1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

208 CHAPTER 6 • COMPLEX INTEGRATION

y y

• l+i l+i

.+-<1o---+---..._..x

• I - I
  (a} The semicirc ular path. (b) The polygonal path.
  Figure 6.9 The two contours C1 and C2 joining - 1 to l.

Applying Theorem 6.1, we have f (z) = z, so

f(z(t)) = z(t)=(- cost+isint) =- cost-isint and

r zdz = r (- cost- isint)(sin t+icost)dt

le, lo

= -i 1" ( cos^2 t + sin^2 t) dt = -7ri.

We parametrize C2 in three parts, one for each line segment:

z1 (t) = - 1 +it, dz1 = i dt, and f (z1 (t)) = - 1 -it;

z2(t) = -1+2t+i, dz2=2dt, and f(z2 (t))=-1+2t- i;

z3(t) = 1 +i(l -t), dz3 = -i dt, and f (z3(t))=1-i(l -t),

where 0 ~ t ~ 1 in each case. We get our answer by adding the three integrals

along the three segments:

r zdz= r' (-1-it)idt+ f

1

(-1+2t-i)2dt+ [1[1-i(l-t)](-i)dt.

~ ~ ~ ~

Separating the right side of this equation into its real and imaginary parts gives

r z dz = 1

1

(6t -3) dt + i 1

1

(-4) dt = -4i.

le, o o

Note that the value of the contour integral a.Jong C 1 isn't t he same as the value

of the contour integral along C2, although both integrals have t he same initial

and terminal points.

Contour integrals have properties that a.re similar to those of integrals of a

complex function of a real variable, which you studied in Section 6.1. If C is

given by Equation (6-10), then the integral for the opposite contour - C is

1

J(z)dz = 1-a f (z(- r )) [- z' (- r )Jdr.

• e -b