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Analytic function

  1. Nov 29, 2013 #1
    How can I find all analytic functions f=u+iv with u(x,y)=(x^2)+(y^2)

    Thanks for the help. I appreciate it.
     
  2. jcsd
  3. Nov 29, 2013 #2

    lurflurf

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    Require that v satisfy the Cauchy–Riemann equations
    $$\dfrac{\partial v}{\partial x}=-\dfrac{\partial u}{\partial y}\\
    \dfrac{\partial v}{\partial y}=\phantom{-}\dfrac{\partial u}{\partial x}$$
     
  4. Nov 29, 2013 #3
    but it doesn't satisfy except (0,0)
     
  5. Nov 30, 2013 #4

    mathwonk

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    note that the cauchy riemann equations imply that ∂^2(u)/∂x^2 + ∂^2(u)/∂y^2 = 0. but that is false for your example, so there are no such analytic functions. i.e. both u and v must be "harmonic" functions in order for u + iv to be analytic, and your u is not harmonic. try u = X^2 - Y^2.
     
  6. Dec 1, 2013 #5
    is it anaytic at (0,0) ?
     
  7. Dec 1, 2013 #6

    HallsofIvy

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    Is what analytic at (0,0)? You asked about a function u+ iv, with [itex]u= x^2+ y^2[/itex].

    As lurflurf said, use the Cauchy-Riemann equations- if f(z)= u(x,y)+ iv(x,y), z= x+ iy is analytic then
    [tex]\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}[/tex]
    [tex]\frac{\partial v}{\partial x}= -\frac{\partial u}{\partial y}[/tex]

    Here, [itex]\partial u/\partial x= 2x[/itex] and [itex]\partial u/\partial y= 2y[/itex] so we must have
    [tex]\frac{\partial v}{\partial y}= 2x[/tex]
    [tex]\frac{\partial v}{\partial x}= -2y[/tex]
    From the second equation, [itex]v= -2xy+ f(x)[/itex] for some function, f, of x alone. Differentiating that with respect to x, [itex]v_x= -2y+ f'(x)= 2x[/itex] which is impossible. There cannot be an analytic function with real part [itex]x^2+ y^2[/itex].
     
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