200 CHAPTER 6 • COMPLEX lNTECRATION
If C is a smooth curve, then ds, the differential of arc length, is given by
ds = V[x' (t)]^2 + [y' (t)]^2 dt = lz' (t)I dt.
The function s (t) = Vix' (t)]^2 + [y' (t)]^2 is continuous, as x' and y' are
continuous functions, so the length L ( C) of the curve C is
(6-11)
Now, consider C to be a curve with parametrization
C: z1 (t) =x(t)+iy(t), for a~ t ~ b.
The opposite curve -C traces out the same set of points in the plane, but
in the reverse order, and has the parametrization
-C: z2(t) = x(- t) + iy(- t) , for - b ~ t ~ - a.
Since z 2 (t) = z 1 (- t), - C is merely C traversed in the opposite sense, as
illustrated in Figure 6.3.
A curve C that is constructed by joining finitely many smooth curves end to
end is called a contou r. Let C 1 , C 2 , .•. , C,,. denote n smooth curves such that
the terminal point of the curve Ck coincides with the initial point of Ck+1 1 for
k = 1, 2, ... , n - 1. We express the contour C by the equation
A synonym for contour is p ath.
- EXAMPLE 6.5 Find a parametrization of t he polygonal path C from - 1 +i
to 3 - i shown in Figure 6.4.
S olution We express C as three smooth curves, or C = C 1 + C2 + C3. If we
set z 0 = -1 + i and z 1 = -1, we can use Equation (1-48) to get a formula for
the straight-line segment joining two points:
C1: zi(t) = Zo + t(z1 -zo) = (- 1 + i) + t [-1-(-1 + i)], for 0 ~ t ~ 1.
When simplified, this formula becomes
C 1 : z1 (t) = - 1 + i (1 - t), for 0 ~ t ~ 1.
Figure 6 .3 The curve C and its opposite curve -C.