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
A blog is that is all about mathematics and calculators, two of my passions in life.
Tuesday, November 25, 2014
Complex Analysis: Line Integral
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