = 1 (w z 0 ) (z z 0 ) = 1 ( z z 0 ) k. w z 0 w z 0 For the integral f(w) 1 w z dw, use 1 w z = 1 (w z 0 ) (z z 0 ) = 1 z z 0 k=0 (w z 0 ) k. z z 0 The integration and the summation may be switched because of the uniform convergence of the series on the paths of integration. Proposition 44. (Uniqueness of Laurent Series Expansion) Let the notation be the same as in the last theorem. If there are complex numbers A k and B k so that f(z) = A k (z z 0 ) k + k=0 k=1 k=0 B k (z z 0 ) k for all z Ann(z 0, r 1, r 2 ), then A k = a k, k 0, and B k = b k, k 1. Sketch of proof. The series must converge compactly on Ann(z 0, r 1, r 2 ). The 1 f(w) integrals 2πi dw and 1 (w z 0 ) k+1 2πi f(w)(w z 0) k 1 dw may be computed by term by term integration. Definition. If f is analytic on the punctured disc B (z 0, r) and has Laurent series expansion f(z) = a k (z z 0 ) k b k + (z z 0 ) k, k=0 k=1

16 MA3111S COMPLEX ANALYSIS I then b 1 is called the residue of f at z 0, denoted by Res(f, z 0 ). k=1 is called the principal part of f at z 0. b k (z z 0 ) k Theorem 45. Suppose that f is analytic on the punctured disc B (z 0, r) and has Laurent series expansion given by (2). Then the isolated singularity z 0 is (1) removable if and only if b k = 0 for all k 1; (2) a pole of order m if and only if b m 0 and b k = 0 for all k > m; (3) an essential singularity if b k 0 for infinitely many k. 9. The Residue Theorem and Applications Definition. Let be a piecewise smooth closed path and let w be a point not belonging to {}. The winding number of around w is defined to be n(, w) = 1 1 2πi z w dz. Proposition 46. The winding number n(, ) : C\{} C is a continuous integer-valued function. Sketch of proof. n(, w) is a continuous (even differentiable) function of w by Proposition 23. Define h(t) = t a (s) (s) w ds for t [a, b], the domain of. Then e h(t) ((t) w) 1. Then n(, w) = h(b) 2πi Z. Remark. Note that the proof shows that h(t) is a continuous branch of log((t) w) and thus Im h(t) is a continuous branch of arg((t) w). Also, n(, w) = 1 2πi (h(b) h(a)). This suggests the geometric meaning of the winding number. Theorem 47. (Cauchy Integral Formula, General Version) Let U be an open set and let be a piecewise smooth closed path in U such that n(, w) = 0 for all w / U. If w U \ {} and f is analytic on U, then f(w)n(, w) = 1 2πi f(z) z w dz. Sketch of proof. (John Dixon) Consider the function ϕ : U U C defined by ϕ(ζ, z) = { f(ζ) f(z) ζ z if ζ z f (z) if ζ = z.

MA3111S COMPLEX ANALYSIS I 17 Let V = {z C : n(, z) = 0}. Then V is an open set in C such that U V = C. The function { ϕ(ζ, z) dζ if z U g(z) = if z V f(ζ) ζ z dζ is well defined. g is analytic on V by Proposition 23. It is analytic on U by Morera s Theorem. So g is entire. When z is large, z V. So lim z g(z) = 0. In particular, g is bounded. Thus g is identically 0. The theorem follows from the fact that g(w) = 0. Corollary 48. (Cauchy s Theorem, General Version) Let U be an open set and let be a piecewise smooth closed path in U such that n(, w) = 0 for all w / U. Then f(z) dz = 0 for all functions f analytic on U. Corollary 49. (Cauchy Integral Formula for Derivatives, General Version) Let U be an open set and let be a piecewise smooth closed path in U such that n(, w) = 0 for all w / U. If w / {} and k N, then f (k) (w)n(, w) = k! 2πi f(z) dz (z w) k+1 Corollary 50. (Residue Theorem) Let U be an open set and suppose that f is analytic on U \ P, where P is a finite subset of U. Assume that is a piecewise smooth closed path in U \ P so that n(, w) = 0 for all w / U. Then 1 2πi f(z) dz = w P Res(f, w)n(, w). Sketch of proof. For each p P, let g p be the principal part of f at p. Apply Cauchy s Theorem to the function f p P g p. Applications of the Residue Theorem in the evaluation of integrals Let P and Q be polynomials with real coefficients. 1. The integral P (x)/q(x) dx. Evaluate this by considering r P (z)/q(z) dz, where r = [ r, r] + C r, C r : [0, π] C, C r (t) = re it. The path r 2. The integrals P (x) Q(x) cos x dx and P (x) Q(x) sin x dx. Consider r P (z) Q(z) eiz dz, with the same path r as in part 1.

