9.4 Exercises Exercise 9.1 Consider two functions, f(x, y) and g(x, y). They are said to be functionally dependent if there is a an h(g) such that f(x, y) = h(g(x, y)). f and g will be functionally dependent if and only if their Jacobian vanishes. If f and g are functionally dependent, then the derivatives of f are f x = h  (g)g x f y = h  (g)g y . Thus we have ∂(f, g) ∂(x, y) =     f x f y g x g y     = f x g y − f y g x = h  (g)g x g y − h  (g)g y g x = 0. If the Jacobian of f and g vanishes, then f x g y − f y g x = 0. This is a first order partial differential equation for f that has the general solution f(x, y) = h(g(x, y)). Prove that an analytic function u(x, y) + ıv(x, y) can be written in terms of a function of a complex variable, f(z) = u(x, y) + ıv(x, y). Exercise 9.2 Which of the following functions are the real part of an analytic function? For those that are, find the harmonic conjugate, v(x, y), and find the analytic function f(z) = u(x, y) + ıv(x, y) as a function of z. 1. x 3 − 3xy 2 − 2xy + y 2. e x sinh y 454 3. e x (sin x cos y cosh y −cos x sin y sinh y) Exercise 9.3 For an analytic function, f(z) = u(r, θ) + ıv(r, θ) prove that under suitable restrictions: f(z) = 2u  z 1/2 , − ı 2 log z  + const. 455 9.5 Hints Hint 9.1 Show that u(x, y) + ıv(x, y) is functionally dependent on x + ıy so that you can write f(z) = f(x + ıy) = u(x, y) + ıv(x, y). Hint 9.2 Hint 9.3 Check out the derivation of Equation 9.2. 456 9.6 Solutions Solution 9.1 u(x, y) + ıv(x, y) is functionally dependent on z = x + ıy if and only if ∂(u + ıv, x + ıy) ∂(x, y) = 0. ∂(u + ıv, x + ıy) ∂(x, y) =     u x + ıv x u y + ıv y 1 ı     = −v x − u y + ı (u x − v y ) Since u and v satisfy the Cauchy-Riemann equations, this vanishes. = 0 Thus we see that u(x, y) + ıv(x, y) is functionally dependent on x + ıy so we can write f(z) = f(x + ıy) = u(x, y) + ıv(x, y). Solution 9.2 1. Consider u(x, y) = x 3 − 3xy 2 − 2xy + y. The Laplacian of this function is ∆u ≡ u xx + u yy = 6x −6x = 0 Since the function is harmonic, it is the real part of an analytic function. Clearly the analytic function is of the form, az 3 + bz 2 + cz + ıd, 457 with a, b and c complex-valued constants and d a real constant. Substituting z = x + ıy and expanding products yields, a  x 3 + ı3x 2 y − 3xy 2 − ıy 3  + b  x 2 + ı2xy − y 2  + c(x + ıy) + ıd. By inspection, we see that the analytic function is f(z) = z 3 + ız 2 − ız + ıd. The harmonic conjugate of u is the imaginary part of f(z), v(x, y) = 3x 2 y − y 3 + x 2 − y 2 − x + d. We can also do this problem with analytic continuation. The derivatives of u are u x = 3x 2 − 3y 2 − 2y, u y = −6xy − 2x + 1. The derivative of f(z) is f  (z) = u x − ıu y = 3x 2 − 2y 2 − 2y + ı(6xy − 2x + 1). On the real axis we have f  (z = x) = 3x 2 − ı2x + ı. Using analytic continuation, we see that f  (z) = 3z 2 − ı2z + ı. Integration yields f(z) = z 3 − ız 2 + ız + const 458 2. Consider u(x, y) = e x sinh y. The Laplacian of this function is ∆u = e x sinh y + e x sinh y = 2 e x sinh y. Since the function is not harmonic, it is not the real part of an analytic function. 