EE2007: Engineering Mathematics II Complex Analysis

EE2007: Engineering Mathematics II omplex Analysis Ling KV School of EEE, NTU ekvling@ntu.edu.sg V4.2: Ling KV, August 6, 2006 V4.1: Ling KV, Jul 2005 EE2007 V4.0: Ling KV, Jan 2005, EE2007 V3.1: Ling KV, Jul 2003, G264 V3.0: Ling KV, Jul 2002, G264 V2.0: Wen Y, Jan 2001, E208 V1.0: Ling KV, Jan 1996, E208 EE2007/Ling KV/Aug 2006 EE2007: Engineering Mathematics II omplex Variables week 7 omplex functions and the auchy-riemann equations. week 8 Integration in the complex plane and auchy integral theorem. week 9 The auchy integral formula and evaluation of real integrals. Website: In edventure.ntu.edu.sg which also contains additional materials. Text: Kreyszig, E. (1999). Advanced Engineering Mathematics, hapters 12-15, 8th Ed. John Wiley & Sons, Inc. [NTU Library, QA401.K92] Reference: 1. Brown and hurchill (1996). omplex Variables and Applications, 6th Ed., McGraw-Hill. [NTU Library, QA331.7.B878] 2. Mathews and Howell (2001). omplex Analysis for Mathematics and Engineering, 4th Ed., Jones and Bartlett Publishers. [NTU Library, Red spot, QA331.7M439] EE2007/Ling KV/Aug 2006 1

Introduction This module provides the tools needed for using functions of a complex variable and their applications. We first revisit complex arithmetic and provide geometric insight for working with complex numbers in both rectangular and polar forms. Then, a systematic study of complex functions follows. We then extend the differential calculus to complex functions, define the derivative of a complex function, obtain the auchy-riemann partial differential equations, and specify the category of analytic functions. Integration of a complex function takes place in the complex plane, as a contour integral, the natural analog of the line integral of Vector alculus 1. The auchy Integral Theorem and its application to evaluate contour integrals will be discussed in detail. 1 Vector alculus will be covered in week 10 EE2007/Ling KV/Aug 2006 2 Revision: omplex Numbers A complex number z is defined as z = x + iy where i = 1 Geometrically, a complex number is a point in the complex plane (or the Argand diagram) and can be considered as a vector in the plane. EE2007/Ling KV/Aug 2006 3

It can be seen from the figure that x = r cos θ, y = r sin θ r = z = x 2 + y 2 = z = zz θ = arg(z) = arctan y x radians = Arg(z) + 2nπ, n = 0, ±1, ±2,... where Arg(z) is the principal values of arg(z) and satisfies π < Arg(z) π From Euler s formula e iθ = cos θ + i sin θ for any real value of θ, we can write the polar form of a complex number as z = x + iy = r cos θ + i r sin θ = = EE2007/Ling KV/Aug 2006 4 From Euler s formula Euler s Formula e iθ = cos θ + i sin θ, and e iθ = cos θ i sin θ We have cos θ = eiθ +e iθ 2, and sin θ = eiθ e iθ 2i The Euler s formula is also valid in complex: e iz = cos z + i sin z This suggests the following definition for complex trigonometric functions: cos z = eiz + e iz, and sin z = eiz e iz 2 2i EE2007/Ling KV/Aug 2006 5

Algebraic Rules Let z 1 = x 1 + iy 1 = r 1 θ 1, z 2 = x 2 + iy 2 = r 2 θ 2. Then Addition and Subtraction z 1 ± z 2 = (x 1 + x 2 ) ± i(y 1 + y 2 ) Multiplication z 1 z 2 = (x 1 x 2 y 1 y 2 ) + i(x 1 y 2 + x 2 y 1 ) Division z 1 z 2 = x 1x 2 + y 1 y 2 x 2 2 + y2 2 + i x 2y 1 x 1 y 2 x 2 2 + y2 2 Quiz Time EE2007/Ling KV/Aug 2006 6 loze Test: The Algebra of omplex Numbers omplex number can be defined as pair z = (x, y) of numbers x and y. The numbers x and y are called the parts of z respectively and we write and x =, y = We identify the pair with real number x. This means that real numbers can be thought of as a of complex numbers. omplex numbers of the form are said to be imaginary numbers. Two complex numbers are if they have the real and imaginary parts. Thus (x 1, y 1 ) = (x 2, y 2 ) if and only if and EE2007/Ling KV/Aug 2006 7

The and of z 1 = (x 1, y 1 ) and z 2 = (x 2, y 2 ) are defined as and z 1 + z 2 = (x 1, y 1 ) + (x 2, y 2 ) = z 1 z 2 = (x 1, y 1 ) (x 2, y 2 ) = (x 1 x 2 y 1 y 2, y 1 x 2 + x 1 y 2 ) The last definition looks pretty weird, but we shall shortly see the reason for it. With these definitions of addition and multiplication of complex numbers, we have (x, y) = (x, 0) + (y, 0) EE2007/Ling KV/Aug 2006 8 Thus, setting gives z = (x, y) = x + iy and we have i 2 = (0, 1) (0, 1) =, or i =. Addition and multiplication can now be re-written in the more usual forms: and (x 1 + iy 1 ) + (x 2 + iy 2 ) = + i (x 1 + iy 1 ) (x 2 + iy 2 ) = + i The above derivation of i is firmly based on numbers. There is no about 1, contrary to the use of the historical word. EE2007/Ling KV/Aug 2006 9

