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4. Complex Integration: Cauchy Integral Theorem and Cauchy Integral Formulas

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4. Complex integration: Cauchy integral theorem and Cauchy integral formulas Definite integral of a complex-valued function of a real variable Consider a complex valued function f(t) of a real variable t: f(t)= u(t)+ iv(t), which is assumed to be a piecewise continuous function defined in the closed interval a t b. The integral of f(t) from t = a to ≤ ≤ t = b, is defined as b b b f(t) dt = u(t) dt + i v(t) dt. Za Za Za 1 Properties of a complex integral with real variable of integration 1. b b b Re f(t) dt = Re f(t) dt = u(t) dt. Za Za Za 2. b b b Im f(t) dt = Im f(t) dt = v(t) dt. Za Za Za 3. b b b [γ1f1(t)+ γ2f2(t)] dt = γ1 f1(t) dt + γ2 f2(t) dt, Za Za Za where γ1 and γ2 are any complex constants. 4. b b f(t) dt f(t) dt. Za ≤ Za | | 2 To prove (4), we consider b b b f(t) dt = e iφ f(t) dt = e iφf(t) dt, − − Za Za Za b b where φ = Arg f(t) dt . Since f(t) dt is real, we deduce that Za ! Za b b b f(t) dt = Re e iφf(t) dt = Re [e iφf(t)] dt − − Za Za Za b b iφ e− f(t) dt = f(t) dt. ≤ Za | | Za | | 3 Example Suppose α is real, show that e2απi 1 2π α . | − |≤ | | Solution Let f(t)= eiαt, α and t are real. We obtain 2π 2π eiαt dt eiαt dt = 2π. Z0 ≤ Z0 | | The left-hand side of the above inequality is equal to 2π 2π eiαt e2απi 1 eiαt dt = = | − |. Z0 iα α 0 | | Combining the results, we obtain e2απi 1 2π α , α is real. | − |≤ | | 4 Definition of a contour integral Consider a curve C which is a set of points z = (x,y) in the complex plane defined by x = x(t), y = y(t), a t b, ≤ ≤ where x(t) and y(t) are continuous functions of the real parameter t. One may write z(t)= x(t)+ iy(t), a t b. ≤ ≤ The curve is said to be smooth if z(t) has continuous derivative • z (t) = 0 for all points along the curve. ′ 6 A contour is defined as a curve consisting of a finite number • of smooth curves joined end to end. A contour is said to be a simple closed contour if the initial and final values of z(t) are the same and the contour does not cross itself. 5 Let f(z) be any complex function defined in a domain in the • D complex plane and let C be any contour contained in with D initial point z0 and terminal point z. We divide the contour C into n subarcs by discrete points z , z , z , • 0 1 2 ...,zn 1,zn = z arranged consecutively along the direction of in- − creasing t. Let ζ be an arbitrary point in the subarc z z and form the • k k k+1 sum n 1 − f(ζ )(z z ). k k+1 − k kX=0 6 Subdivision of the contour into n subarcs by discrete points z ,z , , 0 1 ··· zn 1,zn = z. − 7 We write zk = zk+1 zk. Let λ = max zk and take the limit △ − k |△ | n 1 − lim f(ζk) zk. λ 0 △ n → kX=0 →∞ The above limit is defined to be the contour integral of f(z) along the contour C. If the above limit exists, then the function f(z) is said to be inte- grable along the contour C. If we write dz(t) dx(t) dy(t) = + i , a t b, dt dt dt ≤ ≤ then b dz(t) f(z) dz = f(z(t)) dt. ZC Za dt 8 Writing f(z)= u(x,y)+ iv(x,y) and dz = dx + idy, we have f(z) dz = u dx v dy + i u dy + v dx ZC ZC − ZC b dx(t) dy(t) = u(x(t),y(t)) v(x(t),y(t)) dt Za " dt − dt # b dy(t) dx(t) + i u(x(t),y(t)) + v(x(t),y(t)) dt. Za " dt dt # The usual properties of real line integrals are carried over to their complex counterparts. Some of these properties are: (i) f(z) dz is independent of the parameterization of C; ZC (ii) f(z) dz = f(z) dz, where C is the opposite curve of C; C − C − Z− Z (iii) The integrals of f(z) along a string of contours is equal to the sum of the integrals of f(z) along each of these contours. 