3. Consider u(x, y) = e x (sin x cos y cosh y − cos x sin y sinh y). The Laplacian of the function is ∆u = ∂ ∂x ( e x (sin x cos y cosh y − cos x sin y sinh y + cos x cos y cosh y + sin x sin y sinh y)) + ∂ ∂y ( e x (−sin x sin y cosh y − cos x cos y sinh y + sin x cos y sinh y − cos x sin y cosh y)) = 2 e x (cos x cos y cosh y + sin x sin y sinh y) −2 e x (cos x cos y cosh y + sin x sin y sinh y) = 0. Thus u is the real part of an analytic function. The derivative of the analytic function is f  (z) = u x + ıv x = u x − ıu y From the derivatives of u we computed before, we have f(z) = ( e x (sin x cos y cosh y − cos x sin y sinh y + cos x cos y cosh y + sin x sin y sinh y)) − ı ( e x (−sin x sin y cosh y − cos x cos y sinh y + sin x cos y sinh y − cos x sin y cosh y)) Along the real axis, f  (z) has the value, f  (z = x) = e x (sin x + cos x). By analytic continuation, f  (z) is f  (z) = e z (sin z + cos z) 459 We obtain f(z) by integrating. f(z) = e z sin z + const. u is the real part of the analytic function f(z) = e z sin z + ıc, where c is a real constant. We find the harmonic conjugate of u by taking the imaginary part of f . f(z) = e x (cosy + ı sin y)(sin x cosh y + ı cos x sinh y) + ıc v(x, y) = e x sin x sin y cosh y + cos x cos y sinh y + c Solution 9.3 We consider the analytic function: f(z) = u(r, θ) + ıv(r, θ). Recall that the complex derivative in terms of polar coordinates is d dz = e −ıθ ∂ ∂r = − ı r e −ıθ ∂ ∂θ . The Cauchy-Riemann equations are u r = 1 r v θ , v r = − 1 r u θ . We differentiate f(z) and use the partial derivative in r for the right side. f  (z) = e −ıθ (u r + ıv r ) We use the Cauchy-Riemann equations to right f  (z) in terms of the derivatives of u. f  (z) = e −ıθ  u r − ı 1 r u θ  (9.6) Now consider the function of a complex variable, g(ζ): g(ζ) = e −ıζ  u r (r, ζ) − ı 1 r u θ (r, ζ)  = e ψ−ıξ  u r (r, ξ + ıψ) − ı 1 r u θ (r, ξ + ıψ)  460 This function is analytic where f(ζ) is analytic. It is a simple calculus exercise to show that the complex derivative in the ξ direction, ∂ ∂ξ , and the complex derivative in the ψ direction, −ı ∂ ∂ψ , are equal. Since these partial derivatives are equal and continuous, g(ζ) is analytic. We evaluate the function g(ζ) at ζ = −ı log r. (Substitute θ = −ı log r into Equation 9.6.) f   r e ı(−ı log r)  = e −ı(−ı log r)  u r (r, −ı log r) −ı 1 r u θ (r, −ı log r)  rf   r 2  = u r (r, −ı log r) −ı 1 r u θ (r, −ı log r) If the expression is non-singular, then it defines the analytic function, f  (z), on a curve. The analytic continuation to the complex plane is zf   z 2  = u r (z, −ı log z) −ı 1 z u θ (z, −ı log z). We integrate to obtain an expression for f (z 2 ). 