6.2 • CONTOURS AND CONTOUR INTEGRALS 207y y(a) The line segment. (b) The portion oflhe parabola.Figure 6.8 T he two contours C 1 and C 2 joining - 1 -i to 3 + i.
We now multiply out the integrand and put it into its real and imaginary parts:r zdz = 1
1
(3t+2)dt+ij1
(4t+l)dt= 4 +2i.le, - 1 - 1
Similarly, we can parametrize the portion of the parabola x = y^2 + 2y joining
(- 1, - 1) to (3, 1) by y = t and x = t^2 + 2t so that
C 2 :z(t)=t^2 + 2t +it and dz=(2t+ 2 +i)dt, for - l:::;t:::;l.
Along C 2 we have f (z (t)) = t^2 + 2t + it. Theore m 6.1 now givesr z dz = j
1
(t^2 + 2t +it) (2t + 2 + i) dtle, -1
= [
1
1(2t^3 + 6t^2 + 3t) dt + i [11 (3t^2 + 4t) dt = 4 + 2i.
In Example 6.9, the value of t he two integrals is the same. T his outcome
doesn't hold in general, as Example 6.10 shows.
- EXAMPLE 6. 10 Show that
r z dz = -1fi but that
le,
r z dz= -4i,
le,
where C 1 is the semicircular path from - 1 to 1 and C 2 is the polygonal path
from - 1 to 1, re spectively, shown in Figure 6.9.Solution We parametrize the semicircle C 1 as
C 1 : z (t)=- cost+isint and dz= (sint+icost)dt, for 0 :::; t :::; n.