ComplexVariable

ComplexVariable - Functions of a Complex Variable: Contour...

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Functions of a Complex Variable: Contour Integration and Steepest Descent Michael Fowler 9/19/08 Analytic Functions Suppose we have a complex function f = u + iv of a complex variable z = x + iy , defined in some region of the complex plane, where u , v , x , y are real. That is to say, () (, ) , f z uxy i vxy = + with u ( x , y ) and v ( x , y ) real functions in the plane. We now assume that in this region f ( z ) is differentiable, that is to say, 0 ( ) lim z df z f z z f z dz z Δ→ + Δ− = Δ is well-defined. What does this tell us about the functions u ( x , y ) and v ( x , y ), the real and imaginary parts of f ( z )? In fact, the property of differentiability for a function of a complex variable tells us a lot! It does not just mean that the function is reasonably smooth. The crucial difference from a function of a real variable is that Δ z can approach zero from any direction in the complex plane, and the limit in these different directions must be the same . Of course, there are only two independent directions, so what we are really saying is ()() , f xi y fxi y y +∂ + = ∂∂ which we can write in terms of u , v : ( , )( , ) ( , , . u x yv x y u x x y ii ) x y i y += + Equating real and imaginary parts of this equation we find: , . uv v u x yx y == These are called the Cauchy-Riemann equations . It immediately follows that both u ( x , y ) and v ( x , y ) must satisfy the two-dimensional Laplacian
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2 equation, 22 (, ) 0, that is, 0 and 0. uxy uv xy ∂∂ += = = Notice that this implies (just as for an electrostatic potential) that u ( x , y ) cannot have an absolute minimum or maximum inside the region of analyticity. If df ( z )/ dz = 0, but the second-order partial derivatives are nonzero, then they must have opposite sign, signaling a saddlepoint. In the general case, a two-dimensional version of Gauss’ theorem can be used to show there is no local extremum. Furthermore, ., . , uu vv ⎛⎞ ∂∂ ∂∂ ∇∇= = ⎜⎟ ⎝⎠ 0 . That is to say, the contour lines of constant u(x,y) are everywhere orthogonal to the contour lines of constant v(x,y) . (The gradient being orthogonal to the contour lines everywhere.) The important point is that just requiring differentiability of a function of a complex variable imposes a strong constraint on its real and imaginary parts, the functions u ( x , y ) and v ( x , y ). A Simple Example: f ( z ) = z 2 . It is worthwhile building a clear picture of the real and imaginary parts of the function z 2 . The real part is x 2 y 2 , and its contour lines in the square 1 to 1 are shown below. The darker shades are the lower ground . At the origin, there is a saddlepoint with higher ground in both directions of the real axis, lower ground in the pure imaginary directions. The lines x = y , x = y (not shown) are contours at the same level (zero) as the origin. What about the imaginary part? Im z 2 = 2 xy has contours:
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3 Putting the two sets of contour lines on the same diagram it is clear that they always cut each other orthogonally: (Incidentally, this picture has a physical realization. It represents the field lines and equipotentials of a quadrupole magnet, used for focusing beams of charged particles.) Another Example: f ( z ) = 1/ z The definition of differentiation above can be used to show that 2 11 d dz z z = −
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This note was uploaded on 12/07/2011 for the course PHYSICS 751 taught by Professor Michaelfowler during the Fall '07 term at UVA.

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ComplexVariable - Functions of a Complex Variable: Contour...

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