1 2 f  z 2  = u(z, −ı log z) + const We make a change of variables and solve for f(z). f(z) = 2u  z 1/2 , − ı 2 log z  + const. Assuming that the above expression is non-singular, we have found a formula for writing the analytic function in terms of its real part, u(r, θ). With the same method, we can find how to write an analytic function in terms of its imaginary part, v(r, θ). 461 Chapter 10 Contour Integration and the Cauchy-Goursat Theorem Between two evils, I always pick the one I never tried before. - Mae West 10.1 Line Integrals In this section we will recall the definition of a line integral in the Cartesian plane. In the next section we will use this to define the contour integral in the complex plane. Limit Sum Definition. First we develop a limit sum definition of a line integral. Consider a curve C in the Cartesian plane joining the points (a 0 , b 0 ) and (a 1 , b 1 ). We partition the curve into n segments with the points (x 0 , y 0 ), . . . , (x n , y n ) where the first and last points are at the endpoints of the curve. We define the differences, ∆x k = x k+1 − x k and ∆y k = y k+1 − y k , and let (ξ k , ψ k ) be points on the curve between (x k , y k ) and (x k+1 , y k+1 ). This is shown pictorially in Figure 10.1. 462 [...]... |dθ| = −dθ 1 0 z 2 dz = C e 2 ı eıθ dθ π 0 ı e 3 dθ = π = 1 3 e 3 0 π 1 ı0 e − e 3 = 3 1 = (1 − (−1)) 3 2 = 3 4 83 2 0 | e 2 |ı eıθ dθ 2 |z | dz = π C 0 ı eıθ dθ = = π 0 eıθ π = 1 − (−1) =2 3 0 e 2 |ı eıθ dθ| z 2 |dz| = C π 0 − e 2 dθ = π ı 2 0 e 2 π ı = (1 − 1) 2 =0 = 4 0 |z 2 | |dz| = C | e 2 ||ı eıθ dθ| π 0 −dθ = π = [−θ]0 π =π 484 Solution 10 .2 ∞ e−(ax I= 2 +bx) dx −∞ First we complete... ∞ ∞ 1 1 2 2 +ı − e−ax ıω eıωx I = −ı − e−ax eıωx 2a 2a −∞ −∞ ∞ ω 2 e−ax eıωx dx I= 2a −∞ 486 dx ∞ 2 x e−ax sin(ωx) dx = 2 0 ω 2a π − 2 /(4a) e a Solution 10.4 1 We parameterize the contour and do the integration z − z0 = eıθ , θ ∈ [0 2 ) 2 eınθ ı eıθ dθ (z − z0 )n dz = C = 0   2 eı(n+1)θ n+1 0 [ıθ ]2 0 for n = −1 = for n = −1 0 2 for n = −1 for n = −1 2 We parameterize the contour and do the... integrand vanishes and the lengths of the paths of integration are finite Taking the limit as R → ∞ we have, ∞+b/(2a) ∞ 2 e−az dz ≡ −∞+b/(2a) −∞ ∞ 2 /(4a) 2 e−ax dx I = eb We make the change of variables ξ = −∞ ax 2 /(4a) I = eb ∞ e−(ax 1 √ a 2 +bx) ∞ −∞ dx = −∞ 485 2 e−ξ dξ π b2 /(4a) e a 2 e−ax dx −∞ Now we have √ ∞ 2 e−a(x+b/(2a)) dx = Solution 10 .3 Consider ∞ 2 e−ax cos(ωx) dx I =2 0 Since the integrand... the integration z − z0 = 2 + eıθ , 2 n 2 + eıθ (z − z0 ) dz = C = 0   ( 2+ eıθ )n+1  n ı eıθ dθ 2 for n = −1 n+1   log 2 + θ ∈ [0 2 ) 0 2 eıθ 0 for n = −1 3 We parameterize the contour and do the integration z = r eıθ , r = 2 − sin2 487 θ 4 , θ ∈ [0 4π) =0 1 -1 1 -1 Figure 10 .2: The contour: r = 2 − sin2 θ 4 The contour encircles the origin twice See Figure 10 .2 4π z −1 dz = C 0 4π =... (z(t))z (t) dt t0 Example 10 .2. 