Complex Analysis PDF - MA203(CA)

Complex Analysis Contents 1 The Complex Numbers 3 1.1 Some Topological Definitions.............................

Complex Analysis Contents 1 The Complex Numbers 3 1.1 Some Topological Definitions.......................... 6 2 Complex Functions 9 3 Analytic Functions 10 3.1 Differentiability.................................. 10 3.2 Analyticity.................................... 14 3.3 Applications of C-R Equations......................... 15 3.4 Harmonic Functions............................... 16 4 Elementary Functions 18 4.1 Exponential Function.............................. 18 4.2 Logarithmic Function.............................. 20 4.3 Trigonometric and Hyperbolic Functions.................... 21 5 Complex Integration 24 5.1 Integral of a Complex Valued Function of a Real Variable.......... 24 5.2 Contour Integral................................. 25 5.3 Cauchy Integral Theorem............................ 32 5.4 Cauchy’s Integral Formula............................ 37 5.5 Converse of Cauchy-Goursat’s Theorem: Morera’s Theorem......... 43 5.6 Consequences of the Cauchy Integral Formula................. 44 6 Sequence and Series 45 6.1 Sequence...................................... 45 6.2 Series....................................... 46 7 Power Series 47 8 Taylor and Laurent Series 50 8.1 Taylor Series................................... 50 8.2 Laurent Series................................... 52 9 Zeros of Analytic Functions 55 9.1 Maximum Modulus Principle.......................... 60 CONTENTS M. A. Khan 10 Classification of Singular points: Poles 62 11 Residues and Residue Theorem 68 12 Evaluation of Real Integrals by Residue Methods Z 2π 73 12.1 Integral f (cos θ, sin θ) dθ.......................... 73 Z0∞ 12.2 Integral f (x) dx............................... 74 Z−∞ ∞ Z ∞ 12.3 Integral f (x) cos sx dx, f (x) sin sx dx................ 75 −∞ −∞ 2 The Complex Numbers M. A. Khan 1 The Complex Numbers Definition 1.1 (Complex Number). A complex number denoted by z is an ordered pair (x, y) with x ∈ R, y ∈ R. For z = (x, y), x and y are respectively called the real and imaginary part of z. In symbol we write Rez = x, and Imz = y. The set of all complex numbers is denoted by C. We write x for the complex number (x, 0) and i for (0, 1). In fact, the mapping x 7→ (x, 0) defines a field isomorphism of R onto a subset of C, and hence we may consider R as a subset of C. Addition and multiplication of complex numbers are defined as follows: (a, b) + (c, d) := (a + c, b + d) (a, b)(c, d) := (ac − bd, bc + ad). The following are easy to check directly from definitions: 1. z1 + z2 = z2 + z1. 2. z1 z2 = z2 z1. 3. z1 (z2 + z3 ) = z1 z2 + z1 z3. Note that z = (x, y) can be written as z = x + iy by identifying (x, 0) and (y, 0) with x and y respectively. Moreover, i2 = −1. Example 1.1. Find a root of the equation z 2 + 1 = 0 in C. We can define division of complex numbers also. If z 6= 0, then we define 1 1 x − iy = = 2. z x + iy x + y2 From this we get x2 − iy2 (x1 x2 + y1 y2 ) + i(x2 y1 − x1 y2 )   x1 + iy1 = (x1 + iy1 ) =. x2 + iy2 x22 + y22 x22 + y22 Equal Complex Numbers: Two complex numbers (x1 , y1 ) and (x2 , y2 ) are said to be equal if both their real parts and imaginary parts are equal, that is, (x1 , y1 ) = (x2 , y2 ) if and only if x1 = x2 and y1 = y2. Complex Plane: A complex number z = x + iy is defined by the pair of real numbers x and y, so it is natural to assume a one-to-one correspondence between the complex number z = x + iy and the point (x, y) in the xy-plane. We refer to that plane as the complex plane or the z-plane. In the complex plane, any point that lies on the x-axis represents a real number. Therefore, the x-axis is termed the real axis. Similarly, any point on the y-axis represents an imaginary number, so the y-axis is called the imaginary axis. Complex Conjugate: The complex conjugate z of a given complex number z = x + iy is defined by z = x − iy. Conjugation has the following properties which follows easily from the definition: 3 The Complex Numbers M. A. Khan 1. Rez = 12 (z + z) and Imz = 1 2i (z − z). 2. z1 + z2 = z1 + z2. 3. z1 z2 = z1 z2 , and hence for a real number α, αz = αz. Polar Form of Complex Numbers: Let z = x + iy be a non-zero complex number. Then there exist unique r ∈ (0, ∞), and θ ∈ (−π, π] such that z = r(cos θ + i sin θ). r and θ are related to z by the relations p y r = x2 + y 2 and θ = tan−1. x Thus r is the distance of z = (x, y) from the origin and θ is the angle between the positive x-axis and the line segment joining the origin and the point (x, y) (see Figure 1). r is called the modulus of z, and is denoted by |z|. θ is called the principal argument of z and is usually written as θ = Arg z. Note that if φ = Arg z + 2kπ, k being an integer, then z = |z|(cos φ + i sin φ). The representation |z|(cos φ + i sin φ) of z is called the polar representation of z, and φ is called an argument of z. Note that if θ is an argument of z then so is θ + 2kπ, k being an integer, and Arg z is an argument of z lying in the interval (−π, π]. y z = (x, y) |z| arg z x Figure 1: Remark 1.1. As a note of caution, tan−1 xy returns a value in
the interval (− π2 , π2 ].
Therefore, we adjust the value for Arg z by adding π to tan−1 xy if (x, y) lies in the second quadrant or subtracting π from tan−1 xy if (x, y) lies in the third quadrant.
3π −π Example 1.2. Arg 1 = 0, Arg (−1) = π, Arg (−1 + i) = 4 , Arg (1 − i) = 4. Problem 1.1. Let zk = rk (cos θk + i sin θk ), k = 1, 2,... , n. Show that 1. z1 z2 = r1 r2 [cos(θ1 + θ2 ) + i sin(θ1 + θ2 )]. 2. z1 z2 · · · zn = r1 r2 · · · rn [cos(θ1 + θ2 + · · · + θn ) + i sin(θ1 + θ2 + · · · + θn )]. 3. De Moivers formula: z n = [r(cos θ + i sin θ)]n = rn (cos nθ + i sin nθ), where n is an integer. 4 The Complex Numbers M. A. Khan Remark 1.2. De Moivers formula fails when n is not an integer. For instance, consider 1 r = 1, θ = 2π and n = 12. Then De Moivre’s formula gives (1) 2 = −1, which is not true. Problem 1.2. Use polar representation to explain why product of two negative real numbers is a positive real number. Problem 1.3 (nth Root). Given a nonzero complex number z0 and a natural number n ∈ N, find all distinct complex numbers w such that z0 = wn. Solution. Let z0 = r(cos θ + i sin θ). Let w = ρ(cos α + i sin α), and we determine ρ and α in terms of r and θ. By De Moivers formula, we have wn = ρn (cos nα + i sin α). Since 1 |z0 | = |wn |, we obtain ρ = r n. Moreover, as r(cos θ + i sin θ) = r(cos nα + i sin nα), we obtain nα = θ + 2kπ θ + 2kπ ⇒ α= , k = 0, ±1, ±2, ±3,.... n Now we notice that the distinct values of w is given by   1 θ + 2kπ θ + 2kπ |z0 | n cos + i sin , k = 0, 1, 2,... , n − 1. n n That is, other values of k will give complex numbers which is already obtained. 1 Note that the n many complex numbers we got lie on the circle of radius |z0 | n about the origin and constitute the vertices of a regular polygon of n sides. Problem 1.4. Find the curve or region in the complex plane represented by each of the following equations or inequalities: 1. Rez = Imz 2. |z| = 2. 3. |z − z0 | = 2. 4. |z − z0 | < 2. 5. |z − z0 | ≤ 2. 6. 