How to prove that a complex function is differentiable at a point and where is it analytic in its domain

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Suppose $z_0 \in \mathbb{C}$. We say that a function $f$ is complex differentiable at $z_0$ if $\lim_{z \to z_0} \frac{f(z) - f(z_0}{z - z_0}$ exists and is finite. Implicit in this definition is the fact that complex differentiation is not defined at a single point $z_0$ but for all points in a open set $B(z_0, r)$ about $z_0$. What I am saying is that a function cannot be complex differentiable at just one point $z_0$ without being complex differentiable in an open set (neighbourhood) about that point. You stumbled upon a very good question that many students in complex analysis never think about.

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Updated on February 12, 2020

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  • Josh
    Josh over 3 years

    I know the definition of complex differentiable functions and I know the Cauchy-Reimann equations that are used to prove that a function is differentiable in general, but my question is how to prove that a complex function is differentiable at a point?

    • YoTengoUnLCD
      YoTengoUnLCD over 7 years
      It is analytic whenever it's differentiable.
  • Josh
    Josh over 7 years
    Thank you @Mustafa.. I tried to prove it using the definition you just wrote. I substituted z = x+iy in the function and used that form in the definition assuming that when z approaches 0 then x and y approach 0, but I ended up with 0/0 which requires L'hopital rule, but how do you do that with 2 variables. Do you think that I should not have used the form x+iy and use the original form given with f(z) so I can apply the rule and eliminate the undefined 0/0 ? I kinda understand that the function has to be differentiable in a neighborhood, in this case however, I am only trying to examine z=0
  • Mustafa Said
    Mustafa Said over 7 years
    @Josh, one way to determine if $f(z)$ is differentiable at $z=0$ is to apply the operator $\partial{\overline{z}} = \frac{1}{2}(\partial{x} + i \partial{y})$ to $f(z)$ and evaluate at $z=0$. If the result is non-zero, which I suspect it is, then $f(z)$ is not complex differentiable at $z=0$.
  • Josh
    Josh over 7 years
    I don't know this method. Is there another way to prove it using the definition? I also remember a theorem: Let z=x+iy and f(z)=u(x,y)+iv(x,y) on some region G containing the point z0. If f(z) satisfies the Cauchy-Riemann equations and has continuous first partial derivatives in the neighborhood of z0, then f^'(z0) exists and is given by f^'(z0)=lim_(z->z0)(f(z)-f(z0))/(z-z0), and the function is said to be complex differentiable (or, equivalently, analytic or holomorphic).
  • Josh
    Josh over 7 years
    This function is complex differentiable at a neighborhood of z = 0 and it is analytic (not sure where), but I also need to find the value of the derivative at z = 0.
  • Steven Gubkin
    Steven Gubkin over 7 years
    @Mustafa: A function can be complex differentiable at one point, and no other points. For example, $f(x+iy) = x^4+y^4$ is only complex differentiable at $0$. You are also using the notation $\partial$ and $\bar{\partial}$ incorrectly. $\bar{\partial}f = \frac{\partial f}{\partial \bar{z}} d \bar{z} = \frac{1}{2}(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y})(dx-idy)$. $\partial \bar{z}$ is not an operator, it is a $1$-form. $\partial \bar{z} =0$ since $\bar{z}$ is complex antilinear.
  • Steven Gubkin
    Steven Gubkin over 7 years
    A function is at $p$ iff there is a neightborhood of $p$ where $f$ is complex differentiable. So in my example of $f(x+iy) = x^4+y^4$ it is nowhere complex analytic, but is complex differentiable at the origin.