18 MA3111S COMPLEX ANALYSIS I 3. Integrals of the forms 0 P (x) Q(x) dx a P (x) Q(x) dx 0 x α 1 P (x) dx. The Pacman paths For the first and third integrals, consider P (z) Log z dz Q(z) and z α 1 P (z) dz for the path as shown on the left. For the second integral, consider P (z) Log(z a) dz Q(z) for the path shown on the right, with the center of the inner arc at a. 4. Integrals of the form 2π 0 R(cos θ, sin θ) dθ, where R is a rational function (= quotient of polynomials) in two variables. Transform the integral into R( z2 + 1 2z where is the path (t) = e it, 0 t 2π., z2 1 i2z ) dz iz, Applications of the Residue Theorem in the evaluation of series 1. Sums of the form n= f(n) is related to the integral f(z) cot πz dz, where is the square with side 2N + 1 centered at 0, i.e., taking a = N + 1 2, = [a(1 i), a(1 + i)] + [a(1 + i), a( 1 + i)] + [a( 1 + i), a( 1 i)] + [a( 1 i), a(1 i)]. 2. Sums involving binomial coefficients may sometimes be evaluated by making use of the observation ( ) n = Res ( (1 + z) n k z k+1, 0 ).

MA3111S COMPLEX ANALYSIS I 19 10. Analytic Functions as Mappings Proposition 51. Suppose that f is analytic on an open set U and that is a piecewise smooth closed path in U. For any w / f({}), 1 f (z) dz = n(f, w). 2πi f(z) w Theorem 52. (Local Mapping Theorem) Suppose that f an analytic function on an open set U and z 0 U. Assume that f f(z 0 ) has an isolated zero of order m at z 0. Then f is locally m-to-1 at z 0 in the following sense: There exist r, R > 0 so that every w B (f(z 0 ), R) is the image of exactly m distinct points in B(z 0, r). Sketch of proof. Choose r > 0, B(z 0, r) U, so that f(z) f(z 0 ) and f (z) 0 for all z B(z 0, r) \ {z 0 }. Let (t) = z 0 + re it, π t π. Choose R > 0 so that B(f(z 0 ), R) {f } =. Let w B (f(z 0 ), R). Then n(f, w) = n(f, z 0 ) = m, where the last equality follows from Proposition 51. Consider W = {z B(z 0, r) : f(z) = w}. By the Weierstrass Identity Theorem, W is a finite set. Note also that z 0 / W. Say W = {z 1,..., z n }. For each j, f w has a simple zero (zero of order one) at z j since f (z j ) 0. Now m = n(f, w) = 1 2πi f (z) f(z) w dz = n f Res( f w, z j)n(, z j ) by the Residue Theorem. Since each z j is a simple zero of f w, each of the residues in the sum is 1. Also n(, z j ) = 1 for each j. Thus n = m. j=1 An open connected subset of C is called a domain. Corollary 53. (Open Mapping Theorem) A nonconstant analytic function on a domain U maps open subsets of U onto open sets. Corollary 54. Suppose that f is a one-to-one analytic function on an open set U. Then f (z) 0 for all z U. Moreover, f 1 is analytic on the open set f(u) and (f 1 ) 1 (w) = f (f 1 for all w f(u). (w)) Proposition 55. Suppose that (f k ) k=1 is a sequence of one-to-one analytic functions defined on a domain U and that (f k ) k=1 converges compactly on U to a function f. Then either f is constant on U, or f is a one-to-one analytic function on U. Sketch of proof. Suppose that z 1 z 2 and f(z 1 ) = w = f(z 2 ). We may assume that z 1 and z 2 are isolated zeros of f w. Then Res( f f w, z 1) and Res( f f w, z 2) are nonzero. Let j be a small circle around z j, j = 1, 2. By