1 < |z| < 3. Solution. (1) Rez = Imz ⇔ x = y, where z = x + iy. The desired subset p consists of the points of the straight line y = x (Figure 2). (2) |z| = 2 ⇔ x2 + y 2 = 2. The desired subset consists of the points on the circle with center at (0, 0) and radius r = 2. (3) Let z0 = (xp 0 , y0 ). |z − z0 | = 2 ⇔ (x − x0 )2 + (y − y0 )2 = 2. The desired subset consists of the points on the circle with center at z0 = (x0 , y0 ) and radius r = 2. 5 The Complex Numbers M. A. Khan Figure 2: Figure 3: (4) The desired subset consists of the points inside the circle with center at z0 = (x0 , y0 ) and radius r = 2. (5) The desired subset consists of the points inside and on the circle with center at z0 = (x0 , y0 ) and radius r = 2. (6) The desired subset is the annulus between the circles |z| = 1 and |z| = 3 without these circles (Figure 3). 1.1 Some Topological Definitions Neighborhood: As we have seen above, the set of points z such that |z − z0 | < , where z0 ∈ C,  ∈ R contains points that are inside the circle centered at z0 and with radius . We call it a neighborhood of z0 and denote it by N (z0 ; ). A deleted neighborhood of z0 is the set N (z0 ; ) \ {z0 }. We write it as N b (z0 ; ). Limit Point: Let S be a subset of C. A point z0 in C is called an accumulation point or a limit point of S if every neighborhood of z0 contains a point of S other than z0. That b (z0 ; ) ∩ S 6= ∅ for any  > 0. is, N Example 1.3. Consider the set {z : |z| < 2}. The limit points are points on and inside the circle |z| = 2. All the points of the set S = {x + iy : x = y} are limit points of S (see Figure 2). Zero is the only limit point of the set { ni : n ∈ Z \ {0}}. Be aware that a limit point z0 of S may or may not belong to the set S. Problem 1.5. Let z0 be a limit point of S. Then prove that every neighbourhood of z0 contains infinitely many points of S. Solution. Left as an exercise. Interior Point: Let S be a subset of C. A point z0 in S is called an interior point of S if there exists a neighborhood of z0 , all points of which belong to S. Example 1.4. The interior points of the set {z : |z| < 2} and {z : |z| ≤ 2} are points inside the circle |z| = 2. The set {x + iy : x = y} (cf. Figure 2) does not have any interior point. Boundary Point and Boundary: Let S be a subset of C. If every neighborhood of z0 ∈ C contains points of S and also points not belonging to S, then z0 is called a boundary point. The set of all boundary points of the set S is called the boundary of S. 6 The Complex Numbers M. A. Khan Example 1.5. The boundary of the set {z : |z| < 2} and {z : |z| ≤ 2} are given by the circle |z| = 2. The boundary of the set S = {x + iy : x = y} is S itself. Open Set: Let S ⊆ C. S is said to be an open set if each point of S is an interior point of S. Closed Set: A subset S ⊆ C is said to be closed if it contains all its boundary points. Example 1.6. The sets {z : |z| < 2} and {z : Rez > 0} are open set while the sets {z : |z| ≤ 2} and {x + iy : x = y} are closed sets. The set {z : |z| < 2} ∪ {2} is neither open nor closed. C is both open and closed. Proposition 1.1. A subset S ⊆ C is open if and only if its complement S c = {z : z ∈ / S} is closed. Closure: Let S ⊆ C. The closure of S, denoted as S, is the closed set that contains all points in S together with the whole boundary of S. Example 1.7. The closure of a closed set S is S itself. The closed set {z : |z| ≤ 2} is the closure of the set {z : |z| < 2}. Bounded and Unbounded Set: A bounded set is one that can be contained in a large enough circle centered at the origin. That is, there exists a sufficiently large real constant M such that S ⊆ {z : |z| < M }. An unbounded set is one that is not bounded. Example 1.8. The set { n1 + i n1 : n ∈ Z \ {0}} is bounded as it is contained in the circle √ |z| = 2. The set {x + iy : x = y} is not bounded. Definition 1.2. Connected Set: A set S is said to be connected if any two points of S can be joined by a continuous curve lying entirely inside S. For example, a neighborhood N (z0 ; ) is connected. Domain: An open connected set is called an domain. For example, the set {z : Rez ≥ 0} is not a domain since it is not open. The set {z : 0 < Re < 1 or 2 < Re < 3} is also not a domain since it is open but not connected. Region: The set obtained from an open set by joining some, none, or all of its boundary points, is called a region. Definition 1.3. Curve: A curve C in the complex plane C is given by a function γ : [a, b] → C, γ(t) = x(t) + iy(t) with x, y : [a, b] → R being continuous. The curve C is then the set C = {γ(t) : t ∈ [a, b]}. The function γ : [0, 2π] → C given by γ(t) = z0 + r(cos t + i sin t), where z0 is a fixed complex number, gives the circle with center at z0 and radius r. The function γ : [0, 1] → C given by γ(t) = tz1 + (1 − t)z0 gives the line segment joining z0 and z1. 7 The Complex Numbers M. A. Khan Smooth Curve: A curve C is called a smooth curve if γ 0 (t) = x0 (t) + iy 0 (t) is continuous and nonzero for all t. Geometrically this means that C has everywhere a continuously turning tangent. Closed Curve: A curve C is called closed if its terminal point coincides with its initial point, that is γ(a) = γ(b). Contour: A contour is a curve that is obtained by joining finitely many smooth curves end to end. Simple Contour: A contour is called simple if it does not cross itself (if initial point and the final point are same they are not considered as non simple). For instance, a circle is simple, but the curve of the shape ‘8’ is not simple (cf. Figure 4). Curve with Reverse Orientation: Let C : γ : [a, b] → C be a curve. Then the curve with the reverse orientation, denoted as −C : γ− : [a, b] → C and is defined as γ− (t) = γ(b + a − t). Thus for a contour C, the contour with the negative orientation −C make sense. Figure 4: Notation 1.1. Let C1 , C2 ,... , Cn are contours such that the terminal point of Ck coin- cides with the initial point of Ck+1. Then C = C1 + C2 + · · · + Cn will be used to denote the contour obtained by joining the contours C1 , C2 ,... , Cn end to end. Definition 1.4. Simply connected domain: A domain D is called simply connected if every simple closed contour within it encloses points of D only. A domain D is called multiply connected if it is not simply connected. Figure 5: 8 Complex Functions M. A. Khan According to the definition, it is then always possible to construct some simple closed contour inside a multiply connected domain in such a manner that one or more points inside the contour do not belong to the domain. Intuitively, there are holes contained inside some simple closed contour lying completely in the domain. We have one hole in a doubly connected domain and two holes in a triply connected domain (see Figure 5). For example, the domain {z : 1 < |z| < 2} is doubly connected. 