1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

6.2 • CONTOURS AND CONTOUR INTEGRALS 209

Using the change of variable t = - r in this last equation and the property

that J: f (t) dt = - Jb" f (t) dt, we obtain

J_ - C f(z)dz=-ff(z)dz. Jc (6-17)

If two functions f and g can be integrated over the same path of integration
C, then their sum can be integrated over C, and we have the familiar result

fc [f (z) + g(z)]dz = Lf(z)dz+ fcg(z)dz.

Constant multiples also behave as we would expect:

fc (a+ib)f(z)dz = (a+ib) fc! (z)dz.

If two contours C 1 and C 2 are placed end to end so that the terminal point
of C1 coincides with the initial point of C2, then the contour C = Ci + C2 is a
continuation of C 1 , and

r f(z)dz= r f(z)dz+ r f(z)dz.

~.~ k, ~

If the contour C has two parametrizations

C : zt(t) = X 1 (t) + iyi(t),

C: z2(r) = X2 (T) +iy2(r),

for as; ts; b,

for c s; 'T s; d,

and

and there exists a differentiable function 'T = </> ( t) such that

c =</>(a), d=(b), and </>'(t)>O, for a< t < b,

(6-18)

(6- 19 )

then we say that z 2 ( r) is a reparametrization of the contour C. If f is

continuous on C, then we have

(6-20)

Equation (6-20) shows that the value of a contour integral is invariant under
a change in the parametric representation of its contour if the reparametrization
satisfies E<iuations (6-19).

We now give two important inequalities relating to complex integrals.