Multiplication and Division Using Polar Form It is sometimes more convenient to do multiplication and division in the polar form z 1 z 1 z 2 = r 1 r 2 (θ 1 + θ 2 ), = r 1 θ 1 = z 2 r 2 θ 2 The operations of adding or multiplying two complex numbers could be interpreted geometrically as shown on the right learly, z 1 + z 2 z 1 + z 2 and this is known as the triangular inequality which can be extended to z 1 + z 2 + + z n z 1 + z 2 + + z n EE2007/Ling KV/Aug 2006 10 omplex onjugate Given z = x + iy, the complex conjugate of z is defined as z = x iy Thus, we can write Re (z) = 1 2 (z + z) Im (z) = 1 2i (z z) zz = x 2 + y 2 = z 2 z 1 z 2 = z 1z 2 z 2 2 (z 1 ± z 2 ) = z 1 ± z 2, z 1 z 2 = z 1 z 2, ( z 1 z 2 ) = z 1 z 2 Quiz Time EE2007/Ling KV/Aug 2006 11

Principal Value Let z = 1 + i. Then r = z = 1 + 1 = 2 arg z = arctan 1 1 = π 4 ± 2nπ, n = 0, 1, 2,... The principal value of the argument is π 4. If z = 1 i, then arg z = arctan 1 1 = π 4 The principal value of the argument is π 4. ± 2nπ, n = 0, 1, 2,... EE2007/Ling KV/Aug 2006 12 De Moivre s Formula Let z = x + iy = r(cos θ + i sin θ) = r θ Then, for any integer n, z n = r n (cos θ + i sin θ) n = z.z... z }{{} = r.r... r }{{} n n (θ + θ +... + θ) }{{} n This gives the De Moivre s Formula (cos θ + i sin θ) n = cos nθ + i sin nθ = r n (nθ) which is useful in deriving certain trigonometric identities. EE2007/Ling KV/Aug 2006 13

Example: Express cos 2θ and sin 2θ in terms sin θ and cos θ. (cos θ + i sin θ) 2 = = Therefore cos 2θ = cos 2 θ sin 2 θ, and sin 2θ = 2 cos θ sin θ Example: Express cos 4 θ in terms of multiples of θ. From slide 5(Euler s Formula), we know that 2 cos θ = e iθ + e iθ, 2 4 cos 4 θ = = = 2 cos 4θ + 8 cos 2θ + 6 cos 4 θ = 1 [cos 4θ + 4 cos 2θ + 3]. 8 EE2007/Ling KV/Aug 2006 14 Example: Without using a calculator, compute ( 3 + i) 7 and express your answer in rectangular form. This example shows the usefulness of polar form in simplifying calculation ( 3 + i) 7 = = = = 64( 3 + i) EE2007/Ling KV/Aug 2006 15

Roots of omplex Numbers onsider z = w n, n = 1, 2,... For a given z 0, the solution of w in the above equation is called the nth root of z and is denoted by w = n z Given that z 0, w can be found as follows: Let z = r (θ + 2kπ) and w = R φ. Then, z = w n gives r (θ + 2kπ) = (R φ) n = R n (nφ) Thus we have, R =, and φ =, k = 0, 1,..., (n 1) EE2007/Ling KV/Aug 2006 16 In summary w k = n z = n r ( θ + 2kπ ), k = 0, 1,..., (n 1) n Geometrically, the entire set of roots lie at the vertices of a regular polygon of n sides inscribed in a circle of radius n r. EE2007/Ling KV/Aug 2006 17

Example Let us find all values of ( 8i) 1/3, i.e 3 8i. Solution: In this example, we want to find w = ( 8i) 1/3, i.e. our z = 8i. We write z in polar form 8i = 8 ( π 2 + 2kπ), k = 0, ±1, ±2,... and we see that the desired roots are w k = 2 ( π 6 + 2kπ ), k = 0, 1, 2 3 The roots lie at the vertices of an equilateral triangle, inscribed in the circle z = 2 and are equally spaced around that circle every 2π/3 radians, starting with the principal root w 0 = 2 ( π 6 ) = 3 i EE2007/Ling KV/Aug 2006 18 Example Solve z 2 + (1 + i)z + 5i = 0. [1 2i, 2 + i] EE2007/Ling KV/Aug 2006 19