2 Complex Functions Let S be a set of complex numbers in the complex plane. For every point z = x + iy ∈ S, we specify the rule for assigning a corresponding complex number w = u + iv. This defines a function of the complex variable z, and the function is denoted by w = f (z). The set S is called the domain of definition of the function f and the collection of all values of w is called the range of f. A complex function of the complex variable z may be visualized as a pair of real functions of the two real variables x and y, where z = x + iy. Let u(x, y) and v(x, y) be the real and imaginary parts of f (z), respectively. We may write f (z) = u(x, y) + iv(x, y), z = x + iy. For example, consider the function f (z) = z 2 = (x + iy)2 = x2 − y 2 + 2ixy; its real and imaginary parts are the real functions u(x, y) = x2 − y 2 and v(x, y) = 2xy, respectively. Definition 2.1 (Limit). Let z0 ∈ C be a limit point of D. A function f : D → C is said to have the limit l as z approaches z0 , written lim f (z) = l, if for a given  > 0, there z→z0 exists a δ > 0 such that |f (z) − l| <  whenever z ∈ D and 0 < |z − z0 | < δ. The limit l, if exists, must be unique. The value of l is independent of the direction along z which z → z0. For example, consider the limit lim. z→0 z z Along the positive direction of x-axis, we obtain lim = 1. z→0 z z Along the positive direction of y-axis, we obtain lim = −1. z→0 z z Thus, it follows that lim does not exist. z→0 z 9 Analytic Functions M. A. Khan Figure 6: Limit Definition 2.2 (Continuity). A function f : D → C is said to be continuous at z0 ∈ D if lim f (z) = f (z0 ). z→z0 A complex function is said to be continuous in a region R if it is continuous at every point in R. It is not difficult to prove the following. Proposition 2.1. Let f = u + iv, that is u(x, y) and v(x, y) be the real and imaginary parts of the function f (z). Then f is continuous at z0 = x0 + iy0 if and only if u(x, y) and v(x, y) are both continuous at (x0 , y0 ). Example 2.1. Consider the function f (x+iy) = ex cos y+iex sin y. Then f = u+iv, where u(x, y) = ex cos y, and v(x, y) = ex sin y. Since both u(x, y) and v(x, y) are continuous at any point (x0 , y0 ) in the xy-plane, we conclude that f (z) is continuous at any point z0 = x0 + iy0 in C. Consider the function f (z) = |z|2. Then f = u+iv, where u(x, y) = x2 +y 2 and v(x, y) = 0. Here u and v are continuous at every point in the xy-plane, and hence f (z) is continuous everywhere on C. Theorems on real continuous functions can be extended to complex continuous func- tions. For instance, if two complex functions are continuous at a point, then their sum, difference and product are also continuous at that point; and their quotient is continuous at any point where the denominator is non-zero. 3 Analytic Functions 3.1 Differentiability Definition 3.1. Let f : D → C be a function, and z0 ∈ D be a limit point of D. Suppose that the limit f (z) − f (z0 ) lim , z→z0 z − z0 10 Analytic Functions M. A. Khan or equivalently f (z0 + h) − f (z0 ) lim , h→0 h exists. Then we say that f is differentiable at z0 and the limit denoted by f 0 (z0 ) is called the derivative of f at z0. The following results about derivatives follow exactly as in the case of reals. Proposition 3.1. d n 1. Derivative of a constant function is zero and z = nz n−1 , n ∈ Z. dz 2. If α, β ∈ C, then (αf + βg)0 = αf 0 + βg 0. d 3. (Chain Rule) f (g(z)) = f 0 (g(z))g 0 (z) whenever all the terms make sense. dz Like real calculus, we have the following. Proposition 3.2. If f is differentiable at z0 then f is continuous at z0. f (z) − f (z0 ) Proof. Since f (z) is differentiable at z0 , the limit lim exists. Therefore, z→z0 z − z0 f (z) − f (z0 ) lim f (z) = lim (z − z0 ) + f (z0 ) = f (z0 ). z→z0 z→z0 z − z0 Remark 3.1. Continuity of f (z) at a point z0 may not imply the differentiability of f (z) at z0. For example, consider the function f (z) = |z|. p f (z) is continuous at z = 0: Note that f = u+iv, where u = x2 + y 2 , and v(x, y) = 0. As u(x, y) and v(x, y) are continuous at (0, 0), it follows that f (z) is continuous at z = 0. f (0 + h) − f (0) f (z) is NOT differentiable at z = 0: We will show that the limit lim h→0 h does not exist. Suppose h approaches 0 along the straight line y = mx, x > 0, then in this case f (0 + h) − f (0) |h| lim = lim h→0 h h→0h p h21 + h22 = lim , h = h1 + ih2 h→0 h1 + ih2 p h21 + m2 h21 = lim h1 →0 h1 + imh1 (putting h2 = mh1 as h approaches 0 along y = mx) √ 1 + m2 =. 1 + im f (0 + h) − f (0) Since the limit depends on m, lim does not exist, and hence f (z) is not h→0 h differentiable at z = 0. 11 Analytic Functions M. A. Khan Problem 3.1. Consider the function f (z) = |z|2 = x2 +y 2 , z = x+iy. The function f can also be thought of as a function from R2 to R mapping (x, y) to x2 + y 2. Moreover, since the partial derivatives of f are continuous throughout R2 , it follows that f is differentiable everywhere on R2. Show that f (z) is not complex differentiable at any non-zero point z0. Solution. Left as an exercise. Theorem 3.3 (C-R Equations). Let f (z) = u(x, y) + iv(x, y), z = x + iy, be differentiable at z0 = x0 + iy0. Then the partial derivatives of u and v exist at the point (x0 , y0 ), and ux (x0 , y0 ) = vy (x0 , y0 ) and vx (x0 , y0 ) = −uy (x0 , y0 ). (3.1) Further, f 0 (z0 ) = ux (x0 , y0 ) + ivx (x0 , y0 ) = vy (x0 , y0 ) − iuy (x0 , y0 ). (3.2) The Equations in (3.2) are called Cauchy-Riemann equations (C-R equations). Example 3.1. Consider the function f (z) = z 2 +i. Let f = u+iv. Then u(x, y) = x2 −y 2 , v(x, y) = 1 + 2xy. f is differentiable at every z ∈ C. Moreover, for z0 = x0 + iy0 , f 0 (z0 ) = ux (x0 , y0 ) + ivx (x0 , y0 ) = 2x0 + i2y0 = 2z0. Problem 3.2. Use C-R equations to prove that f (z) = |z|2 is not differentiable at any non-zero point z0. Solution. Here f = u + iv, where u(x, y) = x2 + y 2 , and v(x, y) = 0. Then ux (x0 , y0 ) = 2x0 vx (x0 , y0 ) = 0 uy (x0 , y0 ) = 2y0 vy (x0 , y0 ). Therefore, it follows that the C-R equations ux (x0 , y0 ) = vy (x0 , y0 ) and vx (x0 , y0 ) = −uy (x0 , y0 ) are satisfied ONLY at the origin (0, 0), and therefore by Theorem 3.3, f (z) = |z|2 is not differentiable at any non-zero point z0. Example 3.2 (Example of a function which satisfies C-R equations at a p point, but is not differentiable at that point). Consider the function f (z) = f (x + iy) = |xy|. Then f = u + iv, where u(x, y) = |xy|, and v(x, y) = 0 for all (x, y) ∈ R2. We note the p following fact: u and v satisfies C-R equations at the origin: Since v(x, y) = 0 for all (x, y) ∈ R2 , we have vx (0, 0) = vy (0, 0) = 0. Also, u(h, 0) − u(0, 0) ux (0, 0) = lim =0 h→0 h u(0, h) − u(0, 0) uy (0, 0) = lim = 0. h→0 h 12 Analytic Functions M. A. Khan Thus u and v satisfies the Cauchy-Riemann equations. From this fact, can we conclude that f (z) is differentiable at the origin? Answer is no. In fact, f (z) is not differentiable at the origin as shown below. f (z) − f (0) f (z) is not differentiable at the origin: We will show that the limit lim z→0 z−0 does not exist. Suppose z approaches 0 along the straight line y = mx, x > 0, m 6= 0. Then in this case p f (z) − f (0) |xy| lim = lim , z = x + iy z→0 z−0 z→0 x + iy p |mx2 | = lim z→0 x + imx (putting y = mx as z approaches 0 along y = mx) p |m| =. 