Contour Integrals
Let the complex function f(z) be defined as f(z) = u(x,y) + i*v(x,y)
Using the definition of the integral:
∫ f(z) dz = lim n → ∞ [ Σ f(z_k) * Δz_k from k = 1 to n ]
= lim n → ∞ [ Σ ( u(x_k, y_k) + i*v(x_k, y_k) ) * ( Δx_k + i*Δy_k ) from k = 1 to n ]
= lim n → ∞ Σ [ (u(x_k, y_k) * Δx_k - v(x_k, y_k) * Δy_k) + i*(v(x_k, y_k) * Δx_k
+ u(x_k, y_k) * Δy_k) ]
= ( ∫ u(x,y) dx - ∫ v(x,y) dy ) + i*( ∫ u(x,y) dy + ∫ v(x,y) dx )
In summary:
∫ f(z) dz = ( ∫ u(x,y) dx - ∫ v(x,y) dy ) + i*( ∫ u(x,y) dy + ∫ v(x,y) dx )
What is needed:
1. A contour curve f = y(x). This is where you get your end points.
2. We need f(z) to integrate. Separate f(z) into its parametric parts.
Let's look at a couple examples.
Examples
1. Contour: y = x + 1 from (3,4) to (4,5). Integrate ∫ z^2 dz.
Since y = x + 1, x = y - 1
z^2 = (x + i*y)^2 = (x^2 - y^2) + i*(2*x*y)
u = x^2 - y^2
v = 2*x*y
u = x^2 - y^2
u = x^2 - (x + 1)^2
u = -2*x - 1
∫ u dx from 3 to 4 = -8
v = 2 * x * y
v = 2 * (y - 1) * y
∫ v dy from 4 to 5 = 95/3
u = (y - 1)^2 - y^2
u = -2 * y + 1
∫ u dy from 4 to 5 = -8
v = 2 * x * y
v = 2 * x * (x + 1)
∫ v dx from 3 to 4 = 95/3
∫ z^2 dz from (3 + 4*i) to (4 + 5*i) = (-8 - 95/3) + i*(-8 + 95/3) ≈ -39.666667 + 23.666667*i
2. Contour: y = x^2 - 1 from (0, -1) to (2, 1). Integrate ∫ z^2 + 1 dz.
y = x^2 - 1, x = √(y + 1)
z^2 + 1 = (x + i*y)^2 + 1 = x^2 + 2*i*x*y - y^2 + 1
u = x^2 - y^2 + 1
v = 2 * x * y
u = x^2 - y^2 + 1
u = x^2 - (x^2 - 1)^2 + 1
∫ u dx from 0 to 2 = 8/5
v = 2 * x * y
v = 2 * √(y + 1) * y
∫ v dy from -1 to 1 = 8/15 * √2
u = x^2 - y^2 + 1
u = y + 1 - y^2 + 1
u = -y^2 + y + 2
∫ u dy from -1 to 1 = 10/3
v = 2 * x * y
v = 2 * x * (x^2 - 1)
∫ v dx from 0 to 2 = 4
∫ z^2 + 1 dz from -i to 2+i = (8/5 - 8/15 * √2) + i*(4 + 10/3) ≈ 0.845753 + 7.333333*i
This is the basics of line integrals.
Source: Wunsch, A. David. Complex Variables with Applications. 2nd Edition. Addison-Wesley Publishing Company. 1994.
Take care, Eddie
This blog is property of Edward Shore. 2014
Showing posts with label line integrals. Show all posts
Showing posts with label line integrals. Show all posts
Tuesday, November 25, 2014
Complex Analysis: Line Integral
Subscribe to:
Posts (Atom)
Python: Drawing a Ring of Circles
Python: Drawing a Ring of Circles The goal: Draw a number of circles that revolve around the origin (0,0). The center of each circle i...