Is defined as The Exponential Function e z e z = n=0 1 n! zn = e x (cos y + i sin y) If x = 0, we have the so-called Euler formula: e iy = cos y + i sin y. Hence the polar form of a complex number may be written as z = r(cos θ + i sin θ) = re iθ If z = e ix = cos x + i sin x, then sin x = 1 2i (eix e ix ) = 1 (z z), 2i cos x = 1 2 (eix + e ix ) = 1 (z + z) 2 It is also geometrically obvious that e iπ = 1, e iπ/2 = i, and e i4π = 1. EE2007/Ling KV/Aug 2006 20 The omplex Logarithm and General Powers The natural logarithm of z = x + iy is denoted by ln z and is defined as the inverse of the exponential function. That is, w = ln z is defined for z 0 by the relation e w = z. So, if z = re iθ, r > 0, then ln z = ln r + iθ Note that the complex logarithm is infinitely many-valued. The general power of a complex number, z c, can be derived as follows: Let y = z c, ln y = c ln z, y = z c = e c ln z, z 0 EE2007/Ling KV/Aug 2006 21

Examples Evaluate ln(3 4i). Solution: ln(3 4i) = ln 3 4i + i arg(3 4i) = 1.609 i(0.927 ± 2nπ) n = 0, 1,... Principal value: when n = 0. Solve ln z = 2 3 2 i. Solution: z = e 2 3 2 i = e 2 e i3 2 = e 2 (cos 3 2 i sin 3 2 ) = 0.010 i0.135 EE2007/Ling KV/Aug 2006 22 Example: Find the principal value of (1 + i) i. Solution: Let y = (1 + i) i. Then ln y = i ln(1 + i), or y = e i ln(1+i) Hence, (1 + i) i = e i ln(1+i). But, ln(1 + i) = ln( 2e i(π/4+2kπ) ) = ln 2 + i(π/4 + 2kπ), k = 0, ±1,... and the principle value is when k = 0. Therefore, e i ln(1+i) = e i(ln 2+iπ/4) = e π 4 +i(ln 2) EE2007/Ling KV/Aug 2006 23

omplex Functions omplex analysis is concerned with complex functions that are differentiable in some domain. Hence we shall first say what we mean by a complex function and then define the concepts of limits and derivative in complex analogous to calculus. A complex function f is a rule (or mapping) that assigns to every complex number z in S a complex number w in T. Mathematically, we write w = f(z) The set S is called the domain of f and the set T is called the range of f. If z = x + iy and w = u + iv, then we may write w = f(z) = u(x, y) + iv(x, y) EE2007/Ling KV/Aug 2006 24 Example Let w = f(z) = z 2 + 3z. Find u and v and calculate the value of f at z = 1 + 3i. Solution: Let z = x + iy. Then Hence and w = z 2 + 3z = (x + iy) 2 + 3(x + iy) = x 2 y 2 + i2xy + 3x + i3y u = Re (w) = x 2 y 2 + 3x v = Im (w) = 2xy + 3y f(1 + i3) = u(1, 3) + v(1, 3) = 5 + i15. Try using polar form, z = r answer. θ, and check if you get the same EE2007/Ling KV/Aug 2006 25

omplex Functions as Transformations Example: Translation The function w = f(z) = z + c, where c = c 1 + ic 2 is a constant, maps each z = x + iy to w = (x + c 1 ) + i(y + c 2 ). Example: Rotation The function w = f(z) = bz, where b = b β, maps each z = r θ to w = r. b (θ + β). Thus w = bz is a rotation about the origin through β and an enlargement through b. The figure on the right depicts the function w = (1 + i π 3)z = (2 3 )z In this example, we may also write w = (x 3y) + i(y + 3x). EE2007/Ling KV/Aug 2006 26 The Function w = z 2 In polar coordinates, w = z 2 = (r θ) 2 = r 2 (2θ). We can visualize the mapping w = z 2 as shown on the right. To find the image of line Re (z) = 1 under the mapping f(z) = z 2, we use the rectangular coordinates: w = (x + iy) 2 = (x 2 y 2 ) + i2xy = u(x, y) + iv(x, y) Now Re (z) = x = 1. So, we get u = 1 y 2, and v = 2y which are parametric equations of a curve in the w-plane. To see the curve, eliminate y to get u = 1 v 2 /4. EE2007/Ling KV/Aug 2006 27

Limit A function f(z) is said to have the limit L as z approaches a point z o if 1. f(z) is defined in the neighbourhood of z o (except perhaps at z o itself), and 2. f(z) approaches the same complex number L as z z o from all directions within its neighourhood. Mathematically, we write lim f(z) = L z z o if given ɛ > 0, there exists δ > 0, such that f(z) L < ɛ, 0 < z z o < δ In words, the above means that the point f(z) can be made arbitrarily close to the point L if we choose the point z sufficiently close to, but not equal to, the point z o. EE2007/Ling KV/Aug 2006 28 Let f(z) = Example: [Limit] xy x 2 +y 2 + ixy. Does lim x 0,y 0 f(z) exist? Solution: Let y = kx where k is an arbitrary constant. Then kx 2 lim f(z) = lim x 0,y 0 x 0,y=kx x 2 + k 2 x 2 + ikx2 = lim x 0 = k 1 + k 2 k 1 + k 2 + ikx2 The limit depends on k (the direction in which x, y approach zero), hence the limit does not exist. EE2007/Ling KV/Aug 2006 29