1 + im f (z) − f (0) Since the limit depends on m, lim does not exist, and hence f (z) is not z→0 z−0 differentiable at z = 0. Remark 3.2. Example 3.2 shows that the satisfaction of the C-R equations by the real and imaginary parts of a function f at a point z0 does not give the differentiability of the function f at z0. But, in addition of satisfying the C-R equations, if partial derivatives of the real and imaginary parts of f are also continuous at z0 , then we will obtain the differentiability of f at z0. More precisely, we have the following result. Theorem 3.4 (Converse of C-R Equations). Suppose f = u + iv is defined on some neighborhood N (z0 ; ) of z0 = x0 + iy0 such that ux , uy , vx , vy exist on N (z0 ; ) and are continuous at (x0 , y0 ). If u, v satisfies the C-R equations at (x0 , y0 ), then f 0 exist at z0 and f 0 (z0 ) = ux (x0 , y0 ) + ivx (x0 , y0 ). Problem 3.3. Using Theorem 3.4 show that the following functions are differentiable everywhere in the complex plane: 1. f (x + iy) = x3 − 3xy 2 + i(3x2 y − y 3 ) 2. f (x + iy) = e−y cos x + ie−y sin x. Solution. (1) Here f = u + iv where u(x, y) = x3 − 3xy 2 , and v(x, y) = 3x2 y − y 3. Moreover, ux (x, y) = 3x2 − 3y 2 uy (x, y) = −6xy 2 2 vy (x, y) = 3x − 3y vx (x, y) = 6xy. Therefore, ux , uy , vx , vy are all continuous on R2 , and u, v satisfies the C-R equations ux (x, y) = vy (x, y), uy (x, y) = −vx (x, y) at each point (x, y) ∈ R2. Therefore, by Theorem 3.4, f is differentiable everywhere in the complex plane. Moreover, f 0 (z) = ux (x, y) + ivx (x, y) = 3(x2 − y 2 ) + i6xy, z = x + iy. (2) Left as an exercise. 13 Analytic Functions M. A. Khan Problem 3.4. Find all the points where the function f (x + iy) = x2 + iy 2 is differentiable. Solution. Here f = u + iv, where u(x, y) = x2 , and v(x, y) = y 2. Therefore, ux (x, y) = 2x uy (x, y) = 0 vy (x, y) = 2y vx (x, y) = 0. Then ux = vy and uy = −vx gives x = y. Thus, C-R equations are satisfied only when x = y. Therefore, by Theorem 3.3, f is not differentiable at a point which does not lie on the straight line x = y. Moreover, as ux , uy , vx , vy are all continuous at each point on the line x = y, it follows from Theorem 3.4 that f is differentiable at each point on the line x = y. Thus f is differentiable only at the points on the line x = y. Problem 3.5. Using x = r cos θ, y = r sin θ and the chain rule, prove that the C-R equation is equivalent to 1 1 ur = vθ , vr = − uθ. (3.3) r r Equations in (3.3) are the C-R equations in polar form. Solution. Left as an exercise. 3.2 Analyticity Definition 3.2. A function f is said to be analytic at the point z0 if there exists a neigh- borhood N (z0 , ) of z0 ,  > 0 such that f is differentiable at every point z ∈ N (z0 , ). Similarly, f is said to be analytic (or, regular, or holomorphic) on a set D if it is differen- tiable at every point of some open set containing D. We note the following obvious facts: 1. If f is differentiable at all points of an open set D, then f is analytic on D. 2. If f is differentiable at all points of a set D, then it does not mean that f will be analytic on D (see Example 3.3). 3. If f is analytic at all points of a set D, then f is analytic on D. Compare the Item 2 with the Items 1 and 3 and note the differences. Example 3.3. Consider the function f (x + iy) = x2 + iy 2 , and the set D := {x + iy ∈ C : x = y}. As shown in Problem 3.4, f is differentiable at all points of D, but f is not differentiable at any point lying outside D. Therefore, f is not analytic on D as any open set containing D will contain a point lying outside D, and hence contain a point where f is not be differentiable. Suppose two complex functions are analytic in some domain D. Then their sum, difference and product are all analytic in D. Also, their quotient is analytic in D given that the denominator function is non-zero at all points in D. The composition of two analytic functions is also analytic. Problem 3.6. Show that polynomials are analytic on the set C of all complex numbers. 14 Analytic Functions M. A. Khan Solution. Left as an exercise. Theorem 3.5 (Necessary and Sufficient Condition for Analyticity). A function f = u + iv is analytic in a domain1 D if and only if u, v satisfies C-R equations in D, and ux , uy , vx , vy are continuous in D. Example 3.4. Consider the function f = u + iv defined on the set D := {x + iy ∈ C : x = y}, where u(x, y) = x2 , v(x, y) = y 2. Note that u, v satisfies the C-R equations in D, and ux , uy , vx , vy are also continuous in D. But, as shown in Example 3.3, f is not analytic on D. Does this fact contradict Theorem 3.5? If not, explain why. Definition 3.3 (Entire Function). A function analytic on the entire complex plane is called an entire function. Problem 3.7. Suppose f1 (z) is analytic at z0 , while f2 (z) is non-analytic at z0. Then show that f1 (z) + f2 (z) is not analytic at z0. Give an example to show that sum of two non-analytic function can be analytic. Solution. Left as an exercise. 3.3 Applications of C-R Equations Proposition 3.6. Suppose f is analytic in a domain D. If f 0 (z) = 0 for all z ∈ D, then f is constant on D. Solution. Let f = u + iv. Since f is analytic in D, we have f 0 (x + iy) = ux (x, y) + ivx (x, y) = vy (x, y) − iuy (x, y), x + iy ∈ D. Therefore, f 0 (z) = 0 for all z ∈ D gives ux (x, y) = 0 and uy (x, y) = 0 for all (x, y) ∈ D ⇒ u is a constant function in D vx (x, y) = 0 and vy (x, y) = 0 for all (x, y) ∈ D ⇒ v is a constant function in D. This implies that f = u + iv is a constant function in D. Proposition 3.7. Suppose f is analytic in a domain D. If any of Ref , Imf is constant in D, then f is constant in D. Solution. Let f = u + iv. First we assume that Ref is constant on D, that is u(x, y) is a constant function in D. Thus, ux (x, y) = uy (x, y) = 0 for all (x, y) ∈ D. Now, since f is analytic in D, we must have for all (x, y) ∈ D, vy (x, y) = ux (x, y) = 0 and vx (x, y) = −uy (x, y) = 0. This implies that v(x, y) is a constant function on D. Therefore, f is a constant function in D as both u and v is obtained as constant functions in D. 1 Recall that domain is an open connected set 15 Analytic Functions M. A. Khan Problem 3.8. Suppose f is analytic in a domain D such that |f | is constant in D. Then show that f is a constant in D. Solution. Left as an exercise. Example 3.5. The functions |z|, |z|2 , Re z, Im z are non-constant real functions (i.e. imaginary part of f is zero) defined on C. Hence, by Proposition 3.7 they are nowhere analytic. Problem 3.9. Let f = u + iv be a non-constant function such that f = u − iv be analytic in a domain D. Show that f cannot be analytic in D. Solution. Left as an exercise. Problem 3.10. Given an analytic function w = f (z) = u(x, y) + iv(x, y), z = x + iy, the equations u(x, y) = α and v(x, y) = β, α and β are constants, define two families of curves in the complex plane. Show that the two families are mutually orthogonal to each other. Solution. Left as an exercise. 3.4 Harmonic Functions Definition 3.4 (Harmonic Function). A real-valued function φ(x, y) of two real variables x and y is said to be harmonic in a given domain D in the xy-plane if φ has continuous partial derivatives up to the second order in D and satisfies the Laplace equation φxx (x, y) + φyy (x, y) = 0. Definition 3.5 (Conjugate Harmonic Function). If two harmonic functions φ(x, y), ψ(x, y) satisfy C-R equations, namely φx = ψy and φy = −ψx in a domain D, then ψ is called conjugate harmonic function of φ. Remark 3.3. Note that harmonic conjugacy is not a symmetric relation, that is, if ψ is conjugate harmonic function of φ, then it does not mean that φ will be conjugate harmonic function of ψ. This is due to the minus sign in the second CauchyRiemann relation. The following theorem gives the close link between analyticity and harmonic conjugacy. Theorem 3.8. A complex function f (z) = u(x, y) + iv(x, y), z = x + iy, is analytic in a domain D if and only if v is a harmonic conjugate of u in D. Proof. We need one result to prove the above claim. It will be shown in Section 5 that if a complex function is analytic at a point, then its real and imaginary parts have continuous partial derivatives of all orders at that point (cf. Remark 5.2). 