1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

204 CHAPTER 6 • COMPLEX INTEGRATION

last expression should equal J: f ( z ( t)) z ' ( t) dt, as defined in Section 6.1. Of
course, if we're to have any hope of this happening, we would have to get the
same limit regardless of how we parametrize the contour C. As T heorem 6.1
states, this is indeed the case.

We omit the proof of Theorem 6.1 because it involves ideas (e.g., the theory
of the Riemann-Stieltjes integral) that are beyond the scope of this book. A
more rigorous development of the contour integral based on Riemann sums is
presented in advanced texts such as L. V. Ahlfors, Complex Analysis, 3rd ed.
(New York: McGraw-Hill, 19 79).
Two important facets of Theorem 6.1 are worth mentioning. First, Theorem
6.1 makes the problem of evaluating complex-valued functions along contours
easy, as it reduces the task to the evaluat ion of complex-valued functions over
real intervals-a procedure that you studied in Section 6.1. Second, according
to Theorem 6.1, this transformation yields the same answer regardless of the
parametrization we choose for C.

• EXAMPLE 6.7 Give an exact calculation of the integral in Example 6.6.

Solution We must compute f c exp z dz, where C is the line segment joining

A = 0 to B = 2 + i~. According to Equation (1-48), we can parametrize C by

z (t) = (2 +ii) t, for 0::; t::; 1. As z' (t) = (2 +ii), Theorem 6.1 guarantees

that

fc expzdz =fol exp[z(t)]z' (t)dt

= fo

1

exp[(2+i~)t] (2+i~)dt

= ( 2 + i~) fo

1

e^2 teif'dt

= (2+i~) fo\^2 t (cos :t +isin :t) dt

= ( 2 + i~) (fo

1

e^2 t cos :t dt +i fo

1

e^2 t sin :t dt).