Ch4 Gg Jkgjd
Integration
4.1. Introduction. If : D C is simply a function on a real interval D , , then the integral tdt is, of course, simply an ordered pair of everyday 3 rd grade calculus integrals:
tdt xtdt i ytdt,
where t xt iyt. Thus, for example,
t 2 1 it 3 dt
0
1
4 i. 4 3
Nothingreally new here. The excitement begins when we consider the idea of an integral of an honest-to-goodness complex function f : D C, where D is a subset of the complex plane. Let’s define the integral of such things; it is pretty much a straight-forward extension to two dimensions of what we did in one dimension back in Mrs. Turner’s class. Suppose f is a complex-valued function on a subset of thecomplex plane and suppose a and b are complex numbers in the domain of f. In one dimension, there is just one way to get from one number to the other; here we must also specify a path from a to b. Let C be a path from a to b, and we must also require that C be a subset of the domain of f.
4.1
Note we do not even require that a b; but in case a b, we must specify an orientation for theclosed path C. We call a path, or curve, closed in case the initial and terminal points are the same, and a simple closed path is one in which no other points coincide. Next, let P be a partition of the curve; that is, P z 0 , z 1 , z 2 , , z n is a finite subset of C, such that a z 0 , b z n , and such that z j comes immediately after z j1 as we travel along C from a to b.
A Riemannsum associated with the partition P is just what it is in the real case: SP
fz z j , j
j1
n
where z is a point on the arc between z j1 and z j , and z j z j z j1 . (Note that for a j given partition P, there are many SP—depending on how the points z are chosen.) If j there is a number L so that given any 0, there is a partition P of C such that |SP L| whenever P P , then f is said to be integrable on C and the number L is called the integral of f on C. This number L is usually written fzdz.
C
Some properties of integrals are more or less evident from looking at Riemann sums:
cfzdz c fzdz
C C
for any complex constant c.
4.2
fz gzdz fzdz gzdz
C C C
4.2 Evaluating integrals. Now, how onEarth do we ever find such an integral? Let : , C be a complex description of the curve C. We partition C by partitioning the interval , in the usual way: t 0 t 1 t 2 t n . Then a , t 1 , t 2 , , b is partition of C. (Recall we assume that t 0 for a complex description of a curve C.) A corresponding Riemann sum looks like
SP
ft t j t j1 . j
j1
n
We have chosen the points z t , where t j1 t t j . Next, multiply each term in the j j j sum by 1 in disguise: ft t jtj tt j1 t j t j1 . j j1
j1 n
SP
I hope it is now reasonably convincing that ”in the limit”, we have
fzdz ft tdt.
C
(We are, of course, assumingthat the derivative exists.)
Example We shall find the integral of fz x 2 y ixy from a 0 to b 1 i along three different paths, or contours, as some call them. First, let C 1 be the part of the parabola y x 2 connecting the two points. A complex description of C 1 is 1 t t it 2 , 0 t 1:
4.3
1 0.8 0.6 0.4 0.2
0
0.2
0.4
x
0.6
0.8
1Now, 1 t 1 2ti, and f 1 t t 2 t 2 itt 2 2t 2 it 3 . Hence,
fzdz f 1 t 1 tdt
C1 0
1
2t 2 it 3 1 2tidt
0
1
2t 2 2t 4 5t 3 idt
0
1
4 5i 15 4 Next, let’s integrate along the straight line segment C 2 joining 0 and 1 i.
1 0.8 0.6 0.4 0.2
0
0.2
0.4
x
0.6
0.8
1
Here we have 2 t...
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