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Analytic Functions and Equality of their Complex Derivatives

  1. May 26, 2014 #1
    1. The problem statement, all variables and given/known data

    If [itex]f[/itex] and [itex]g[/itex] are both analytic in a domain [itex]D[/itex] and if [itex]f'(z)=g'(z)[/itex] for every [itex]z\in{D}[/itex], show that [itex]f[/itex] and [itex]g[/itex] differ by a constant in [itex]D[/itex]

    2. Relevant equations

    Cauchy-Riemann Equations

    Possibly Mean Value Theorem

    3. The attempt at a solution

    I'm pretty sure I'm making a mountain out of a mole hill here, this question has stumped me for a while, and I'm certain that I'm missing some small insight that would make the answer evident.

    The central premise of the question as I understand it is, for [itex]f(x,y)=u(x,y)+iv(x,y)[/itex] and [itex]g(x,y)=a(x,y)+ib(x,y)[/itex], we should have:

    [itex]u_x(x,y)=v_y(x,y)=c_x(x,y)=d_y(x,y)[/itex]
    and
    [itex]u_y(x,y)=-v_x(x,y)=c_y(x,y)=-d_x(x,y)[/itex].

    I think that I am able to integrate these functions as one normally would, for example:

    [itex]u(x,y)=\int{u_{x}(x,y)dx}+A(y)+B[/itex]
    [itex]u(x,y)=\int{u_{y}(x,y)dy}+C(x)+D[/itex]

    where I've taken the liberty of adding the function and constant of integration outside of the integral. My impression from this is that I should be able to use this idea and the Cauchy-Riemann equations to show that all of the functions that appear from the integral vanish (or at least become constant) and that I am just left with the sum of two constants. Unfortunately I haven't actually gotten anywhere with that.


    Another approach I've tried was to consider an open disk contained [itex]D[/itex] and centered at [itex]a+ib[/itex] with an arbitrary point given by [itex]z=x+iy[/itex] and to apply the Mean Value Theorem by writing:

    [itex]u_{x}(x,y)-u_{x}(a,y)+u_{x}(a,y)-u_{x}(a,b)=(x-a)u_{xx}(X)+(y-b)u_{xy}(Y)[/itex]

    and doing this for each of [itex]u_x(x,y)=v_y(x,y)=c_x(x,y)=d_y(x,y)[/itex] and [itex]u_y(x,y)=-v_x(x,y)=c_y(x,y)=-d_x(x,y)[/itex], however, I haven't seen any path to take with this either that would be relevant to the question at hand.

    This has been bugging me, and I'm convinced there is a straightforward way of going about this.
     
  2. jcsd
  3. May 26, 2014 #2

    CAF123

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    Gold Member

    f'(z) = g'(z) on D and the derivatives exist everywhere within D. Then this means f'(z) - g'(z) vanishes everywhere within D. So...?
     
  4. May 26, 2014 #3
    Hahaha! I knew it was going to be something I would want to smack myself about!

    So let me see if I got this then:

    [itex]f'(z)-g'(z)=0[/itex], hence [itex]\int{f'(z)-g'(z)}dz=\int{(0)}dz\rightarrow{f(z)-g(z)}=C[/itex] for some complex number [itex]C[/itex], right?
     
  5. May 26, 2014 #4

    CAF123

    User Avatar
    Gold Member

    Yeah, looks fine.
     
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