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Cauchy Riemann relations

  1. Oct 28, 2006 #1
    The cauchy Riemann relations can be written:

    [tex]\frac{\partial f}{\partial \bar{z}}=0[/tex]

    Is there an 'easy to see reason' why a function should not depend on the independent variable [itex]\bar{z}[/tex] to be differentiable?
     
  2. jcsd
  3. Oct 28, 2006 #2
    What exactly do you mean? Do you want to take the derivative of the complex part?

    The Cauch-Riemann equations can be used to see the following: A holomorphic function [itex]f:\mathbb{C}\rightarrow\mathbb{C}[/itex] is defined by its real part, plus some constant function.

    Proof: Let's look at two holomorphic functions [itex]f,g[/itex]. Let [itex]Re(f)=Re(g)[/itex], then look at the new function [itex]h[/itex] defined by [itex]h:=f-g[/itex]. From the properties of complex differentiation we know that this function [itex]h[/itex] also is complex differentiable and because of [itex]Re(f)=Re(g)[/itex], we know that [itex]Re(h)=0[/itex]. Let [itex]h=u+iv[/itex] for some real functions [itex]u,v[/itex] (You should know that this can be done, otherwise consult a complex analysis book like Rudin). In our special case we know that [itex]u=0[/itex], and therefore any derivative of [itex]u[/itex] is also equal to 0. The Cauchy-Riemann equations now state that [itex]0=\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}[/itex] and [itex]0=\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}[/itex]. That's it. Both partial derivatives of [itex]v[/itex] are equal to 0, so the imaginary part of [itex]h[/itex] is constant, and thus the difference of [itex]f[/itex] and [itex]g[/itex] is constant.

    Is it clear?
     
    Last edited: Oct 28, 2006
  4. Oct 28, 2006 #3

    matt grime

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    It is a definition, Weimster. I'm not sure what you're asking. It is the definition of analytic.
     
  5. Oct 28, 2006 #4
    Is it? I thought from the text that a function is called differentiable in z_0 if the limit

    [tex]lim _{\Delta z \rightarrow 0} \frac{f(z_0+\Delta z)-f(z_0)}{\Delta z}[/tex]

    exists and is independent of how you approach z_0

    So I considered the CR equations not as the definition of , but as a necessary condition for differentiablility. In that sense, if a function satisfies CR for a certain z_0 this imples that the above limit exists and is independent of how z-) is approached.

    Now my question was how this is related to the given formulation of CR.

    Why is the limit independent of approach when the function is independent of [tex]\bar{z}[/itex]?
     
  6. Oct 28, 2006 #5
    It is clear, but can you also use this argumentation to show that an analytic function is a function of z plus some constant, to show that

    [tex]\frac{\partial f}{\partial \bar{z}}=0[/tex]
     
  7. Oct 28, 2006 #6

    matt grime

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    I wish you wouldn't say 'indepedent of z bar'. That isn't how to think of it. z and z bar are not independent variables - if I give you z you know what z bar is. Intuition from real analysis isn't much help in complex analysis. If a function is once complex differentiable it is infinitely complex differentiable, for instance, which is definitely not true in real analysis. Analytic functions correspond to the subset of harmonic functions.

    In any case, it is normally proved, when the book decides to use the limit definition above, that this is if and only if the CR equations are satisfied. Doesn't it do so in your book? I imagine a google search will produce the link.
     
    Last edited: Oct 28, 2006
  8. Oct 28, 2006 #7

    matt grime

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  9. Oct 28, 2006 #8
    It says in my book that they can be treated as independent variables!
     
  10. Oct 28, 2006 #9

    matt grime

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    Here's an example of why you shouldn't think like that, or indeed several examples.

    consider f(z) = |z|^2, is that analytic? There's not z bar in there is there? It is 'independent of z bar'... oh, wait, |z|^2=zz* (again, star for conjugate), so actually, it is 'dependent on z bar'...

    Now consider sin(z*)*. That depends on z bar right? But it's actually complex differentiable, so it can't be.... Go figure.
     
    Last edited: Oct 28, 2006
  11. Nov 6, 2006 #10

    mathwonk

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    the zbar part of the derivative is the conjugate lienar part of the real lienar approximation,a nd the zm part is the compklex linear part. by definition, a conplex function is compklex differentiqble if it has a compelx linear approximation, hence the compelx conjugate part of the linear approximation should be zero.

    riemann himself gavea slightly more intuitive explantion which i forget at the moment. (what does he know?)
     
  12. Nov 7, 2006 #11

    mathwonk

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    i.e. zbar and z are both real linear functions from C to C. and they span the vector space of such real linear functions as a complex vector space.

    A differentiable function f from C to C has a real linear derivative, which is a real linear map C-->C, and hence can be expressed as a linear combination of zbar and z. The coefficient of zbar is called df/dzbar, while the coefficient of z is called df/dz.

    Hence if df/dzbar = 0 then the derivative is complex linear.

    the point is that df/dzbar is not defined as a limit wrt some "variable" zbar, but as a coefficient of a basis vector in a linear space.
     
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