9 Example Evaluate the integral 1 dz, C z z I − 0 where C is a circle centered at z0 and of any radius. The path is traced out once in the anticlockwise direction. Solution The circle can be parameterized by z(t)= z + reit, 0 t 2π, 0 ≤ ≤ where r is any positive real number. The contour integral becomes 1 2π 1 dz(t) 2π ireit dz = dt = dt = 2πi. C z z 0 z(t) z dt 0 reit I − 0 Z − 0 Z The value of the integral is independent of the radius r. 10 Example Evaluate the integral 1 (i) 2 and (ii) , z dz 2 dz ZC| | ZC z where the contour C is (a) the line segment with initial point 1 and final point i; − (b) the arc of the unit circle Im z 0 with initial point 1 and final ≥ − point i. Do the two results agree? 11 Solution (i) Consider z 2 dz, ZC | | (a) Parameterize the line segment by z = 1+(1+ i)t, 0 t 1, − ≤ ≤ so that z 2 = ( 1+ t)2 + t2 and dz = (1+ i) dt. | | − The value of the integral becomes 1 2 z 2 dz = (2t2 2t + 1)(1 + i) dt = (1 + i). ZC| | Z0 − 3 12 (b) Along the unit circle, z = 1 and z = eiθ, dz = ieiθdθ. The initial | | point and the final point of the path correspond to θ = π and π θ = 2, respectively. The contour integral can be evaluated as π π 2 2 z 2 dz = ieiθ dθ = eiθ =1+ i. ZC| | Zπ π The results in (a) and (b) do not agree. Hence, the value of this contour integral does depend on the path of integration. 13 1 (ii) Consider . 2 dz ZC z (a) line segment from 1 to i − 1 1 1 1+ i 1 1 dz = dt = = 1 = 1+i. C z2 0 [ 1+(1+ i)t]2 − 1+(1+ i)t − − i − Z Z 0 − − (b) subarc from 1 to i − π π 1 2 1 2 dz = ieiθ dθ = e iθ = 1+ i. 2 2iθ − ZC z Zπ e − π − 14 Estimation of the absolute value of a complex integral The upper bound for the absolute value of a complex integral can be related to the length of the contour C and the absolute value of f(z) along C. In fact, f(z) dz ML, ZC ≤ where M is the upper bound of f(z ) along C and L is the arc length | | of the contour C. 15 We consider b dz(t) f(z) dz = f(z(t)) dt ZC Za dt b dz(t) f(z(t)) dt ≤ Za | | dt b dz(t) M dt ≤ Za dt 2 2 b dx (t) dy(t) = M v + dt = ML. a u dt ! dt ! Z u t 16 Example Show that 1 dz 2, where C is the line segment joining 1+ i and 1+ i. 2 ZC z ≤ − Solution Along the contour C, we have z = x + i, 1 x 1, so that 1 1 1 − ≤ ≤ 1 ≤ z √2. Correspondingly, 1. Here, M = max = 1 | | ≤ 2 ≤ z 2 ≤ z C z 2 and the arc length L = 2. We have| | ∈ | | 1 dz ML = 2. 2 ZC z ≤ 17 Example Estimate an upper bound of the modulus of the integral Log z I = dz C z 4i Z − where C is the circle z = 3. | | ln z + Arg z Log z | | | | Now, so that z 4i ≤ z 4i − || |−| || Log z ln3+ π max =ln3+ π; L = (2π)(3) = 6π. z C z 4i ≤ 3 4 ∈ − | − | Log z Hence, dz 6π(π + ln 3). ZC z 4i ≤ − 18 Example Find an upper bound for ez/(z2 + 1) dz , where Γ is the circle Γ Z = 2 traversed once in the counterclockwise direction. z | | Solution The path of integration has length L = 4π. Next we seek an upper bound M for the function ez/(z2 + 1) when z = 2. Writing z = | | x + iy, we have ez = ex+iy = ex e2, for z = x2 + y2 = 2, | | | | ≤ | | q and by the triangle inequality z2 + 1 z 2 1 = 4 1 = 3 for z = 2. | |≥| | − − | | Hence, ez/(z2 + 1) e2/3 for z = 2, and so | |≤ | | ez e2 dz 4π. Γ z2 + 1 ≤ 3 · Z 19 Path independence Under what conditions that f(z) dz = f(z) dz, ZC1 ZC2 where C and C are two contours in a domain with the same 1 2 D initial and final points and f(z) is piecewise continuous inside . D The property of path independence is valid for f(z)= 1 but it fails z2 when f(z) = z 2. The above query is equivalent to the question: | | When does f(z) dz = 0 IC hold, where C is any closed contour lying completely inside ? The D equivalence is revealed if we treat C as C C .
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