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Is f(z)=x differentiable with respect to z?

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data

    The only thing given is f(z)=x. However, I am under the assumption that z is a complex variable where z=x+iy. I'm also assuming that x is a real variable.

    In this example, I know that f(z)=x is not differentiable with respect to z because it does not satisfy the Cauchy-Riemann Equations but I need to prove this using the limit definition of the derivative.

    2. Relevant equations

    I used the definition of the limit i.e. limit as h→0 [f(x+h)-f(x)]/h however I'm not quite sure what f(x+h) translates into in this problem.

    3. The attempt at a solution

    this may be a straight forward question that I'm over thinking but I get lim Δz→0 [Δx/Δz] which does not exist. Is that correct?

    This is my first post. I don't know how to enter math notation that is easier to read. Help with that would also be appreciated.
     
  2. jcsd
  3. Dec 5, 2011 #2

    HallsofIvy

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    Remember that z is a complex number so the limits are taken in the complex plane. That is two-dimensional so there are an infinite number of ways of taking a limit, not just "from below" and "from above".

    The derivative at, say, [itex]z_0= x_0+ iy_0[/itex], is given by [itex]\lim_{h\to 0}\frac{f(z_0+ h)- f(z_0)}{h}[/itex] and h itself is a complex number.

    Suppose we take the limit as h approaches 0 along the a line parallel to the real axis. Write f(z)= u(x,y)+ iv(x,y), we have u(x,y)= x and v(x,y)= 0.

    [tex]\lim{h\to 0}\frac{u(x_0+h,y_0)+ iv(x_0+h,y_0)- u(x_0,y_0)- iv(x_0,y_0}{h}= \lim_{h\to 0}\frac{x_0+ h- x_0}{h}= 1[/tex]

    Now take the limit as h goes to 0 parallel to the imaginary axis: h is now ih:
    [tex]\lim_{h\to 0}\frac{u(x_0, y_0+ h)+ iv(x_0, y_0+h)- u(x_0,y_0)- iv(x_0,y_0)}{ih}= \lim_{h\to 0}\frac{x_0- x_0}{ih}= 0[/tex]
    since those are not the same, the limit as h goes to 0 along different paths is different. The limit itself does not exist.
     
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