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Then look at the real and imaginary parts of fz i

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Unformatted text preview: =∂y ∂x If f: DÆC is analytic, then the partial derivatives I ∂u ∂v = ∂x ∂y and Conversely, if u(x, y) and v(x, y) are have continuous first-order partial derivatives in D and satisfy the Cauchy-Riemann conditions on D, then f is analytic in D with ∂u ∂v ∂u ∂u ∂v ∂u ∂v ∂v f'(z) = +i = -i = -i = +i ∂x ∂x ∂x ∂y ∂y ∂y ∂y ∂x Note that the second equation just above says that f'(z) is the complex conjugate of the gradient of u(x, y) Proof Suppose f: DÆ C is analytic. Then look at the real and imaginary parts of f'(z) I using ∆z = ∆x, and ∆z = i∆y. We find: ∂u ∂v ∆z = ∆x: f'(z) = +i ∂x ∂x ∂v ∂u ∆z = i∆y f'(z) = -i ∂y ∂y Equating coefficients gives us the result. Proving the converse is beyond the scope of this course. (Basically, one proves that the above formula for f'(z) works as a derivative.) Examples 2 2 Show that f(z) = x - y i is nowhere analytic. Now let us fiddle with the CR equations. Start with ∂u ∂v ∂u ∂v = and =∂x ∂y ∂y ∂x and take ∂/∂x of both sides of the first, and ∂/∂y of the second: 2 2 2 2 ∂u ∂v ∂u ∂v and 2 = ∂x∂y 2 = - ∂x∂y ∂x ∂y Combining these gives 2 ∂u 2 ∂u 2+ 2=1 ∂x ∂y u is harmonic Similarly, we see that v is harmo...
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