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The strength of the source is equal to the total flux

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Unformatted text preview: )/2 + i (1 - cos ø)/2 ] = [r + r cos ø]/2 + i [r - r cos ø]/2 2 2 1/2 2 2 1/2 = [(x +y ) + x]/2 + i [(x +y ) - x]/2 1/2 (This is only valid in the upper half plane, since both of z coords are positive..) So now take B = 1 for simplicity (!) and look at 2 1/2 22 1/2 (z - 4) = (x -y - 4 + i2xy) = 2 2 2 2 2 1/2 (([x -y -4] + 4x y ) 2 2 2 2 2 + x -y - 4)/2 2 2 1/2 2 2 + i (([x -y -4] + 4x y ) - x +y + 4)/2 -1 adding this to z = x+iy and dividing by 2 now gives F (z) in terms of Cartesian coordinates. The flow lines we want are horizontal lines in H: z(t) = t + iK, where K is the height of the line. For simplicity (!) let us take B = 1, and then compute the real and imaginary -1 parts of F (t + iK): 44 2 2 2 2 2 1/2 t + ({[t -K -4] + 4t K } Real part = 2 2 2 2 2 2 + t - K - 4)/2 2 2 1/2 2 2 K + ({[t -K -4] + 4t K } - t + K +4)/2 Imaginary part = 2 To plot them technologically, use the following formulas for x and y: x = ( t + S QRT((((t^2-k^2-4)^2+4*t^2*k^2)^.5+t^2-k^24)/2))/2 y = (k + SQRT((((t^2...
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