Does lim ( z z 0 z z z ) Example: [Limit] exist? Solution: Let z = re iθ. Then z 0 when r 0. (re iθ lim f(z) = lim z 0 r 0 re iθ = e i2θ e i2θ So, the limit does not exist. reiθ re iθ) EE2007/Ling KV/Aug 2006 30 ontinuity A function f(z) is said to be continuous at z = z o if 1. f(z o ) exists, 2. lim z zo f(z) exists, and 3. lim z zo f(z) = f(z o ) Note: Just writing statement (3) implies the truth of (1) and (2). We say that f is a continuous function if f is continuous for all z in the domain S. Example f(z) = z2 1 z 1 is not continuous at z = 1 as it is not defined there. Question: What is lim z 1 z 2 1 z 1? EE2007/Ling KV/Aug 2006 31

Example: Let f(0) = 0, and for z 0, f(z) = Re(z 2 )/ z 2. Determine whether f(z) is continuous at the origin. Solution: lim z 0 Re(z2 )/ z 2 x 2 y 2 = lim z 0 x 2 + y 2 = Hence, f is not continuous at the origin. Alternatively, { 1 if y = 0 1 if x = 0 lim z 0 Re(z2 )/ z 2 r 2 cos 2θ = lim r 0 r 2 = cos 2θ The limit does not exists because it depends on the direction of approach to the origin. EE2007/Ling KV/Aug 2006 32 Example Is f(z) = Im (z) 1 + z z continuous at z = 0? Solution: f(z = 0) = First, the value of f(z = 0) is Im (0) 1 + 0 = 0 Second, let z = re iθ. Then z 0 when r 0. Im (re iθ ) lim f(z) = lim z 0 r 0 1 + re iθ r iθ = lim r 0 r sin θ 1 + r 2 = 0 = f(0) So, f(z) is continuous at z = 0. EE2007/Ling KV/Aug 2006 33

Example Is f(z) = { zre (z) z, z 0 0, z = 0 continuous at z = 0? Solution: Let z = re iθ. Then z 0 when r 0. lim z 0 = lim r 0 re iθ Re (re iθ ) re iθ = lim r 0 re iθ r cos θ re iθ = lim r 0 re i2θ cos θ = 0 = f(z = 0) So, f(z) is continous at z = 0. Quiz Time EE2007/Ling KV/Aug 2006 34 Derivatives of omplex Functions The derivative of a complex function f at a point z o is written as f (z o ) and is defined as f (z o ) = lim z zo f(z) f(z o ) z z o or, by substituting z = z o + z Example f f(z o + z) f(z o ) (z o ) = lim z 0 z d (z + z) 2 z 2 dz (z2 ) = lim z 0 z Thus f(z) = z 2 is differentiable for all z. if the limit exists. = lim (2z + z) = 2z. z 0 EE2007/Ling KV/Aug 2006 35

The usual differentiation formulae hold as for real variables. For example d d d (c) = 0, (z) = 1, dz dz dz (zn ) = nz n 1 and d dz (2z2 + i) 5 = 5(2z 2 + i) 4 4z = 20z(2z 2 + i) 4 However, care is required for more unusual functions. For example, f(x) = x 2 = x 2 f (x) = 2x x but f(z) = z 2 f (z) exists only at z = 0 which we shall show in the next example. EE2007/Ling KV/Aug 2006 36 Differentiability of f(z) = z 2 f z + z 2 z 2 (z) = lim z 0 z = lim z 0 (z + z)(z + z) zz z = lim (z + z + z z z 0 z ) = lim r 0 (z + re iθ + z re iθ re iθ ) = z + ze 2iθ For the limit to exist, it must be independent of θ. So f (z) exists only when z = 0. EE2007/Ling KV/Aug 2006 37

Example Discuss the differentiability of z. Solution: Again, with more unusual functions, we begin from the definition. Let f(z) = z, then f z + z z (z) = lim z 0 z z = lim z 0 z Now, consider z = re iθ. Then z 0 from all directions when r 0. Thus, we could determine the limit as follows: f z (z) = lim z 0 z = lim re iθ r 0 re iθ = e i2θ The limit depends on θ so it doesn t exist. Hence, f(z) = z is not differentiable anywhere. EE2007/Ling KV/Aug 2006 38 Proof: Write f(z) as Differentiability ontinuity f(z) = f(z) f(z o) z z o (z z o ) + f(z o ) Then, apply limit on the above equation gives lim f(z) = lim z 0 z 0 f(z) f(z o ) (z z o ) + lim f(z o ) z z o z 0 If f is differentiable, the derivative of f at z o exists, i.e. So f(z) f(z o ) lim = f (z o ) z z o z z o lim f(z) = f (z o ) 0 + f(z o ) lim f(z) = f(z o ) z z o z zo i.e. f(z) is continuous at z o if f is differentiable. EE2007/Ling KV/Aug 2006 39

ontinuity Differentiability Note that the converse is not true. For example, f(z) = z is not differentiable at all (see page 38) but continuous everywhere in the complex plane. Also, as shown in page 37, f(z) = z 2 is only differentiable at z = 0 but it is continuous everywhere. an you show that f(z) = z 2 is continuous everywhere? Quiz Time EE2007/Ling KV/Aug 2006 40 The auchy-riemann Equations So far, we determined the differentiability of complex functions and showed how to compute their derivatives (if exists) by using the definition f (z o ) = lim z 0 f(z o + z) f(z o ) z. Often, we encounter complex function written in the form f(z) = u(x, y) + iv(x, y), e.g. f(z) = (x 3 3xy 2 ) + i(3x 2 y y 3 ) Is there other ways to determine whether f(z) is differentiable, and if so, to find the derivative f (z)? The answer is yes, thanks to the French mathematician A.L. auchy 2 and the German mathematician G.F.B. Riemann. 2 auchy is not pronounced as kaushee, but kōshē. (see Mathews and Howell, p.102) EE2007/Ling KV/Aug 2006 41

