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Unformatted text preview: 2 Since dh dy = 3 y 2 , then h ( y ) = y 3 + c . Now we have the harmonic conjugate, v ( x, y ) =-3 x 2 y + y 3 + c (where c is just a constant). For fun, we can write down an expression for the corresponding analytic function, f ( z ) : f ( z ) = u + iv = 3 xy 2-x 3-i (3 x 2 y ) + i ( y 3 ) + c , which can be condensed as f ( z ) =-z 3 + c for z = x + iy . (Hint on condensing: since our expression for f ( z ) has terms that go like x 3 and y 3 , a natural place to start is to compare our expression to that of z 3 . If you factor out z 3 carefully, youll get z 3 = z 2 * z = (( x 2-y 2 ) + i (2 xy )) * ( x + iy ) = x 3-xy 2 + i ( x 2 )-y 3 ) + i (2 x 2 y )-2 xy 2 = x 3-3 xy 2-i ( y 3 ) + i (3 x 2 y ) , which is almost exactly our expression for f ( z ) , but just off by a - sign.)...
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