1 Let C be the positively oriented unit circle about the origin in the complex plane Evaluate: 1 C z dz 2 1 C z dz 3 1 C z |dz| 465 In each case we parameterize the contour and then do the integral 1 dz = ı eıθ dθ z = eıθ , 2 eıθ ı eıθ dθ z dz = 0 C 2 1 2 e = 2 0 1 ı4π 1 ı0 e − e = 2 2 =0 2 C 2 1 dz = z 0 1 ıθ ı e dθ = ı eıθ 2 dθ = 2 0 3 |dz| = ı eıθ dθ = ı eıθ |dθ|... ∞ 2 e−ax cos(ωx) dx I= −∞ Since 2 e−ax sin(ωx) is an odd function, ∞ 2 e−ax eıωx dx I= −∞ We evaluate this integral with the result of Exercise 10 .2 ∞ 2 e−ax cos(ωx) dx = 2 0 Consider ∞ π − 2 /(4a) e a 2 x e−ax sin(ωx) dx I =2 0 Since the integrand is an even function, ∞ 2 x e−ax sin(ωx) dx I= −∞ Since x 2 e−ax cos(ωx) is an odd function, ∞ 2 x e−ax eıωx dx I = −ı −∞ We add a dash of integration by parts... 1 z−a=r eı(θ +2 ) dz = [log(z − a)]z−a=r eıθ C z−a = Log r + ı(θ + 2 ) − (Log r + ıθ) = 2 477 10.9 Exercises Exercise 10.1 C is the arc corresponding to the unit semi-circle, |z| = 1, (z) ≥ 0, directed from z = −1 to z = 1 Evaluate z 2 dz 1 C z 2 dz 2 C z 2 |dz| 3 C z 2 |dz| 4 C Hint, Solution Exercise 10 .2 Evaluate ∞ e−(ax 2 +bx) dx, −∞ where a, b ∈ C and (a) > 0 Use the fact that ∞ 2 e−x dx = √... exponential I=e ∞ b2 /(4a) 2 e−a(x+b/(2a)) dx −∞ 2 Consider the parallelogram in the complex plane with corners at ±R and ±R + b/(2a) The integral of e−az on this contour vanishes as it is an entire function We relate the integral along one side of the parallelogram to the integrals along the other three sides R+b/(2a) −R 2 R e−az dz = + −R+b/(2a) R+b/(2a) −R+b/(2a) −R 2 e−az dz + R The first and third integrals... by the polar equation r = 2 − sin2 θ 4 Is this result compatible with the results of part (a)? Hint, Solution Exercise 10.5 1 Use bounding arguments to show that lim R→∞ CR z + Log z dz = 0 z3 + 1 where CR is the positive closed contour |z| = R 2 Place a bound on Log z dz C where C is the arc of the circle |z| = 2 from − 2 to 2 479 3 Deduce that C z2 − 1 R2 + 1 dz ≤ πr 2 z2 + 1 R −1 where C is a semicircle... Exercise 10 .3 Evaluate ∞ 2 e−ax cos(ωx) dx, 2 ∞ 0 0 478 2 x e−ax sin(ωx)dx, and 2 where (a) > 0 and ω ∈ R Hint, Solution Exercise 10.4 Use an admissible parameterization to evaluate (z − z0 )n dz, n∈Z C for the following cases: 1 C is the circle |z − z0 | = 1 traversed in the counterclockwise direction 2 C is the circle |z − z0 − 2| = 1 traversed in the counterclockwise direction 3 z0 = 0, n = −1 and C . the form, az 3 + bz 2 + cz + ıd, 457 with a, b and c complex-valued constants and d a real constant. Substituting z = x + ıy and expanding products yields, a  x 3 + ı3x 2 y − 3xy 2 − ıy 3  +. ıu y = 3x 2 − 2y 2 − 2y + ı(6xy − 2x + 1). On the real axis we have f  (z = x) = 3x 2 − ı2x + ı. Using analytic continuation, we see that f  (z) = 3z 2 − ı2z + ı. Integration yields f(z) = z 3 −. = e ıθ , dz = ı e ıθ dθ  C z dz =  2 0 e ıθ ı e ıθ dθ =  1 2 e 2  2 0 =  1 2 e ı4π − 1 2 e ı0  = 0 2.  C 1 z dz =  2 0 1 e ıθ ı e ıθ dθ = ı  2 0 dθ = 2 3. |dz| =   ı e ıθ dθ   =   ı e ıθ   |dθ|