1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

206 CHAPTER 6 • COMPLEX INTEGRATION

Suppose that f (z) = u (z) + iv (z) and that z (t) = x (t) + iy (t) is a

parametrization for the contour C. Then

lf(z)dz =lb f(z(t))z'(t)dt

= lb [u(z(t))+iv(z(t))][x'(t)+iy'(t)]dt

=lb [u(z(t))x' (t)-v(z(t))y' (t)]dt

+ i lb [v(z(t))x' (t) +u(z(t))y' (t)]dt

= lb ( UX f - vy^1 ) dt + i lb ( VX I + UY^1 ) dt,

where we are equating u with u(z(t)), x' with x'(t), and so on.

(6-15)

If we use the differentials given in Equation (6-13), then we can write Equa-

tion (6-15) in terms ofline integrals of the real-valued functions u and v , giving

[t(z)dz = l u dx-v dy+ i l v dx+ u dy, (6-16)

which is easy to remember if we recall that symbolically

J (z)dz = (u+iv)(dx+ i dy).

We emphasize that Equation ( 6-16) is merely a notational device for applying

Theorem 6.1. You should carefully apply Theorem 6.1, as illustrated in Examples

6.7 and 6.8, before using any shortcuts suggested by Equation (6-16).

• EXAMPLE 6.9 Show that

r z dz = r z dz = 4 + 2i,

le, le.

where 01 is the line segment from - 1 - i to 3 + i and C2 is the portion of the

parabola x = y^2 + 2y joining - 1 - i to 3 + i , as indicated in Figure 6.8.

The line segment joining (- 1, -1) to (3, 1) is given by the slope-intercept

formula y = ~x - ~, which can be written as x = 2y + 1. If we choose the

parametrization y = t and x = 2t + 1, we can write segment 01 as

01 : z(t) = 2t + 1 +it and dz= (2 +i) dt, for -^1 :=:; t :=:; 1.

Along 01 we have f (z (t)) = 2t + 1 +it. Applying Theorem 6. 1 gives

r z dz = f

1

(2t + 1 + it)(2 + i) dt.

le, l -1