Derivation of the -R Equations Recall that the derivatives of a complex function f at a point z o is defined as f (z) = lim z 0 f(z+ z) f(z) z = lim x, y 0 u(x+ x,y+ y)+iv(x+ x,y+ y) u(x,y) iv(x,y) x+i y onsider along the x-axis, i.e. y = 0, we have f (z) = lim x 0 u(x+ x,y)+iv(x+ x,y) u(x,y) iv(x,y) x = lim x 0 u(x+ x,y) u(x,y)+i(v(x+ x,y) v(x,y)) x = u x + i v x EE2007/Ling KV/Aug 2006 42 Similarly, along y-axis, i.e. x = 0, we have f u(x,y+ y)+iv(x,y+ y) u(x,y) iv(x,y) (z) = lim y 0 i y u(x,y+ y) u(x,y)+i(v(x,y+ y) v(x,y)) = lim y 0 i y = i u y + v y For the derivative to exist, the two limits must agree, i.e. u x = v y, v x = u y Thus the -R equations are u x = v y and v x = u y. It can be shown that, when z 0, the -R equations in polar coordinates are 3 u r = 1 r v θ and v r = 1 r u θ 3 see, for example, Brown and hurchill (1996), pp.52 53. EE2007/Ling KV/Aug 2006 43

Derivatives of omplex Functions If f(z) = u(x, y) + iv(x, y) and f (z) exists, then f (z) = u x + iv x = v y iu y = u x iu y = v y + iv x In polar form, if f(z) = u(r, θ) + iv(r, θ), and f (z) exists, then f (z) = e iθ (u r + iv r ) = 1 r e iθ (v θ iu θ ) EE2007/Ling KV/Aug 2006 44 -R onditions for Differentiability 4 Theorem 1. Let the complex function f(z) = u(x, y) + iv(x, y) be a continuous function that is defined in some neighbourhood of the point z o = x o + iy o. If all the partial derivatives u x, u y, v x and v y are continuous at the point (x o, y o ) and if the -R equations u x = v y and v x = u y are satisfied, the f(z) is differentiable at z o and the derivative f (z o ) can be computed with formulas given in p. 44 4 Mathews and Howell, p.106 EE2007/Ling KV/Aug 2006 45

Example Show that the function f(z) = (x 3 3xy 2 ) + i(3x 2 y y 3 ) is differentiable for all z and find its derivative. Solution: In this case, we have u(x, y) = x 3 3xy 2 and v(x, y) = 3x 2 y y 3. We then calculate u x = 3x 2 3y 2 = v y and u y = 6xy = v x So the -R equations are satisfied. Moreover, u, v, u x, u y, v x and v y are continuous everywhere. Hence, f is differentiable everywhere and f (z) = u x + iv x = 3x 2 3y 2 + i6xy = 3(x 2 y 2 + i2xy) = 3z 2 This result isn t surprising because (x + iy) 3 = x 3 3xy 2 + i(3x 2 y y 3 ) and so the function f is really our old friend f(z) = z 3. EE2007/Ling KV/Aug 2006 46 Analytic Functions A function f(z) is said to be analytic at a point z o if its derivative exists not only at z o but also at each point z in some neighbourhood of z o. A function f(z) is said to be analytic in a domain D if it is analytic at each point in D. Hence analyticity implies differentiability and continuity. The point z = z o where f(z) ceases to be analytic is called the singular point or singularity 5 of f(z). Examples (see p. 35 and 38) f(z) = z 2 is analytic everywhere in the complex plane. f(z) = z is NOT analytic at any point. 5 We shall restrict ourselves to only singular point called pole of order m. The general classification of singularity which requires background in Laurent s Series will be omitted due to time constraint. EE2007/Ling KV/Aug 2006 47

A Test for Analyticity Theorem 2. The complex function f(z) = u(x, y) + iv(x, y) is analytic at a point z o if for every point in the neighbourhood of z o 1. u, v and their partial derivatives exist and are continuous, and 2. the auchy-riemann equations u x = v y and v x = u y are satisfied. If the above two conditions are satisfied in some domain D, then the function is analytic in D. EE2007/Ling KV/Aug 2006 48 Example: Verify that f(z) = z is not analytic. Solution: We have seen from p. 38 that z is not differentiable, hence z is not analytic. Alternatively, using -R equations with u(x, y) = x, v(x, y) = y we see that u y = v x = 0, but u x = 1 v y = 1 -R equations are not satisfied, hence the function is not analytic. EE2007/Ling KV/Aug 2006 49

