¿Cuál es el camino más corto para un cable eléctrico? Planta eléctrica ciudad planta eléctrica Ejemplo típico que motiva la necesidad de integrales de contorno Integrales de contorno o de camino

Secciones cónicas Círculo ElipseParábolaHipérbola Ecuación general de una sección cónica: Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0

Longitud de una curva z’(t) = x’(t) + iy’(t) Si x’(t) e y’(t) son continuas en el intervalo a <= t <= b entonces: C es un arco diferenciable y su longitud es: Si z’(t) no es 0 en ningún punto de a < t < b entonces podemos definir un vector tangente: Y decimos que z(t) es un arco suave. Contorno = arco suave a trozos yy xx C: z(t)

Ejemplo: la longitud de camino es independiente de la parametrización a b r x= r cos(t) y= r sin(t) 0  t  π/2 x= r cos(2  ) y= r sin(2  ) 0    π Dos parametrizaciones distintas

a x y 1

a x 10.5

Circulation and Net Flux Let T and N denote the unit tangent vector and the unit normal vector to a positively oriented simple closed contour C. When we interpret the complex function f(z) as a vector, the line integrals (6) (7) have special interpretations.

The line integral (6) is called the circulation around C and (7) is called the net flux across C. Note that and so circulation = (8) net flex = (9)

Given the flow f(z) = (1 + i)z, compute the circulation around and the net flux across the circle C: |z| = 1. Solution

The complex function where k = a + ib and z 1 are complex numbers, gives rise to a flow in the domain z  z 1. If C is a simple closed contour containing z = z 1 in its interior, then we have The circulation around C is 2  b and the net flux across C is 2  a. If z 1 were in the exterior of C both of them would be zero.

Note that when k is real, the circulation around C is zero but the net flux across C is 2  k. The complex number z 1 is called a source when k > 0 and is a sink when k < 0.

Evaluate where C is the circle |z| = 1. Solution This integrand is not analytic at z = 0, −4 but only z = 0 lies within C. Since We get z 0 = 0, n = 2, f(z) = (z + 1)/(z + 4), f  (z) = −6/(z + 4) 3. By (6):

Evaluate Solution Though C is not simple, we can think of it is as the union of two simple closed contours C 1 and C 2 in Fig

For I 1 : z 0 = 0, f(z) = (z 3 + 3)/(z – i) 2 : For I 2 : z 0 = i, n = 1, f(z) = (z 3 + 3)/z, f ’(z) = (2z 3 –3 )/z 2 : We get