16 Analytic Functions M. A. Khan (⇒) We assume that f = u + iv is analytic in D, and we prove that v is a harmonic conjugate of u in D. Since f is analytic in D, by the above mentioned fact, u and v have continuous partial derivatives of all orders in D, and ux (x, y) = vy (x, y), uy (x, y) = −vx (x, y) for all (x, y) ∈ D. Therefore, for all (x, y) ∈ D uxx (x, y) = vyx (x, y) = vxy (x, y) (since the partial derivatives are all continuous) uyy (x, y) = −vxy (x, y) vxx (x, y) = uyx (x, y) = −uxy (x, y) (since the partial derivatives are all continuous) vyy (x, y) = uxy (x, y). This gives ux x(x, y) + uy (x, y) = 0 and vx x(x, y) + vy (x, y) = 0 for all (x, y) ∈ D. Thus, we have shown that v is a harmonic conjugate of u in D. (⇐) We now assume that v is a harmonic conjugate of u in D, and we prove that f = u+iv is analytic in D. In fact, this follows directly from Theorem 3.5. Proposition 3.9. Any two harmonic conjugates v, w of u in a domain D differ by a constant, that is, v(x, y) − w(x, y) = K for all (x, y) ∈ D, where K is a real constant. Proof. By the given conditions, the functions f1 = u + iv, and f2 = u + iw are both analytic in D, and hence f1 − f2 = i(v − w) is analytic in D. Note that Re(f1 − f2 ) = 0, and hence by Proposition 3.7, f1 − f2 is constant in D. This gives v − w as a real constant in D. Problem 3.11. Let v be harmonic conjugate of u in a domain D. Then show that v + K is also a harmonic conjugate of u in D, where K is a real constant. Solution. Left as an exercise. Given harmonic function u(x, y) in a simply connected domain2 D, it is always possible to obtain its harmonic conjugate v(x, y) of u in D as illustrated by the following problem. Problem 3.12. Find a conjugate harmonic function of u(x, y) = x2 − y 2 − y in C. Solution. Let v be harmonic conjugate of u in C. Then for all (x, y) ∈ C, vy (x, y) = ux (x, y) = 2x and (3.4) vx (x, y) = −uy (x, y) = 2y + 1. (3.5) From (3.4), we obtain v(x, y) = 2xy + φ(x) (3.6) Differentiating (3.6) w.r.t x, we obtain vx (x, y) = 2y + φ0 (x) (3.7) 2 Defined in Section 5 17 Elementary Functions M. A. Khan From (3.5) and (3.7), we obtain 2y + φ0 (x) = 2y + 1 ⇒ φ0 (x) = 1 ⇒ φ(x) = x + K Thus for each value of the real constant K (for instance K = 2), we obtain a conjugate harmonic function of u as v(x, y) = 2xy + K. Problem 3.13. Given an analytic function w = f (z) = u(x, y) + iv(x, y), z = x + iy, the equations u(x, y) = α and v(x, y) = β, α and β are constants, define two families of curves in the complex plane. Show that the two families are mutually orthogonal to each other. Solution. Left as an exercise. Problem 3.14. Given the harmonic function u(x, y) = ex cos y + xy, find the family of curves that is orthogonal to the family u(x, y) = α, α is constant. Solution. Left as an exercise. 4 Elementary Functions 4.1 Exponential Function Definition 4.1. Complex exponential function ez is defined by ez = ex (cos y + i sin y), z = x + iy. Therefore, for z = iy, y ∈ R, we obtain eiy = cos y + i sin y. (4.1) Equation (4.1) is known as Euler formula. Using this formula, the polar form of any complex number z = r(cos θ + i sin θ) can be written as z = reiθ. Proposition 4.1 (Properties). 1. |ez | = ex , and arg ez = y ± 2nπ, (n = 0, 1, 2,...), where z = x + iy. d z 2. ez is an entire function, and e = ez. dz 3. ez1 +z2 = ez1 ez2. 18 Elementary Functions M. A. Khan 4. ez 6= 0 for all z. 5. ez is periodic with the fundamental period 2πi, that is ez+2kπi = ez , for any z and integer k. [Thus the complex exponential function is periodic while its real counterpart is not] 6. ez is NOT injective. 7. If H = {x + iy : −π < y ≤ π}. Then ez is bijective from H to C \ {0}. Proof. (1): Obvious. (2): Let u and v be the real and imaginary parts of the function ez. Then u(x, y) = ex cos y, v(x, y) = ex sin y. Note that u and v satisfies the C-R equation ux = vy , uy = −vx for all (x, y) ∈ R2. Moreover, ux , uy , vx , vy are all continuous on R2. Therefore, by Theorem 3.5, it follows that ez is an entire function, and d z e = ux (x, y) + ivx (x, y) (By Theorem 3.3) dz = ex cos y + iex ex sin y = ez. (3): Left as an exercise. (4): Note that |ez | = ex 6= 0 for all z. Therefore, ez 6= 0 for all z. (5): Left as an exercise. (6): As ez is periodic, it follows that ez is not injective. (7): Left as an exercise. Problem 4.1. Let f (z) be a complex function which satisfies the following: 1. f (z) is an entire function, 2. f 0 (z) = f (z) for all z, and 3. f (x) = ex for real x. Then show that f (z) = ez. Solution. Left as an exercise. Problem 4.2. Find all the roots of the equation ez = i. Solution. Left as an exercise. 19 Elementary Functions M. A. Khan 4.2 Logarithmic Function We use C∗ and H to denote the set C \ {0}, and {x + iy : −π < y ≤ π} respectively. Definition 4.2. For z ∈ C∗ , we define log z = ln |z| + iarg z. Here ln |z| stands for the real logarithm of |z|. Since arg z = Arg z + 2kπ, k ∈ Z, it follows that log z is not well defined as a function. In fact, log z is an instance of multi-valued function. Therefore, we have the following definition. Definition 4.3. For z ∈ C∗ the principal value of the logarithm is defined as Log z = ln |z| + iArg z. Note that Log z : C∗ → H is well defined (now it is single valued). The connection between log z and Log z is Log z + i2kπ = log z for some k ∈ Z. Theorem 4.2 (Properties). 1. For z 6= 0, eLog z = z. 2. For z ∈ H, Log ez = z. / H, Log ez 6= z. 3. For z ∈ 4. For real x > 0, Log x = ln x. 5. Log z is not continuous on the negative real axis R− = {z = x + iy : x < 0, y = 0}. [Unlike real logarithm, it is defined there, but useless] 6. Log z is analytic on the set C∗ \ R−. d 1 7. Log z =. dz z 8. The identity Log (z1 z2 ) = Log z1 + Log z2 is true iff Arg z1 + Arg z2 ∈ (−π, π]. Proof. (1): eLog z = eln |z|+iArg z = |z|(cos Arg z + i sin Arg z) = z. (2): Let z = x + iy, −π < y ≤ π. Then Log ez = Log (ex cos y + iex sin y) = ln ex + iy = x + iy. (3): Let z = x + iy ∈ / (−π, π]. Note that Arg (ex cos y + iex sin y) 6= y / H, and hence y ∈ as y ∈ / (−π, π]. Therefore Log ez = Log (ex cos y + iex sin y) = ln ex + iArg (ex cos y + iex sin y) 6= x + iy = z. (4): Left as an exercise. 20 Elementary Functions M. A. Khan (5): Consider an arbitrary point z = −α ∈ R− , α > 0, and we show that Log z is not continuous at α. Consider the sequences 1 1 {an = αei(π− n ) } and {bn = αei(−π+ n ) }. Then lim an = lim an = z, n→∞ n→∞ but 1 lim Log an = lim ln α + i(π − ) = ln α + iπ, and n→∞ n→∞ n 1 lim Log bn = lim ln α + i(−π + ) = ln α − iπ. n→∞ n→∞ n Since lim Log an 6= lim Log bn , it follows that Log z is not continuous at α. n→∞ n→∞ (6): Consider the domain D = C∗ \ R− = {z : z 6= 0, Arg z ∈ (−π, π)}, and let Log z = ln r + iθ = u(r, θ) + iv(r, θ), where z = r(cos θ + i sin θ) ∈ D. Therefor u(r, θ) = ln r, v(r, θ) = θ. Then 1 1 1 ur = vθ = and vr = − uθ = 0. r r r Thus the C-R equations are satisfied. Since ur , uθ , vr , vθ are continuous in D, it follows that Log z is analytic in D. Note: Note that the function f (z) = Arg z is discontinuous at each point α on the nega- tive x-axis. In fact, if z approaches α through the upper half plane, then limz→α f (z) = π, and if z approaches α through the lower half plane, then limz→α f (z) = −π. Therefore, above argument will not give analyticity of Log z in C∗. (7): Left as an exercise. (8): Left as an exercise. 