Example: Is f(z) = z 3 analytic? Solution: In general, polynomials of complex variables are analytic (see p. 52). Here we show using -R equations. Given f(z) = z 3, u(r, θ) = r 3 cos 3θ, v(r, θ) = r 3 sin 3θ Therefore u r = 3r 2 cos 3θ, u θ = 3r 3 sin 3θ v r = 3r 2 sin 3θ, and v θ = 3r 3 cos 3θ Since -R equations u r = 1 r v θ and v r = 1 r u θ are satisfied and the functions u, v and their partial derivatives are continuous, z 3 is analytic. EE2007/Ling KV/Aug 2006 50 Examples: Discuss the analyticity of the function f(z) = x 2 + iy 2. Solution: With u = x 2, v = y 2, we have u x = 2x, v y = 2y v x = 0, u y = 0 Thus, from -R equations, f(z) is differentiable only for those values of z that lie along the straight line x = y. If z o lies on this line, any circle centered at z o will contain points for which f (z) does not exist. Thus f(z) is nowhere analytic. EE2007/Ling KV/Aug 2006 51

Some ommon (and Important) Functions Polynomials, i.e., functions of the form f(z) = c o + c 1 z + c 2 z 2 +... + c n z n where c o,..., c n are complex constants, are analytic in the entire complex plane. Rational functions, i.e. quotient of two polynomials f(z) = g(z) h(z) are analytic except at points where h(z) = 0. Partial fractions of the form f(z) = c (z z o ) m where c and z o complex, m a positive integer are analytic except at z o. Quiz Time EE2007/Ling KV/Aug 2006 52 In the case of a real definite integral omplex Integration b a f(x) dx means we integrate along the x-axis from a to b, and the integrand f(x) is defined for each point between a and b. In the case of a complex definite integral, or line integral, f(z) dz, we integrate along curve, in a given direction, in the complex plane and the integrand is defined for each point on and is called the contour or path of integration. If is a closed contour, we sometimes denote the complex line integral by f(z) dz If is on the real axis, then z = x and the complex integral becomes a real definite integral. EE2007/Ling KV/Aug 2006 53

Parametric Representation of a Path or ontour A contour or path of integration on the complex plane can be represented in the form of z(t) = x(t) + iy(t), a t b where t is a real parameter. This establishes a continuous mapping of the interval a t b into the xy- or z-plane, and the direction of the path is according to the increasing values of t. Example: The path in the figure on the right can be represented by { x + ix, 0 x 1 z = x + i, 1 x 2 EE2007/Ling KV/Aug 2006 54 omplex Line Integration Example: Evaluate z dz where is given by x = 3t, y = t 2, 1 t 4. Solution: Since z = x + iy, we write z(t) = 3t + it 2, which gives dz(t) = (3 + i2t)dt and hence z dz = = 4 1 4 1 (3t it 2 )(3 + i2t) dt (2t 3 + 9t) dt + i 4 1 3t 2 dt = 195 + i65 EE2007/Ling KV/Aug 2006 55

Example: Evaluate 1 z dz where is the unit circle in the complex plane, counter-clockwise. Solution: The path can be represented by z(t) = cos t + i sin t = e it, 0 t 2π and dz(t) = ie it dt = izdt. Hence 1 2π z dz = i dt = 2πi 0 EE2007/Ling KV/Aug 2006 56 Example 6 : Evaluate (z z o) m dz where is a W circle of radius ρ with centre at z o. The path is represented as z(θ) = z o + ρe iθ, 0 θ 2π. Then (z z o ) m = ρ m e imθ, dz = iρe iθ dθ Hence, = (z z o ) m dz = 2π { 2πi (m = 1) 0 (m 1, m integer) 0 2π ρ m e imθ iρe iθ dθ = iρ m+1 e i(m+1)θ dθ 0 6 This is a generalization of the example on pg 56. EE2007/Ling KV/Aug 2006 57

Integration by the use of the path From the previous examples, we arrive at a practical means of evaluating a complex line integral: Theorem 3. [Integration by the use of the path] Let be a piecewise smooth path, represented by z = z(t), where a t b. Let f(z) be a continuous function on. Then Proof: See Text f(z) dz = b a f[z(t)] dz dt dt EE2007/Ling KV/Aug 2006 58 Basic Properties of omplex Line Integrals 1. Linearity [k 1 f 1 (z) + k 2 f 2 (z)] dz = k 1 f 1 (z) dz + k 2 f 2 (z) dz 2. Subdivision of path f(z) dz = f(z) dz + f(z) dz 1 2 3. Sense of integration z2 z 1 f(z) dz = z1 z 2 f(z) dz EE2007/Ling KV/Aug 2006 59

