Examples 41 2 a evaluate z dz where c x 3t y t

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Unformatted text preview: z as dx + idy we obtain Û Û Ù f(z) dz = Ù (u + iv)(dx + idy) ı ı C C Û Û = Ù u dx - v dy + iÙ u dy + v dx ı ı C C where the real and imaginary parts are just ordinary path-integrals, as in Calc 3. In fact, if we think of f(z) as a vector field “u, v‘, then Û Û Û Ù f(z) dz = Ù f(z) .dr - i Ù i f(z) .dr ı ı ı C C C (Note that the is do not cancel since we are thinking of things as vector fields here.) However, to evaluate it, we need not go so far, but instead stay with complex numbers: 12 b Û Û Ù f(z) dz = Ù f(z(t)) z'(t) dt ı ı a C where z(t) = x(t) + iy(t). A consequence of this is that, if f(z) has an antiderivative in D, then Û Ù O f(z) dz = 0 ı C over any closed contour C. Question Why? Answer Write f(z) = F'(z), and so b b Û Û Û Ù O f(z) dz = Ù f(z(t)) z'(t) dt = Ù F'(z(t)) dt ı ı ı a a C = F(z(b)) - F(z(a)) = 0 since z(b) = z(a) for a closed contour. Examples 4.1 Û 2 (A) Evaluate Ù z– dz, where C: x = 3t, y = t ; -1 ≤ t ≤ 4 ı C Û1 Ù (B) Evaluate O dz, where C is the unit circle centered at the origin, traversed counterız C it clockwise. To make it easier, use polar coordinates: Write the curve as z = e with 0 ≤ t it ...
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This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

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