4.3 Trigonometric and Hyperbolic Functions Using the Euler formula eiy = cos y + i sin y the real sine and cosine functions can be expressed in terms of eiy and e−iy as follows: eiy − e−iy eiy + e−iy sin y = and cos y =. 2i 2 It is natural to define the complex sine and cosine functions in terms of the complex exponential functions eiz and e−iz in the same manner as for the real functions. Thus, we have the following definition. Definition 4.4. The functions sin z and cos z are defined as follows: eiz − e−iz eiz + e−iz sin z = and cos z =. 2i 2 21 Elementary Functions M. A. Khan Theorem 4.3 (Properties). 1. sin z and cos z are entire functions. d d 2. sin z = cos z, and cos z = − sin z. dz dz 3. Real and imaginary parts of sin z and cos z are given by sin z = sinx cosh y + i cos x sinh y cos z = cos x cosh y − i sin x sinh y, ex −e−x ex +e−x where sinh x = 2 and cosh x = 2. 4. Modulii of sin z and cos z are given by q | sin z| = sin2 x + sinh2 y q | cos z| = cos2 x + sinh2 y. 5. sin z and cos z are unbounded. [Recall that real sine and cosine functions are bounded. This is an important differ- ence between real and complex sine and cosine functions] 6. sin z = 0 iff z = kπ, and cos z = 0 iff z = kπ + π2 , where k ∈ Z. Proof. (1): The complex sine and cosine functions are entire since they are formed by the linear combination of the entire functions eiz and e−iz. (2): Left as an exercise. (3): Note that for z = x + iy, eiz = e−y+ix = e−y (cos x + i sin x) e−iz = ey−ix = ey (cos x − i sin x). Therefore, eiz − e−iz sin z = 2i i sin x(ey + e−y ) cos x(e−y − ey ) = + 2i 2i = sin x cosh y + cos x sinh y. Similarly, one can prove cos z = cos x cosh y − i sin x sinh y. (4): Left as an exercise. (5): Since sinh y is unbounded at large values of y, from (4) it follows that sin z and cos z are unbounded. 22 Elementary Functions M. A. Khan (6): Let z = x + iy. sin z = 0 eiz − e−iz ⇒ =0 2i ⇒ e2iz = 1 ⇒ e−2y cos 2x = 1 and e−2y sin 2x = 0. Since e−2y 6= 0, we obtain sin 2x = 0, and hence x = 12 nπ, n ∈ Z. Note that for x = 12 nπ, n ∈ Z, we obtain cos 2x ∈ {−1, 1}, and hence e−2y cos 2x = 1 gives e−2y = 1, and cos 2x = 1. This gives y = 0, and x = nπ, k ∈ Z. Therefore z = kπ. Similarly one can prove that cos z = 0 iff z = kπ + π2. The following theorem follows just by applying the definitions. Theorem 4.4. 1. sin(−z) = − sin z and cos(−z) = cos z. 2. sin(z1 + z2 ) = sin z1 cos z2 + cos z1 sin z2. 3. sin 2z = 2 sin z cos z. 4. sin(z + π) = sin z, sin(z + 2π) = sin z. 5. cos(z1 + z2 ) = cos z1 cos z2 − sin z1 sin z2. 6. cos 2z = cos2 z − sin2 z. Now using sine and cosine we can define tan z, sec z, cosec z as in the real case. z −z We can also define complex analogue of the hyperbolic functions sinh z = e −e 2 and ez +e−z cosh z = 2. The functions tan z and sec z are analytic in any domain that does not include points where cos z = 0. Similarly, the functions cot z and cosec z are analytic in any domain excluding those points z such that sin z = 0. Derivative formulas for the complex trigonometric and hyperbolic functions are exactly the same as those for their real counterparts, and can be deduced easily using the derivative of ez. Problem 4.3. Show that cos z = cos z. Solution. Left as an exercise. 23 Complex Integration M. A. Khan 5 Complex Integration 5.1 Integral of a Complex Valued Function of a Real Variable Let f : [a, b] → C be a piecewise continuous function. Then f (t) = u(t) + iv(t) where u, v : [a, b] → R. We then define Z b Z b Z b f (t) dt = u(t) dt + i v(t) dt. a a a Theorem 5.1 (Properties). Let f : [a, b] → C be a piecewise continuous function. 1. If F 0 (t) = f (t) for all t ∈ [a, b], then Z b f (t) dt = F (b) − F (a). a Rb Rb Rb 2. Re a f (t) dt = a Re f (t) dt = a u(t) dt. Rb Rb Rb 3. Im a f (t) dt = a Im f (t) dt = a v(t) dt. Rb Rb Rb 4. a [f (t) + g(t)] dt = a f (t) dt + a g(t) dt. Rb Rb 5. a αf (t) dt = α a f (t) dt, where α is any complex constant. Rb Rc Rb 6. If a < c < b, then a f (t) dt = a f (t) dt + c f (t) dt. Rb Rb 7. a f (t) dt ≤ a |f (t)| dt. Proof. Item (1) follows from the fundamental theorem of calculus. R Proofs  of Items (2)-(6) b are obvious. We provide the proof of Item (7). Let φ = Arg a f (t) dt Z b Z b f (t) dt = e−iφ f (t) dt (∵ |z| = e−iArg z z) a a Z b = e−iφ f (t) dt a Z b Z b Z b = Re e−iφ f (t) dt (∵ e −iφ f (t) dt = f (t) dt is real ) a a a Z b   = Re e−iφ f (t) dt a Note that this is Riemann integral of the real function Re e−iφ f (t)
 Z b ≤ e−iφ f (t) dt a  Using the property of Riemann integral with the fact that   Re e−iφ f (t) ≤ e−iφ f (t)  Z b = |f (t)| dt. a 24 Complex Integration M. A. Khan Problem 5.1. For α ∈ R, show that Z b eiαb − eiαa eiαt dt =. a iα Solution. Left as an exercise. 5.2 Contour Integral Definition 5.1 (Contour Integral). Let C = γ(t), t ∈ [a, b] be a contour contained in a domain D, and f : D → C be a piecewise continuous function R defined on C. Then the contour integral of f (z) along the contour C, denoted as C f (z) dz is defined as Z Z b f γ(t) γ 0 (t) dt.
f (z) dz = (5.1) C a R Proposition 5.2. The contour integral C f (z) dz is independent of the parametrization of the curve C. Proof. Let γ : [a, b] → C and µ : [c, d] → C be two parametrizations of the curve C. Then there exists a bijective and increasing function g : [c, d] → [a, b] such that µ = γ ◦ g. Now, Z Z b f (z) dz = f (γ(u))γ 0 (u) du γ a b dγ(g(t)) 0 Z = f (γ(g(t))) g (t) dt (Taking u = g(t)) a du Z b d d = f (γ(g(t))) (γ(g(t))) dt (∵ (γ(g(t))) = γ 0 (g(t))g 0 (t)) a dt dt Z b = f (µ(t))µ0 (t) dt a 1 R Problem 5.2. Find C z−z 0 dz, where C is a circle centered at z0 and of radius r. The path is traced out once in the anticlockwise direction. 1 Solution. Here f (z) = z−z 0. A parametrization of C is given by γ(θ) = z0 + reiθ , 0 ≤ θ ≤ 2π, where r is the radius of the circle C. Therefore, 1 Z Z dz = f (γ(θ))γ 0 (θ) dθ C z − z0 C Z 2π = f (z0 + reiθ )ireiθ dθ 0 Z 2π 1 = iθ ireiθ dθ 0 re Z 2π = i dθ 0 = 2πi. 25 Complex Integration M. A. Khan Theorem 5.3 (Properties). R R 1. C f (z) dz = − −C f (z) dz. 2. Let C = C1 + C2 + · · · + Cn , where the terminal point of Ck coincides with the initial point of Ck+1 , then Z Z Z Z f (z) dz = f (z) dz + f (z) dz + · · · + f (z) dz. C C1 C2 Cn R R 3. C Kf (z) dz = K f (z) dz, K is a complex constant. C R R R 4. C (f (z) + g(z)) dz = C f (z) dz + C g(z) dz. Proof. We provide the proof of Item 1 only. Rest of the proofs follow straightway from the definitions involved and left as exercise. Let C : γ(t), a ≤ t ≤ b be a parametrization of C. Then a parametrization of −C is given by −C : γ− (t) = γ(b + a − t), a ≤ t ≤ b. Therefore, b d Z Z f (z) dz = f (γ− (t)) (γ− (t)) dt −C a dt b d Z = f (γ(b + a − t)) (γ(b + a − t)) dt a dt b d Z = (γ(s)) dt, s = b + a − t f (γ(s)) a dt Z a d ds = − f (γ(s)) (γ(s)) ds ds dt Z ab d = f (γ(s)) (γ(s)) ds b ds Z b d = − f (γ(s)) (γ(s)) ds ds Za = − f (z) dz. C z 2 dz, where R Problem 5.3. Evaluate I = C 1. C is along x-axis from 0 to 1 and then along the line parallel