long as as n = -1. Therefore, we can use the Fundamental Theorem of Calculus (which works equally well for real integrals of complex-valued functions), to evaluate n+1 1 -1 = n = 2 k + 1 odd, 0, 2 t + i (t - 1) 2 z n dz = = , n = 2 k even. n+1 P t = -1 n+1 Thus, when n 0 is a positive integer, we obtain the same result as before. Interestingly, in this case the complex integral is well-defined even when n is a negative integer because, unlike the real line segment, the parabolic path does not go through the singularity of z n at z = 0. The case n = -1 needs to be done slightly differently, and integration of 1/z along the parabolic path is left as an exercise for the reader -- one that requires some care. We recommend trying the exercise now, and then verifying your answer once we have become a little more familiar with basic complex integration techniques. Finally, let us try integrating around a semi-circular arc, again with the same endpoints -1 and 1. If we parametrize the semi-circle S + by z(t) = e i t , 0 t , we find
z n dz =
S+ 0
zn
dz dt = dt
t=0
e i nt i e i t dt =
0
i e i (n+1)t dt
0 0,
2 , n = 2 k even. n+1 This value is the negative of the previous cases -- but this can be explained by the fact that the circular arc is oriented to go from 1 to -1 whereas the line segment and parabola both go from -1 to 1. Just as with line integrals, the direction of the curve determines the sign of the complex integral; if we reverse direction, replacing t by - t, we end up with the same value as the preceding two complex integrals. Moreover -- again provided n = -1 -- it does not matter whether we use the upper semicircle or lower semicircle to go from -1 to 1 -- the result is exactly the same. However, the case n = -1 is an exception to this "rule". Integrating along the upper semicircle S + from 1 to -1 yields
S+
e i (n+1)t = n+1
1 - e i (n+1) = = - n+1
-1 = n = 2 k + 1 odd,
dz = z
i dt = i ,
0
(7.111)
whereas integrating along the lower semicircle S - from 1 to -1 yields the negative
S-
dz = z
-
0
i dt = - i .
(7.112)
Hence, when integrating the function 1/z, it makes a difference which direction we go around the origin. Integrating z n for any integer n = -1 around an entire circle gives zero -- irrespective of the radius. This can be seen as follows. We parametrize a circle of radius r by z(t) = re i t for 0 t 2 . Then, by the same computation,
2 2
z dz =
C 0
n
(r e
n i nt
)(r i e ) dt =
0
it
ir
n+1 i (n+1)t
e
r n+1 i (n+1)t e dt = n+1
2
= 0,
t=0
(7.113)
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Peter J. Olver
provided n = -1. The circle on the integral sign serves to remind us that we are integrating around a closed curve. The case n = -1 remains special. Integrating once around the circle in the counter-clockwise direction yields a nonzero result dz = z
2
i dt = 2 i .
0
(7.114)
C
Let us note that a complex integral does not depend on the particular parametrization of the curve C. It does, however, depend upon its orientation: if we traverse the curve in the reverse direction, then the complex integral changes its sign: f (z) dz = - f (z) dz.
C
(7.115)
-C
Moreover, if we chop up the curve into two non-overlapping pieces, C = C1 C2 , with a common orientation, then the complex integral can be decomposed into a sum over the pieces: f (z) =
C1 C2 C1
f (z) dz +
C2
f (z) dz.
(7.116)
For instance, the integral (7.114) of 1/z around the circle is the difference of the individual semicircular integrals (7.111, 112); the lower semicircular integral acquires a negative to sign flip its orientation so as to agree with that of the entire circle. All these facts are immediate consequences of the basic properties of line integrals, or can be proved directly from the defining formula (7.108). Note: In complex integration theory, a simple closed curve is often referred to as a contour , and so complex integration is sometimes referred to as contour integration. Unless explicitly stated otherwise, we always go around contours in the counter-clockwise direction. Further experiments lead us to suspect that complex integrals are usually pathindependent, and hence evaluate to zero around closed contours. One must be careful, though, as the integral (7.114) makes clear. Path independence, in fact, follows from the complex version of the Fundamental Theorem of Calculus. Theorem 7.47. Let f (z) = F (z) be the derivative of a single-valued complex function F (z) defined on a domain C. Let C be any curve with initial point and final point . Then f (z) dz =
C C
F (z) dz = F () - F ().
(7.117)
Proof : This follows immediately from the definition (7.108) and the chain rule:
b
F (z) dz =
C a
F (z(t))
dz dt = dt
b a
d F (z(t)) dt = F (z(b)) - F (z(a)) = F () - F (), dt Q.E.D.
c 2012 Peter J. Olver
where = z(a) and = z(b) are the endpoints of the curve. 1/19/12 268
For example, when n = -1, the function f (z) = z n is the derivative of the single1 z n+1 . Hence valued function F (z) = n+1 z n dz =
C
n+1 n+1 - n+1 n+1
whenever C is (almost) any curve connecting to . The only restriction is that, when n < 0, the curve is not allowed to pass through the singularity at the origin z = 0. In contrast, the function f (z) = 1/z is the derivative of the complex logarithm log z = log | z | + i ph z, which is not single-valued on all directly. However, if our curve is does not include the origin, 0 complex logarithm to evaluate the of C \ {0}, and so Theorem 7.47 cannot be applied contained within a simply connected subdomain that C, then we can use any single-valued branch of the integral dz = log - log , z
C
where , are the endpoints of the curve. Since the common multiples of 2 i cancel, the answer does not depend upon which particular branch of the complex logarithm is chosen as long as we are consistent in our choice. For example, on the upper semicircle S + of radius 1 going from 1 to -1,
S+
dz = log(-1) - log 1 = i , z
where we use the branch of log z = log | z | + i ph z with 0 ph z . On the other hand, if we integrate on the lower semi-circle S - going from 1 to -1, we need to adopt a different branch, say that with - ph z 0. With this choice, the integral becomes
S-
dz = log(-1) - log 1 = - i , z
thus reproducing (7.111, 112). Pay particular attention to the different values of log(-1) in the two cases! Cauchy's Theorem The preceding considerations suggest the following fundamental theorem, due in its general form to Cauchy. Before stating it, we introduce the convention that a complex function f (z) is to be called analytic on a domain C provided it is analytic at every point inside and, in addition, remains (at least) continuous on the boundary . When is bounded, its boundary consists of one or more simple closed curves. In general, as in Green's Theorem 6.13, we orient so that the domain is always on our left hand side. This means that the outermost boundary curve is traversed in the counter-clockwise direction, but those around interior holes take on a clockwise orientation. Our convention is depicted in Figure 7.32. 1/19/12 269
c 2012 Peter J. Olver