1550251515-Classical_Complex_Analysis__Gonzalez_

Integration 433

F(z) = (z - a)m+ljm + 1 is still analytic in the region C - {a}, and

F'(z) = (z-a)m. However, form= -1, (7.8-13) does not hold for a closed

path about a because no primitive of (z - a)-^1 can be made analytic in

the whole of C - {a}. For instance, if r: z = a+ reit, 0 :::; t :::; 271", we

obtain, by direct evaluation,

r -^1 - dz = f 27r i dt = 271"i

Jr z - a Jo

Theorem 7.9 Let 'Y be a rectifiable arc joining the points zo and z 1 , and

suppose that the functions g( z) and h(z) are analytic in some region R

containing 'Y*. Then

j g(z)h'(z) dz= g(z1)h(z1) - g(zo)h(zo) - j g^1 (z)h(z) dz

7 7

This is the integration by parts formula for complex integrals.

Proof Let F(z) = g(z)h(z). Then in R we have

F'(z) = g(z)h'(z) + g'(z)h(z)

and (7.8-12) gives

j[g(z)h'(z) + g'(z)h(z)] dz= g(z1)h(z1)-g(zo)h(zo)

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and the conclusion follows easily on applying Theorem 7.2-1.

Corollary 7.5 Let C be a closed rectifiable contour, and suppose that g

and h are analytic in some region containing C*. Then

j g(z)h'(z) dz= - j g'(z)h(z) dz

c c

Proof It follows from the preceding theorem by letting z 1 = z 0 •

Theorem 7.10 The integral of a continuous complex function f(z) =
u(x, y) + iv(x, y) along the rectifiable arc"'(: z = z(t) = x(t) + iy(t), a :::;
t :::; (J, can be expressed in terms of real line integrals as follows:

j f(z)dz= j(udx-vdy)+i j(vdx+udy) (7.8-14)

Proof We recall (see, e.g., [40], pp. 218-219) that real line integrals in

the plane are defined in the following manner: Let P(x, y) and Q(x, y)

be functions defined along the arc "'(: x = x(t), y = y(t), a :::; t :::; (J.