1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

6.2 • CONTOURS AND CONTOUR INTEGRALS 205

Each integral in the last expression can be done using integration by parts.

(There is a simpler way-see Remark 6.1.) We leave as an exercise to show that

the final answer simplifies to exp (2 + i~) -1, as we claimed in Example 6.6.

• EXAMPLE 6.8 Evaluate f ct( 2 ) ,_: 2 dz.

Solution Recall that ct (2) is the upper semicircle with radius 1 centered

at x = 2 oriented in a positive direction (i.e., counterclockwise). The function

z (t) = 2 + eit, 0 ~ t ~ 71', is a parametrization for C. We apply Theorem 6. 1

with f (z) = • .: 2. (Note: f (z \t)) = z(tj_ 2 , and z' (t) = ieit.) Hence

1

--^1 1"^1. 1"

2

dz= (

2

't)

2

ie'tdt= idt=i?T.

c z - o + e' - o

To help convince yourself that the value of the integral is independent of the

parametrization chosen for the given contour, try working through Example 6.8

with z(t) = 2 +ei"^1 , for 0~t~1.

A convenient bookkeeping device can help you remember how to apply The-

orem 6.1. Because fc f (z) dz= J: f (z (t)) z ' (t) dt, you can symbolically equate

z with z (t) and dz with z' (t) dt. These identities should be easy to remember

because z is supposed to be a point on the contour C parametrized by z (t), and

~; = z' (t), according to the Leibniz notation for the derivative.

If z (t) = x (t) + iy (t), then by the preceding paragraph we have

dz= z' (t) dt = [x' (t) + iy' (t)J dt = dx + idy, (6-13)

where dx and dy a.re the differentials for x (t) and y (t), respectively (i.e., dx is

equated with x' (t) dt, etc.). The expression dz is often called the complex dif-

fe rential of z. J ust as dx and dy are intuitively considered to be small segments

along the x-and y-axes in real variables, we can think of dz as representing a

tiny piece of the contour C. Moreover, if we write

ldzl = l[x' (t) + iy' (t)) dtl =Ix' (t) + iy' (t)I dt = Vix' (t))^2 + [y' (t)]^2 dt,

we can put Equation (6- 11 ) into the form

L ( C) =lb Jrx' (t)J

2

+ [y' (t)J

2

dt = fc ldzl. (6-14)

so we can think of ldzl as representing the length of dz.