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CalcIVComplex3 - Informal lecture notes for Dec 4 Now that...

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Informal lecture notes for Dec. 4 Now that we have a solid background in the basics of complex differentiation, it’s 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 0 , t 1 , . . . , t n = b , then the corresponding Riemann sum looks like n 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 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 C f ( z ) dz = b a [ u ( x ( t ) , y ( t )) x ( t ) - v ( x ( t ) , y ( t )) y ( t )] dt + i b a [ v ( x ( t ) , y ( t )) x ( t ) + u ( x ( t ) , y ( t )) y ( t )] dt = C u dx - v dy + i 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 π/ 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
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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 we’re about to integrate this expression. We have C f ( z ) dz = π/ 2 0 ( sin t - 4 sin t cos 2 t ) dt + i π/ 2 0 ( cos t - 4 cos t sin 2 t ) dt = - cos t + 4 3 cos 3 t π/ 2 0 + i sin t - 4 3 sin 3 t π/ 2 0 = - 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 π/ 2. Then we have f ( z ( t )) = e 2 it . Further, we see that dz = ie it dt , and thus C f ( z ) dz = i π/ 2 0 e 3 it dt = i π/ 2 0 cos(3 t ) dt - π/ 2 0 sin(3 t ) dt = i 1 3 sin(3 t ) π/ 2 0 + 1 3 cos(3 t ) π/ 2 0 = - 1 3 i - 1 3 . Note that here we wrote f ( z ( t )) = e 2 it = cos(2 t ) + i sin(2 t ), while above we had that this was equal to cos 2 t - sin 2 t + i (2 cos t sin t ). However, these can be seen to be equal using the double-angle formulas. In fact (to continue this digression one more step), the relationship between the exponential and trig functions in complex analysis provides an efficient way to derive and/or remember various trig identities.
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