Example: Evaluate 1+i 0 Re z dz along (a), (b) 1 and 2. 1. can be represented by z(t) = t + it, 0 t 1 which gives dz = (1 + i)dt. Hence, 1+i 0 Re z dz = 1 0 t(1 + i) dt = 1 (1 + i) 2 2. 1 and 2 are represented by 1 : z(t) = t, 0 t 1 giving dz = dt 2 : z(t) = 1 + it, 0 t 1 giving dz = idt Along 1, Re (z) = t and along 2, Re (z) = 1. Hence, 1+i 0 Re z dz = 1 + 1 = 2 0 t dt + 1 0 1 i dt = 1 2 + i EE2007/Ling KV/Aug 2006 60 Example: Evaluate 1+i 0 z dz along (a), (b) 1 and 2. 1. Along, z is represented by z(t) = t + it, 0 t 1 which gives dz = (1 + i) dt. Hence, 2. 1+i 0 z dz = = 1 0 1 0 (t + it)(1 + i) dt (t t + i2t) dt = it 2 1 0 = i Along 1 : z(t) = t, 0 t 1 and dz = dt Along 2 : z(t) = 1 + it, 0 t 1 and dz = idt Hence, 1+i 0 z dz = 1 + 1 = 2 0 t dt + 1 0 (1 + it) i dt = i EE2007/Ling KV/Aug 2006 61

We have seen from the examples that a line integral of f(z) in the complex plane may or may not depend on the choice of the path itself. Sometimes, the integrals evaluated to the value of zero or 2πi. Under what condition will the integral be independent of the path? Under what condition will the integral be zero? Is there something special about the value 2πi? To answer these questions, we need additional background: oncept of Simple losed Path and Simply onnected Domain auchy s Integral Theorem But, before we continue, let s pause for a quiz. For this quiz, ignore question 4. Quiz Time EE2007/Ling KV/Aug 2006 62 Simple losed Path and Simply onnected Domain A simple closed path is a closed path that does not intersect or touch itself. A simply connected domain D in the complex plane is a domain such that every simple closed path in D encloses only points of D. A domain that is not simply connected is called multiply connected. EE2007/Ling KV/Aug 2006 63

auchy s Integral Theroem Theorem 4. [auchy s Integral Theorem] If f(z) is analytic in a simply connected domain D, then for every simple closed path in D, f(z) dz = 0. Example: e z dz = 0, cos z dz = 0 and z n dz = 0, n = 0, 1,... for any closed path since these functions are entire (analytic for all z). Example: 1 z 2 dz = 0, + 4 : unit circle although the integrand is not analytic at z = ±2i, these points are not enclosed by. EE2007/Ling KV/Aug 2006 64 Independence of Path Now we are ready to answer the question set out on p. 62. Theorem 5. [Independence of Path] If f(z) is analytic in a simply connected domain D, then the integral of f(z) is independent of the path in D. Proof: Let z 1 and z 2 be any point in D. onsider two paths 1 and 2 in D from z 1 to z 2 as shown in the figure below. Denote by 2 the path 2 with the direction reversed. From auchy s theorem, we have Thus f dz + 1 f dz = 1 2 2 f dz = 0 f dz = f dz 2 EE2007/Ling KV/Aug 2006 65

Indefinite Integration of Analytic Functions From auchy s Integral Theorem, we have a simpler, albeit restricted method to evaluate omplex Line Integral: Theorem 6. [Indefinite Integration of Analytic Functions] Let f(z) be analytic in a simply connected domain D. Then there exists an analytic function F (z) such that F (z) = f(z) in D and for all path in D joining two points z o and z 1 in D, we have z1 f(z) dz = F (z 1 ) F (z o ) z o Example: Example: 1+i 0 z 2 dz = 1 3 z3 1+i 0 = 1 3 (1 + i)3 = 2 3 + 2 3 i πi πi cos z dz = sin z πi πi = 2 sin πi EE2007/Ling KV/Aug 2006 66 auchy s Integral Formula The most important consequence of auchy s integral theorem is auchy s integral formula. This formula is useful for evaluating integrals of the form f(z) dz m = 1, 2, 3... (z z o ) m Theorem 7. [auchy s Integral Formula] Let f(z) be analytic in a simply connected domain D. Then for any point z o in D and any simple closed path in D that encloses z o f(z) dz = 2πif(z o ) z z o In general, f(z) 2πi dz = (z z o ) m (m 1)! f (m 1) (z o ) m = 1, 2, 3,... The integration being taken W. (See text for proof.) EE2007/Ling KV/Aug 2006 67

Examples (auchy s Integral Formula) e z z 2 dz = 2πiez z=2 = 2πie 2 for any simple closed path enclosing z o = 2. Evaluate Solution: = z 3 6 2z i dz, Since encloses z = i 2 : unit circle, W which is not analytic, z 3 6 2(z i/2) dz = πi(z3 6) z=i/2 = πi( i 8 6) EE2007/Ling KV/Aug 2006 68 Evaluate cos z dz where is any simple closed path enclosing (z πi) 2 z = πi, W. Solution: = 2πi d dz cos z z=πi = 2πi sin(πi) Evaluate z 4 3z 2 +6 dz where is any simple closed path (z+i) 3 enclosing z = i, W. Solution: = 2πi d 2 2! dz 2(z4 3z 2 + 6) z= i = πi(12z 2 6) z= i = 18πi. EE2007/Ling KV/Aug 2006 69

