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Unformatted text preview: Informal lecture notes for Dec. 4 Now that we have a solid background in the basics of complex differentiation, its time to move on to complex integration. 1 Contour integrals Consider a curve C in the complex plane parametrized by z ( t ) for a z b . Of course, we can write this in terms of its real and imaginary parts as z ( t ) = x ( t )+ iy ( t ). Now suppose we have a (complex) function f ( z ) = u ( x, y )+ iv ( x, y ). We want to integrate f along the curve C . In order to define such an integral we start, as usual, with a Riemann sum approximation. If we subdivide the curve by subdividing the interval at points a = t , t 1 , . . . , t n = b , then the corresponding Riemann sum looks like n X j =1 f ( z ( t )) ( z ( t i )- z ( t i- 1 )) where the dot here refers, of course, to complex multiplication. (This is the difference with the definition of a real line integral, in which both factors are vectors and we take their dot product.) Then the contour integral of f along C , written R C f ( z ) dz , is defined to be the limit as n , assuming that this limit exists. This makes a good definition, but in order to compute with it, we will re- write it in terms of the parametrization and the real and imaginary parts of f and z . A little computation shows that Z C f ( z ) dz = Z b a [ u ( x ( t ) , y ( t )) x ( t )- v ( x ( t ) , y ( t )) y ( t )] dt + i Z b a [ v ( x ( t ) , y ( t )) x ( t ) + u ( x ( t ) , y ( t )) y ( t )] dt = Z C u dx- v dy + i Z C v dx + u dy. An easy way to remember this is by writing the complex differential as dz = dx + i dy and then multiplying by f ( z ) written in terms of its real and imaginary parts. Example: Let C be the quarter-circle from 1 to i and let f ( z ) = z 2 . Then we can parametrize C as z ( t ) = cos t + i sin t for 0 t / 2. In terms of this parametrization, we have that f ( z ( t )) = cos 2 t- sin 2 t + i (2 cos t sin t ). Further, we have dz = dx + i dy = sin t dt + i cos t dt , and thus, after some algebra, we find that f ( z ( t )) dz = sin 3 t- 3 sin t cos 2 t + i ( cos 3 t- 3 cos t sin 2 t ) dt = sin t- 4 sin t cos 2 t + i ( cos t- 4 cos t sin 2 t ) dt. 1 Note that the last line is obtained using cos 2 t = 1- sin 2 t and vice versa, and we have done this with an eye toward the fact that were about to integrate this expression. We have Z C f ( z ) dz = Z / 2 ( sin t- 4 sin t cos 2 t ) dt + i Z / 2 ( cos t- 4 cos t sin 2 t ) dt =- cos t + 4 3 cos 3 t / 2 + i sin t- 4 3 sin 3 t / 2 =- 1 3- 1 3 i. Before leaving this example, it is worth noting that these computations can be streamlined by taking better advantage of complex functions. In particular, this parametrization of the curve C can be re-written as z ( t ) = e it for 0 t / 2. Then we have f ( z ( t )) = e 2 it . Further, we see that dz = ie it dt , and thus Z C f ( z ) dz = i Z / 2 e 3 it dt = i Z / 2 cos(3 t ) dt- Z / 2 sin(3 t ) dt = i 1 3 sin(3 t ) / 2 + 1 3 cos(3 t ) /...
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