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Z=x+iy; f(z)=u+iv; f'(z)=?; Cauchy-Riemann Conditions

  1. Oct 1, 2015 #1
    Mod note: Fixed the broken LaTeX in the following, and edited the difference quotient definition of f'(z).
    The following is from a mathematical introduction to a physics book:

    "The real part u and the imaginary part v of w=u+i v are functions of the two variables x and y. Nevertheless, two arbitrary functions u(x,y) and v(x,y) cannot, in general, be considered to be the real and imaginary parts of a function of a complex variable. If the functions u and v originate in a complex function, they satisfy certain special conditions. If we denote by f'(z) the derivative of the function f with respect to its argument, then, since u+i v=f(x+i y)=f(z),

    [itex] \frac{\partial u}{\partial x}+ i \frac{\partial v}{\partial x}=f'(z)\frac{\partial z}{\partial x}=f'(z) [/itex] ;

    [itex] \frac{\partial u}{\partial y}+\imath\frac{\partial v}{\partial y}=f'(z)\frac{\partial z}{\partial y}=\imath f'(z) [/itex].

    Comparison gives the Cauchy-Riemann Differential Equations

    [itex]\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}[/itex]

    [itex]\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}[/itex]."

    All of this makes sense to me except what [itex]f'(z)[/itex] means.

    Simply doing what I do with functions of real numbers, I write:

    [tex]z=x+\imath y[/tex]
    [tex]f(z)=u+\imath v[/tex]
    [tex]f'(z)=\lim_{\Delta z\to 0} \frac{f(z + \Delta z)-f (z)}{\Delta z}[/tex]

    But [itex]\Delta z[/itex] seems to be an infinite set of small displacements of the form [itex]\Delta x+\imath \Delta y[/itex]. What the book seems to be saying is that a continuously differentiable complex function of a complex argument is not simply a continuously differentiable mapping of a region of [itex]R^2[/itex] onto another region of [itex]R^2[/itex]. There are some other conditions implied.

    So my question is, what is meant by f'(z)?
     
    Last edited by a moderator: Oct 1, 2015
  2. jcsd
  3. Oct 1, 2015 #2
    I can't fix the formatting right now. I hope to return to this tomorrow. Sorry for the mess.
     
  4. Oct 1, 2015 #3

    Mark44

    Staff: Mentor

    I fixed it. You had many of your closing itex and tex tags with the wrong slash. The right ones are [ /itex] and [ /tex] (without the extra spaces). For myself, I find it easier to use a pair of # characters at the beginning and end (for inline LaTeX) or a pair of $ characters at beginning and end (for standalone LaTeX).
     
  5. Oct 1, 2015 #4

    WWGD

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    I am not sure of what you are looking for, but your function will be differentiable as a function from ## \mathbb R^2 \rightarrow \mathbb R^2 ## _and_ will be analytic , in the sense that it will be infinitely - differentiable and it will be analytic, i.e., at each point ## z_0 ## where ## f ## is analytic as a Complex function, it will also be analytic in the sense that ## f(z_0)## will equal (convergence -wise) its Taylor series at that point. f will also, unlike some Real-analytic functions, be an open map.
     
    Last edited: Oct 1, 2015
  6. Oct 1, 2015 #5

    FactChecker

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    The key point is that f'(z) must be a single complex number that represents a local complex multiplication of the image around f(z). That is, the image of a small neighborhood around z gets mapped to a neighborhood of f(z) that is scaled (expanded or shrunk) by the amount |f'(z)| and rotated evenly by the angle arg(f'(z)). The consequence is that the partial derivatives must be related by the Cauchy Riemann equations.
     
  7. Oct 1, 2015 #6

    WWGD

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    Right, just to extend a bit what Fact Checker said, notice that the scaled Jacobian ( scaling by |f'(z)| ) is a rotation matrix.
     
  8. Oct 6, 2015 #7
    [Apologies for the length of this comment.]

    For a function f: ℝ → ℝ, its derivative c = f'(x0) (if f'(x) exists in some open interval containing x0), the value of c tells you that near x0, applying f to a point is close to multiplying by c. More precisely:

    f(x) - f(x0) = c(x-x0), approximately.

    Multiplication by the real number c = f'(x0) tells you the factor of "infinitesimal magnification" of applying f, at the point x0. Also, the sign of c says whether f reverses the direction of a small interval about x0.

    Knowing that f'(x) exists in an open interval containing x0 does not say anything about the existence of higher derivatives, like f''(x), which might fail to exist at any value of x.

    If instead g: ℂ → ℂ has a derivative c = g'(z0) (if g'(z) exists in some open disk containing z0), the value of c tells you that near z0, applying g to a point is close to multiplying by c. More precisely:

    g(z) - g(z0) = c(z-z0), approximately.

    Multiplication by the complex number c, in this case, tells you the infinitesimal magnification of applying g at z0 — given by |c| again. But c also tells you the infinitesimal rotation at z0, given by the unit complex number c/|c|.

    As WWGD mentions above, when g'(z) exists in an open disk D of ℂ, then — amazingly — all higher derivatives g''(z), g'''(z), ..., g(n)(z), ... also exist in that open disk! This is intuitively explained by your comment that "Δz seems to be an infinite set of small displacements of the form Δx+ıΔy." The fact that the definition of the derivative g'(z) exists is much stronger than for a real-valued function like f(x). It even follows that the power series for g(z) must exist and converge in the open disk D.
     
  9. Oct 7, 2015 #8
    OK, I finally gained a fair understanding of what is meant by [itex]f'(z)[/itex]. I haven't seen it written out formally, but this ancient lecture by Dr. Herbert Gross of MIT tells me that the derivative of a complex function of a complex argument is defined to be a unique complex value resulting from

    [itex]f'(z)=\lim_{\Delta z\to 0} \frac{f(z + \Delta z)-f (z)}{\Delta z}[/itex].

    That is, the result is the same, no mater what direction in the complex plane we approach the point [itex]z[/itex] from.



    I'm sure there are far more accurate and exact ways of stating this, but it's a good enough "rough and ready" understanding to allow me to move on.
     
  10. Oct 7, 2015 #9
    My apologies. Since that limit-of-the-difference-quotient formula was in your original post, I thought you already knew that was the definition of f'(z).

    (It's not only in that video, but in virtually all books and notes about complex variables.)

    Actually, it's nice that you apparently rediscovered the correct definition of differentiating a function of a complex variable!
     
    Last edited: Oct 7, 2015
  11. Oct 7, 2015 #10

    WWGD

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    OP: Maybe over the run, after doing some Complex Analysis, you want to find functions that are Real-differentiable but not Complex-analytic. You can also try to find functions that are Real-analytic , meaning each of (f_1,f_2) (think of (f_1,f_2)--> f_1+if_2) is an analytic function , but that are not Complex-Analytic -- meaning they do not satisfy Cauchy-Riemann.
     
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