Evaluate 1 z 2 + 1 dz, : z = 3, W Solution: The integrand is not analytic at z = ±i which are inside. auchy s formula only applies to one singular point inside. Use partial fraction decomposition and apply auchy s formula dz z 2 + 1 = dz (z + i)(z i) = 1 1 [ 2i z i 1 z + i ] dz = 1 [2πi 2πi] = 0 2i Quiz Time EE2007/Ling KV/Aug 2006 70 auchy s Theorem for Multiply onnected Domains Suppose, 1,..., n are simple closed curves with a positive orientation such that 1, 2,..., n are interior to but regions interior to each k, k = 1, 2,..., n, have no points in common. If f is analytic on each contour and at each point interior to but exterior to all the k, k = 1, 2,..., n, then f(z) dz = n k=1 k f(z) dz EE2007/Ling KV/Aug 2006 71

Example 7 : Evaluate dz z 2 +1 where is the circle z = 3. 1 Solution: The integrand is not analytic at z = ±i. Both of these z 2 +1 points lie within the contour. Introduce 1 and 2 as shown in the figure to exclude these points and using the auchy s Theorem to this multiply connected domain dz z 2 + 1 = dz (z + i)(z i) 1/(z + i) 1/(z i) = dz + dz 1 z i 2 z + i 1 = 2πi[ z + i ] 1 z=i + 2πi[ z i ] z= i = 2πi 1 2i + 2πi 1 2i = 0 7 This is the same example in pg 70 where we solved using partial fraction. Here we shall apply auchy s Theorem to Multiply onnected Domains. EE2007/Ling KV/Aug 2006 72 Evaluation of Real Integrals 2π 0 F (cos θ, sin θ) dθ where F (cos θ, sin θ) is a real function of cos θ and sin θ and is finite on the interval of integration. Basic Idea: Let z = e iθ, we have cos θ = z + z 2, sin θ = z z 2i and dz = ie iθ dθ dθ = 1 iz dz This allows us to convert F (cos θ, sin θ) into f(z) and the integration interval of 0 θ 2π is changed to the unit circle. Thus 2π 0 F (cos θ, sin θ) dθ = f(z) 1 iz dz, : unit circle, W EE2007/Ling KV/Aug 2006 73

Example: Evaluate 2π 0 dθ 2 cos θ Solution: Let z = e iθ. Substitute cos θ = 1 2 (z + 1 dz ) and dθ = z iz the real integral becomes dz/iz 2 1 2 (z + 1 z ) = 2 i where is the W unit circle. dz (z ( 2 + 1))(z ( 2 1)) The integrand has simple pole at z = 2 1 inside and z = 2 + 1 outside. Hence, using auchy s Integral Formula, the integral is 2 i dz (z ( 2 + 1))(z ( 2 1)) = 2 i = 2 i (2πi) 1 z ( 2 + 1) z= 2 1 = 2π 1 z ( 2+1) z ( 2 1) dz EE2007/Ling KV/Aug 2006 74 Improper Integrals of Rational Functions f(x) dx = UHP f(z) dz = n k=1 c k in UHP f(z) dz, if 1. f(x) = p(x) q(x) is a real function with p(x) and q(x) no common factors and q(x) 0 for all real x. 2. deg(q(x)) deg(p(x)) + 2. E.g. f(x) = 1 satisfies the above conditions but f(x) = 1+x 4 x3 does not. 1+x 4 EE2007/Ling KV/Aug 2006 75

Proof: : Basic Idea onsider the corresponding contour integral f(z) dz where is the Upper Half Plane (UHP) contour shown on the right R Then f(z) dz = f(z) dz + f(x) dx S R Since f(z) is rational, f(z) has finitely many poles in the UHP. If we choose R large enough, the contour will enclose all these poles. In addition, due to assumption 2, it can be shown that lim R f(z) dz = 0. S Hence, f(x) dx = f(z) dz UHP EE2007/Ling KV/Aug 2006 76 Example: Show that dx = π 0 1+x 4 2. 2 Solution: First, check that f(x) satisfies our assumptions. Now, consider f(z) = 1 which has four simple poles at z = e πi/4, e 3πi/4, e 3πi/4, e πi/4. 1+z 4 Only the first two poles, e πi/4 and e 3πi/4, lie inside the UHP. The corresponding complex integral is 1 1 + z 4 dz = 1 (z e πi/4 )(z e 3πi/4 )(z e 3πi/4 )(z e πi/4 ) dz UHP = c 1 UHP 1 (z e 3πi/4 )(z e 3πi/4 )(z e πi/4 ) z e πi/4 dz + = 2πi( 1 4 eπi/4 + 1 4 e πi/4 ) 1 Since is even, 1+x 4 0 dx 1 + x 4 = 1 2 c 2 1 (z e πi/4 )(z e 3πi/4 )(z e πi/4 ) dz z e 3πi/4 dx 1 + x 4 = πi 4 (eiπ/4 e iπ/4 ) = π 2 sin π 4 = π 2 2 EE2007/Ling KV/Aug 2006 77