20 MA3111S COMPLEX ANALYSIS I Proposition 51 and the Residue Theorem n(f 1, w), n(f 2, w) 0. Using the M-L estimate, we find that n(f k j, w) 0 for large k and j = 1, 2. Now 0 n(f k j, w) = 1 f k (z) 2πi f k (z) w dz implies that f k (z) w must have a zero inside j. This shows that f k is not one-to-one. Proposition 56. For any α B(0, 1), define m α on B(0, 1) by m α (z) = z α 1 αz. Then m α is one-to-one and analytic on B(0, 1) and maps B(0, 1) onto itself. Sketch of proof. m α (z) = 1 if z = 1. Apply the Cauchy Integral Formula to m k α and estimate the integral by the M-L estimate. Deduce that m α (z) 1 if z 1. Now m α m α = m α m α is the identity map on B(0, 1). Thus m α maps B(0, 1) onto B(0, 1) and B(0, 1) onto B(0, 1). 11. Riemann Mapping Theorem A domain U in C is said to be simply connected if n(, w) = 0 for every piecewise smooth closed path in U and every w / U. Two domains U and V are said to be conformally equivalent if there is a function f analytic on U that maps U bijectively onto V. It turns out that there are only two types of simply connected domains up to conformal equivalence; one type is the entire plane C and the other type is conformally equivalent to the unit disc B(0, 1). This result is the famous Riemann Mapping Theorem. Theorem 57. (Existence of Antiderivative) Suppose that f is analytic on a simply connected domain U. Then there exists an analytic function F on U such that F = f. Sketch of proof. Fix z 0 in U and define F (z) = 1 2πi z f(w) dw for some piecewise smooth path z in U that begins at z 0 and ends at z. Take note of the general version of Cauchy s Theorem to show that F = f. Theorem 58. (Existence of an Analytic Square Root) Suppose that f is analytic on a simply connected domain U and is never 0 there. Then there exist an analytic function g on U such that g 2 = f. Sketch of proof. There exists analytic h such that h = f /f. Then e h f is a constant, say C. Take any square root C of C and set g = C e h/2. Definition. Let F be a nonempty family of complex continuous functions defined on an open set U in C. We say that F is (1) compactly bounded if for every closed (in C) and bounded subset K of U, sup sup f(z) < ; z K f F j

MA3111S COMPLEX ANALYSIS I 21 (2) equicontinuous at z 0 U if for all ε > 0, there exists δ > 0 so that f(z) f(z 0 ) < ε for all f F and all z B(z 0, δ); (3) equicontinuous on a subset K of U if for all ε > 0, there exists δ > 0 so that f(z 1 ) f(z 2 ) < ε for all f F and all z 1, z 2 K with z 1 z 2 < δ; (4) compactly equicontinuous if F is equicontinuous on every closed and bounded subset of U. Theorem 59. (Ascoli-Arzela s Theorem) Let F be a compactly bounded, compactly equicontinuous family of continuous complex functions on an open set U in C. Then every sequence in F has a subsequence that converge compactly on U. Sketch of proof. Let (f n ) n=1 be a sequence in F and let K be a closed and bounded subset of U. Given ε > 0, let δ > 0 be chosen according to the definition of equicontinuity. Cover K with finitely many balls B(z j, δ), 1 j m. Choose an infinite subset N of N so that f n1 (z j ) f n2 (z j ) ε converges for each j and all n 1, n 2 N 1. If n 1, n 2 are numbers in N and z K, say z B(z j, δ), then f n1 (z) f n2 (z) f n1 (z) f n1 (z j ) + f n1 (z j ) f n2 (z j ) + f n2 (z j ) f n2 (z) 3ε. Using this argument, we can find infinite sets N N 1 N 2 so that f k (z) f j (z) < 1/m if z K and k, j N m. Pick a subsequence (f nm ) m=1 so that n m N m for all m. Then (f nm ) m=1 converges uniformly on K. Finally, there exists a sequence of closed bounded subsets (K m ) m=1 such that U = m=1 K m. From the above, choose infinite sets N M 1 M 2 so that (f n ) n Nm converges uniformly on M m for each m. Pick a subsequence (f nm ) m=1 so that n m M m for all m. Then (f nm ) m=1 converges uniformly on K k for each k. Since each closed bounded subset of U is contained in some K m, (f nm ) m=1 converges compactly on U. Theorem 60. (Montel s Theorem) Let F be a nonempty family of analytic functions on an open U in C. Then every sequence in F has a compactly convergent subsequence if and only if F is compactly bounded. Remark. A family F in which every sequence has a compactly convergent subsequence is called a normal family in the literature. We will refrain from using this term since the word normal is severely overused. Sketch of proof. If F is a compactly bounded family of analytic functions, then it is also equicontinuous at any z 0. The reason is embodied in the following estimate. Suppose B(z 0, r) U and let be the circle of radius r centered at z 0. Choose M so that f(z) M for all f M and all z {}.

22 MA3111S COMPLEX ANALYSIS I For all z with z z 0 ε r/2, 1 f(z) f(z 0 ) = 2πi 4Mε/r 2. f(ζ) ( 1 ζ z 1 ζ z 0 ) dζ Using the usual covering argument, we further deduce that F is compactly equicontinuous. Theorem 61. (Riemann Mapping Theorem) If U is a simply connected domain that is not C, then there is a one-to-one analytic function on U that maps U onto B(0, 1). The proof of the Riemann Mapping Theorem consists of verifying the following steps. Let F be the family of all one-to-one analytic functions on U such that f(z) < 1 for all z U. Then Step 1 F is nonempty. Step 2 Pick a point z 0 U. There exists f F such that f (z 0 ) g (z 0 ) for all g F. Step 3 The function f maps U ono-one onto B(0, 1) (and z 0 to 0).