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Modulus of z

  1. Feb 1, 2012 #1
    [tex]f(z) = |z|[/tex]

    By the Cauchy-Riemann equations,

    [tex]u_x = \frac{x}{\sqrt{x^2+y^2}}[/tex]

    [tex]v_y = -v_x = 0[/tex]

    [tex]u_y = \frac{y}{\sqrt{x^2+y^2}}[/tex]

    Since the C.R. equations don't work at (0,0), how can show [itex]f(z)[/itex] is not holomorphic at (0,0)?
     
  2. jcsd
  3. Feb 1, 2012 #2

    Dick

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    You mean you are claiming the CR equations DO WORK at z=0, right? Actually, they don't. u isn't differentiable with respect to x at (0,0). u(x,0)=|x|.
     
  4. Feb 1, 2012 #3
    I didn't claim they work at z = 0. I asked how to show they don't.
     
  5. Feb 1, 2012 #4
    A function is complex differentiable if their partial derivatives for u and v exist and they satisfy the C-R-eq. Since the p.d. for u do not exist, f(z) is not complex differentiable (in z=0). This means that f(z) is not holomorphic in z=0.
     
  6. Feb 1, 2012 #5
    So just take the limit of f(z) approaching from the x and y axis to show they limits are different. Thus, f(z) is not differentiable at z = 0?
     
  7. Feb 1, 2012 #6

    Dick

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    That's the way to show f(z) is discontinuous at z=0. It's not. Write down a difference quotient for the x derivative of u(x,y) at z=0. (u(0+h,0)-u(0,0))/h. Show different ways of letting h->0 lead to different limits.
     
  8. Feb 1, 2012 #7
    This

    [tex]\lim_{h\to 0}\frac{\sqrt{(x+h)^2+y^2}-\sqrt{x^2+y^2}}{h}[/tex]

    ??
     
  9. Feb 1, 2012 #8

    Dick

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    Yes. You were wondering what happens at z=0. So put x=0 and y=0.
     
  10. Feb 1, 2012 #9
    That is what I essentially did but my professor wrote, I need to consider (0,0) separately.
     
  11. Feb 1, 2012 #10

    Dick

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    We ARE doing (0,0) separately. Away from (0,0) you can use your formulas for u_x and u_y. At (0,0) we are looking at the difference quotient. BTW what do you conclude from that?
     
  12. Feb 1, 2012 #11
    The limit is [itex]\pm 1[/itex].

    What I was saying is on an assignment that is what my professor wrote even though I showed the [itex]\pm 1[/itex] too.
     
  13. Feb 1, 2012 #12

    Dick

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    Since you can find two different limits that actually means the limit doesn't exist and the function isn't differentiable. Better have a talk with your professor about what is actually required here.
     
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