to y-axis from 1 to 1 + 2i. 2. C is the line segment from 0 to 1 + 2i. 3. C is an arc of unit circle |z| = 1 traversed in the anticlockwise direction with initial point −1 and final point i. 4. C is an arc of unit circle |z| = 1 traversed in the clockwise direction with initial point −1 and final point i. 26 Complex Integration M. A. Khan Solution. (1): Let C1 be x-axis from 0 to 1 and C2 be the line parallel to y-axis from 1 to 1 + 2i. Then C = C1 + C2 (cf. Figure 7). Parametrizations of C1 and C2 are given by C1 : γ1 (t) = t, 0 ≤ t ≤ 1 C2 : γ2 (t) = (1 − t) + t(1 + 2i) = 1 + 2it, 0 ≤ t ≤ 1 Therefore, 1 1 Z Z z 2 dz = t2 dt = (from (5.1) ) C1 0 3 1 2 Z Z z 2 dz = (1 + 2it)2 2i dt = −4 − i. C2 0 3 Hence, 11 + 2i Z Z Z 2 2 z dz = z dz + z 2 dz = −. C C1 C2 3 y 1 + 2i y 1 + 2i C2 C x x O C1 1 O Figure 7: Figure 8: (2): Left as an exercise. C is given by Figure 8. (3): Let C1 , C2 , C3 and C4 are the arc of the unit circle S : |z| = 1 from 1 to i, i to −1, −1 to −i and −i to 1 respectively (cf. Figure 9). Then C = C3 + C4 + C1. Therefore Z Z Z Z Z z 2 dz = z 2 dz + z 2 dz + z 2 dz + z 2 dz S ZC1 Z C2 C3 C4 = z 2 dz + z 2 dz C C2 Parametrizations of S and C2 are given by S: γ1 (t) = eiθ , 0 ≤ θ ≤ 2π π C2 : γ2 (t) = eiθ , ≤ θ ≤ π. 2 27 Complex Integration M. A. Khan Therefore, Z Z Z 2 2 ∴ z dz = − z dz + z 2 dz C C2 S Z π  2 Z 2π  2 iθ iθ = − e ie dθ + eiθ ieiθ dθ π 2 0 1 h 3iθ iπ 1 h 3iθ i2π = − e + e 3 π 2 3 0 1 =(1 − i) 3 π (4): C is shown in Figure 10. Let D : γ(t) = eiθ , 2 ≤ θ ≤ π, be the arc of the unit circle y y i i C2 C C1 x x −1 O 1 −1 O 1 C3 C4 −i −i Figure 9: Figure 10: |z| = 1 from i to −1. Then −C = D. Therefore Z Z 2 z dz = − z 2 dz (by Item 1 of Theorem 5.3) C −C Z π  2 = − eiθ ieiθ dθ π 2 1h iπ = − e3iθ π 3 2 1 = − (i − 1) 3 1 = (1 − i). 3 The reason for obtaining same answer for the problems in Items 3 and 4 of Problem 5.3 is due to the following result. Theorem 5.4 (Integration using Indefinite Integral). Let f be a continuous function defined on an open set D and there exist a function F defined on D such that F 0 = f. Let z1 , z2 ∈ D. Then for any contour C lying in D starting from z1 , and ending at z2 , Z f (z) dz = F (z2 ) − F (z1 ). C 28 Complex Integration M. A. Khan Proof. Suppose that that C is given by a map γ : [a, b] → C. Then d F (γ(t)) = F 0 (γ(t))γ 0 (t). dt Hence Z Z b f (z) dz = f (γ(t))γ 0 (t) dt C a b d Z = F (γ(t)) dt a dt Z b d = G(t) dt, where G(t) = F (γ(t)) a dt = G(b) − G(a) (by Item (1) of Theorem 5.1) = F (γ(b)) − F (γ(a)) = F (z2 ) − F (z1 ). R Remark 5.1. In order to use Theorem 5.4 to evaluate C f (z) dz, we need to find a simply connected domain D such that 1. f is continuous on D, 2. F 0 = f on D, for some F , 3. C lies in D. Compare Theorem 5.4 with the fundamental theorem of calculus. Problem 5.4. Evaluate the following integrals. 1. C1 z 2 dz and C2 z 2 dz where C1 and C2 are arc of unit circle |z| = 1 with initial R R point −1 and final point i traversed in the clockwise and anticlockwise directions respectively. 2. C z 2 dz, where C is any contour starting from z1 and ending at z2. R R 3. C cos z dz, where C is any contour starting from −iπ and ending at iπ. 4. C z1 dz, where C is the straight line from −i to 1 and then from 1 to i. R d z3   Solution. (1): Note that = z 2 for all z ∈ C. Therefore, we take C to be the dz 3 simply connected domain D and then by Theorem 5.4, we obtain Z Z h z 3 ii i3 − (−1)3 1−i 2 z dz = z 2 dz = = =. C1 C2 3 −1 3 3 This problem illustrates why we obtain same answer for the problems in Items 3 and 4 of Problem 5.3. 29 Complex Integration M. A. Khan (2): Again using Theorem 5.4, we obtain Z h z 3 iz2 z 3 − z13 z 2 dz = = 2. C 3 z1 3 d (3): Since sin z = cos z for all z ∈ C, and hence by Theorem 5.4, we obtain dz Z h iiπ cos z dz = sin z = sin(iπ) − sin(−iπ) = 2 sin(iπ). C −iπ d 1 (4): Let us take the domain D to be the set C \ {x + iy : y = 0, x ≤ 0}. Since Log z = dz z for all z ∈ D, and C lies in D, we obtain from Theorem 5.4 1 ii π −π Z h dz = Log z = Log (i) − Log (−i) = i − i = iπ. C z −i 2 2 Problem 5.5. Explain why the following integrals cannot be evaluated using Theorem 5.4: 1. C z1 dz, where C is the straight line from i to −i. R 2. C |z|2 dz, where C is an arc of unit circle |z| = 1 traversed in the clockwise direction R with initial point −1 and final point i. Problem 5.6. Evaluate C |z|2 dz, where C is an arc of unit circle |z| = 1 traversed in R the clockwise direction with initial point −1 and final point i. Solution. Left as an exercise. Definition 5.2 (Arc Length of a Contour). The arc length of a contour C : γ(t) = x(t) + iy(t), a ≤ t ≤ b is given by Z b L(γ) = |γ 0 (t)| dt a Z bp = (x0 (t))2 + (y 0 (t))2 dt a Example 5.1. Consider the circle with center at z0 and radius r given by γ(t) = z0 + reit , 0 ≤ t ≤ 2π. Then Z 2π h i2π L(γ) = |ireit | dt = rt = 2πr. 0 t=0 Theorem 5.5 (L-M Formula). If L is the arc length of a contour C : γ(t), a ≤ t ≤ b and M is a positive number such that |f (z)| ≤ M for all z ∈ C, then Z f (z) dz ≤ M L. C 30 Complex Integration M. A. Khan Proof. Z Z b f (z) dz = f (γ(t))γ 0 (t) dt C a Z b ≤ |f (γ(t))| γ 0 (t) dt a Z b ≤ M γ 0 (t) dt a Z b = M γ 0 (t) dt a = ML 31 Complex Integration M. A. Khan 5.3 Cauchy Integral Theorem Theorem 5.6 (Cauchy Integral Theorem). Let f (z) be a function such that 1. f (z) is analytic on and inside a simple closed contour C, and 2. f 0 (z) be continuous on and inside C. Then Z f (z) dz = 0. (5.2) C Proof. Let f = u + iv, and C : γ(t) = x(t) + iy(t), a ≤ t ≤ b. Then Z Z b f (z) dz = f (γ(t))γ 0 (t) dt C a Z b u x(t), y(t) + iv x(t), y(t) x0 (t) + iy 0 (t) dt

   = a Z bh i u x(t), y(t) x0 (t) − v x(t), y(t) y 0 (t) dt +

= a Z b (v x(t), y(t) x0 (t) + u x(t), y(t) y 0 (t)) dt

i a Z b
   u x(t), y(t) , −v x(t), y(t) · x0 (t), y 0 (t) dt +
= a Z b
   v x(t), y(t) , u x(t), y(t) · x0 (t), y 0 (t) dt
i a h i ‘·’ denotes the dot product, and the integrals here are the line integrals Z Z = u dx − v dy + v dx + u dy ZCZ C ZZ = (−vx − uy ) dx dy + i (ux − vy ) dx dy, D D where D is the closed region enclosed by C h ZZ Z i Using Green’s Theorem: (Nx − My ) dx dy = M dx + N dy D C = 0 (Using C-R equations). In 1903, Goursat was able to obtain the same result as in eq. (5.2) without assuming the continuity of f 0 (z). This stronger version is called the Cauchy-Goursat Theorem or Goursat Theorem, and is stated as follows. Theorem 5.7 (Cauchy-Goursat Theorem). Let f (z) be a function such that f (z) is ana- lytic on and inside a simple closed contour C. Then Z f (z) dz = 0. C The Cauchy-Goursat theorem can be stated in the following alternative form: 32 Complex Integration M. A. Khan Theorem 5.8 (Alternative Form of Cauchy-Goursat Theorem). Let f (z) be a function analytic throughout a simply connected domain D and C be a simple closed contour lying completely inside D (cf. Figure 11). Then Z f (z) dz = 0. C Figure 11: Problem 5.7. Prove that the statements of the Cauchy-Goursat theorem given in Theo- rems 5.7 and 5.8 are equivalent. Problem 5.8. Evaluate the following integrals where C denotes the circle of unit radius with center at zero. R 1. C sin z. n 2. C ez. R 2 eiz R 3. C z 2 +4 dz. n Solution. (1)-(2): Since f (z) = sin z and g(z) = ez are both analytic on and inside C, it follows from Cauchy-Goursat Theorem 5.7 that Z Z n sin z = 0 and ez = 0. C C iz 2 (3): The function f (z) = ze2 +4 is analytic on and inside C. Therefore, from Cauchy- GoursatTheorem 5.7 we have 2 eiz Z 2 dz = 0. C z +4 iz 2 Note that the integrand ze2 +4 is not analytic at z = ±2 but that does not bother us as these points are outside C. Problem 5.9. Evaluate C cosec2 z, where C denotes the circle of unit radius with center R at zero. Explain why Cauchy-Goursat’s Theorem cannot be applied here to evaluate the integral. 