path.htm Paths and Integrals
Paths and Integrals
A C1 path C is a complex valued function
C: z = z(t) = x(t) + i y(t), a ≤ t ≤ b,
where z(t) is continuously differentiable. The path C is represented by its image with an arrow drawn in the direction of increasing t.
picture of a simple path C

pathpic1.gif

We shall assume the path is simple - it does not intersect itself except possibly at its endpoints.
The path is closed if it is simple and the endpoints are the same: z(a) = z(b).
If C is a path, the path − C is the path with the same image but traced in the opposite direction. If C is parameterized by zC(t), 0 ≤ t ≤ 1, then −C may be parameterized by
−C: z = z−C(t) = zC(1 − t), 0 ≤ t ≤ 1.
picture of −C

pathpic2.gif

We shall deal with paths which are continuous and piecewise C1. Such paths can written as a formal sum C1 + C2 +…+ CN, where the terminal point of Cj is the initial point of Cj+1, j = 1, …, N−1. 1
picture of 3 arcs

pathpic3.gif

For our purposes C will consist of a [small] number of arcs and line segments.
Integrals on Paths
Let C be a continuous and [piecewise] C1 path and let f(z) be a continuous function defined on C. Let C be parameterized by
C: z = z(t) = x(t) + i y(t), a ≤ t ≤ b.
Then the integral of f(z)  dz on C is defined as:
•
The Quick Definition

⌠
⌡

C 
 f(z)  dz = ⌠
⌡ 		b

a 
 f(z(t)) dz
dt
  dt,
where C is parameterized by
C: z = z(t) = x(t) + i y(t), a ≤ t ≤ b.
•
The Riemann Sum Definition:
Let Π: a = t0 < t1 < … < tM = b, be a partition of [a,b], and zj′ = z(tj′) be a typical point in the image of [tj,tj+1]; define the Riemann sum
R(f,Π,zj′)
= M−1
∑
j=0 
 f(zj′)  (zj+1 − zj)
≈ M−1
∑
j=0 
 f(zj′)·z′(tj′)·(tj+1 − tj)
=
∑
z along C 
 f(z)  ∆z .
Then
⌠
⌡

C 
 f(z)  dz
=
lim
max|∆z| → 0 
 R(f, Π,zj′)
=
lim
max|∆z| → 0 
∑
z along C 
 f(z)  ∆z.
The "quick" definition gives an effective way to compute ∫Cf(z)  dz. The "Riemann sum" definition emphasizes that ∫Cf(z)  dz is independent of the parameterization of C. Either definition gives that
⌠
⌡

C 
 f(z) ±g(z)  dz
= ⌠
⌡

C 
 f(z)   dz ±⌠
⌡

C 
 g(z)  dz,
⌠
⌡

−C 
 f(z)  dz
= − ⌠
⌡

C 
 f(z)  dz,
⌠
⌡

C1 + C2 
 f(z)  dz
= ⌠
⌡

C1 
 f(z)  dz + ⌠
⌡

C2 
 f(z)  dz.
The last relation says that for fixed f(z), ∫Cf(z)  dz is additive as a map on sums of paths or chains.2
A Version of the Fundamental Theorem of Calculus

Theorem (FTC Version I). Let C be a continuous piecewise C1path and let F(z) be analytic at every point on C. Then
⌠
⌡

C 
 F′(z)  dz = F(z(b)) − F(z(a)).
where
C: z = z(t) = x(t) + i y(t), a ≤ t ≤ b.
is a parameterization of C.
Proof: By the chain rule for differentiation
d
dt
 F(z(t)) = F′(z(t)) dz
dt
 ,
so by FTC (Version I) for functions of a real variable
⌠
⌡

C 
 F′(z)  dz
= ⌠
⌡ 		b

a 
 d
dt
 F(z(t))  dt
= F(z(b)) − F(z(a)).
The Most Important Path Integral
If the curve C is simple and closed and traversed in the counterclockwise direction, we often write
⌠
⌡

C 
 f(z)  dz = ⌠
(⎜)
⌡


C 
 f(z)  dz
The most important integral is the integral of f(z) = [1/z] around a circle containing the origin.

Theorem. Let CR be the circle of radius R, centered at 0, and traversed in the counterclockwise direction. Then
⌠
(⎜)
⌡


CR 
 1
z
  dz = 2 πi.
 Proof: CR can be parameterized by the angle t, 0 ≤ t ≤ 2π:
CR: z(t)
= R eit
= R (cos(t) + i sin(t)),
dz
= i R eit  dt.
Then
⌠
(⎜)
⌡


CR 
 1
z
  dz
= ⌠
⌡ 		2π

0 
 iR eit
R eit
  dt
= ⌠
⌡ 		2π

0 
 i  dt
= 2 πi.
Exercise
Use the above Fundamental Theorem Calculus or the parametric representation to show that for n an integer, n ≠ −1,
⌠
(⎜)
⌡


CR 
 zn  dz = 0.

Footnotes:

1More generally, we can consider a chain : C1 + C2 + …+CN, a formal sum even when the components do not connect. The concept of chain is used in differential geometry.
2 Compare to the calculus result
⌠
⌡ 		b

a 
 f(t)  dt + ⌠
⌡ 		c

b 
 f(t)  dt = ⌠
⌡ 		c

a 
 f(t)  dt.


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On 16 Nov 2013, 20:56.