33 Complex Integration M. A. Khan d Solution. Let us take the domain C∗. Note that cosec2 z is continuous in C∗ ,(− cot z) = dz 2 ∗ ∗ Rcosec z 2for all z ∈ C , and C lies in C. Therefore, we obtain from Theorem 5.4 C cosec z = 0 as initial and final points of C are Rsame. Goursat’s Theorem cannot be applied to evaluate C cosec2 z as the function cosec2 z is not analytic inside C (in fact, cosec2 z is not analytic at origin). Theorem 5.9 (Consequences of Cauchy-Goursat’s Theorem). 1. (Independence of Path): Let D be a simply connected domain and f be an an- alytic function defined in D. Then the integral of f (z) is independent of path in D. That is, if z1 , z2 are two points in D and C1 and C2 be two contours inside D joining z1 and z2 , then we have Z Z f (z) dz = f (z) dz. C1 C2 2. (Existence of Antiderivative): If f is an analytic function on a simply connected domain D then there exists a function F which is analytic on D such that F 0 = f. C1 z1 C1 z1 z2 C2 z2 −C2 Figure 12: Figure 13: Figure 14: Figure 15: Proof. (1): For paths C1 and C2 that have only the endpoints in common (cf. Figure 12), we consider closed curve C := C1 + (−C2 ) obtain by joining C1 and −C2 as shown in 1 1 34 Complex Integration M. A. Khan Figure 13. Then by Cauchy-Goursat’s Theorem, we have Z f (z) dz = 0 ZC Z ⇒ dz + dz = 0 C1 −C2 Z Z ⇒ dz − dz = 0 ZC1 C2 Z ⇒ f (z) dz = f (z) dz. C1 C2 For paths that have finitely many further common points, apply the above argument to each “loop” (portions of C1 and C2 between consecutive common points; four loops in Figure 14. For paths with infinitely many common points we would need additional argumentation not to be presented here. (2): Fix a point z0 ∈ D, and define Z z F (z) = f (w) dw, z0 where the integral is considered as a contour integral over any curve lying in D and joining z with z0. By the first part of this theorem, the integral does not depend on the curve we choose and hence the function F is well defined. We will show that F 0 = f. We keep z fixed. Then we choose z + ∆z in D so that the whole line segment with endpoints z and z + ∆ is in D (Figure 15). This can be done because D is a domain, hence it contains a neighborhood of z. Then Z z+∆z Z z F (z + ∆z) − F (z) = f (w) dw − f (w) dw z0 z0 Z z+∆z = f (w) dw, z where theR curve joining z and z + ∆ can be considered as a straight line. Since z is fixed, z+∆z we have z f (z) dw = ∆z, and hence we obtain z+∆z F (z + ∆z) − F (z) 1 Z   − f (z) = f (w) − f (z) dw. (5.3) ∆z ∆z z Since f is continuous at z, given  > 0, there exist a δ > 0 such that |f (w) − f (z)| <  if |w − z| < δ. Thus for |∆z| < δ, we obtain |f (w) − f (z)| < , for all w lying on the line segment with endpoints z and z + ∆z. Therefore, by L-M formula (cf. Theorem 5.5), we obtain Z z+∆z   f (w) − f (z) dw ≤ ∆z. z 35 Complex Integration M. A. Khan This together with eq. (5.3) gives for |∆z| < δ, F (z + ∆z) − F (z) − f (z) ≤ . ∆z This implies F (z + ∆z) − F (z) F 0 (z) = lim = f (z). ∆z→0 ∆z C C P1 A Q1 D1 L1 E F R1 C2 L2 Q2 H G D2 C1 L3 R2 B P2 Figure 16: Figure 17: Theorem 5.10 (Cauchy-Goursat’s Theorem for Multiply Connected Domain). Let C be a positively oriented simple closed contour and Ck , k = 1, 2,... n denote a finite number of positively oriented simple closed contours all lying wholly within C, but each Ck lies in the exterior of every other whose interior have no points in common (cf. Figure 16 for n = 2). If a function f is analytic throughout the closed region D consisting of all points within and on C except for the points interior to each Ck (gray region of Figure 16 for n = 2). Then Z Z Z Z f (z) dz = f (z) dz + f (z) dz + · · · + f (z) dz. C C1 C2 Cn 1 Proof. We provide the 1 proof for the case when n = 2. Let R1 and R2 be the portions of C1 from G to H and H to G respectively as shown in Figure 17. Similarly, Q1 and Q2 are the portions of C2 from E to F and F to E and P1 and P2 are the portions of C from A to B and B to A. By three cuts L1 , L2 , L3 , we divide the domain D into two simply connected domains D1 and D2 (cf. Figure 17). Domain D1 and D2 are enclosed by the simple closed contours T1 and T2 respectively, where T1 = P1 + L3 + (−R1 ) + L2 + (−Q1 ) + L1 T2 = P2 + (−L1 ) + (−Q2 ) + (−L2 ) + (−R2 ) + (−L3 ). 36 Complex Integration M. A. Khan From Cauchy-Goursat Theorem 5.7, we obtain Z f (z) dz = 0 T1 Z f (z) dz = 0 T2 and hence Z Z 0 = f (z) dz + f (z) dz T1 T2 hZ Z Z Z Z = f (z) dz + f (z) dz + f (z) dz + f (z) dz + f (z) dz P L3 −R1 L2 −Q1 Z1 i h Z Z Z Z + f (z) dz + f (z) dz + f (z) dz + f (z) dz + f (z) dz L1 P2 −L1 −Q2 −L2 Z Z i + f (z) dz + f (z) dz −R2 hZ Z −L3 Z Z Z = f (z) dz + f (z) dz − f (z) dz + f (z) dz − f (z) dz P1 L3 R1 L2 Q1 Z i hZ Z Z Z + f (z) dz + f (z) dz − f (z) dz − f (z) dz − f (z) dz L1 P2 L1 Q2 L2 Z Z i − f (z) dz − f (z) dz Z R2 Z L3 Z Z Z Z = f (z) dz + f (z) dz − f (z) dz − f (z) dz − f (z) dz − f (z) dz P1 P2 R1 R2 Q1 Q2 Z Z Z = f (z) dz − f (z) dz − f (z) dz C C1 C2 This gives Z Z Z f (z) dz = f (z) dz + f (z) dz. C C1 C2 For domains of higher connectivity the idea remains the same. In fact, for n = k, we need k + 1 cuts. Corollary 5.11. Let C1 and C2 be two simple closed contours with same orientation such that C2 completely lies inside C1. Let f be a function analytic in the annulus region between C1 and C2 including the points on the curves. Then Z Z f (z) dz = f (z) dz. C1 C2 5.4 Cauchy’s Integral Formula Theorem 5.12 (Cauchy’s Integral Formula). Let the function f (z) be analytic on and inside a positively oriented simple closed contour C and z0 be any point inside C. Then 1 f (z) Z f (z0 ) = dz. (5.4) 2πi C z − z0 37 Complex Integration y M. A. Khan C Cr × z x Figure 18: Proof. We draw a circle Cr , with radius r, centered at the point z0 , small enough to be f (z) completely inside C (see Figure 18). Since z−z 0 is analytic in the region lying between Cr and C, by Theorem 5.10, we have 1 f (z) 1 f (z) Z Z dz = dz 2πi C z − z0 2πi Cr z − z0 1 f (z) − f (z0 ) f (z0 ) 1 Z Z = dz + dz 2πi Cr z − z0 2πi Cr z − z0 1 f (z) − f (z0 ) Z = dz + f (z0 ) (by Problem 5.2) 2πi Cr z − z0 Therefore, to complete the proof, it suffices to show that 1 f (z) − f (z0 ) Z dz = 0. 2πi Cr z − z0 Since f is continuous at z0 , for each  > 0, there exists δ > 0 such that |f (z) − f (z0 )| <  whenever |z − z0 | < δ. Now, suppose we choose r < δ and thus we ensure that Cr lies completely inside the contour C. Let g(z) = f (z)−f z−z0 (z0 ). Note that f (z) − f (z0 ) f (z) − f (z0 )  |g(z)| = = < for all z ∈ Cr : |z − z0 | = r. z − z0 r r Thus, by LM-formula (Theorem 5.5), we have f (z) − f (z0 )  Z Z dz = g(z) dz ≤ × 2πr = 2π. (5.5) Cr z − z0 Cr r Therefore, 1 f (z) − f (z0 ) 1 f (z) − f (z0 ) Z Z dz = dz 2πi Cr z − z0 2π Cr z − z0 ≤  (by eq. (5.5)) Since the above inequality holds for every , it follows that 1 f (z) − f (z0 ) Z