Prove f(z) = |z| is not analytic

  • Thread starter inversquare
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In summary: I'm not sure what you mean there. IIRC, a function is differentiable at a point if the derivative exists at that point and is continuous at that point. A derivative exists at a point if the limit, from the definition of a derivative, exists. A limit exists iff all one-sided limits exist and are the same value. So a polar form (in 2D case anyways) would consider all paths and, if the limit wrt to the radius exists and is independent of the angle, then the function is differentiable at that point, given that it is also continuous.Right, my bad!In summary, Chad thinks that it is not too difficult to prove the limit does not exist, but he is missing something
  • #1
inversquare
17
0
[tex]z\in\mathbb C[/tex]

I imagine it is not too difficult, I'm just missing something. I need to use the limit definition to prove it,

[tex] lim_{Δz\rightarrow 0} \frac{f(z+Δz)-f(z)}{Δz} [/tex]

Alternatively, using Cauchy-Riemann conditions, am I correct to assume

[tex]u(x,y) = x^2 + y^2[/tex] and [tex] v(x,y) = 0 [/tex]

Then,

[tex]u_x ≠ v_y[/tex] and [tex]u_y ≠ - v_x[/tex]

?

Thanks!

Chad
 
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  • #2
Your u(x) is not correct, but you can use the Cauchy-Riemann equations, or show that the limit does not exist somewhere.
 
  • #3
Right, my bad!

[tex] u(x) = \sqrt{x^2+y^2} [/tex]

Correct?

Any idea how to prove it using the limit approach?

Chad
 
  • #4
Or rather, any idea how to prove it is differentiable nowhere?
 
  • #5
Try rewriting it in polar form centered around an arbitrary point. The limit exists if the value is the same for any path. Hence, the value of the limit should be independent of the angle.
 
  • #6
If you center around 0 you get independent of the angle but it is not differentiable (|r|)

I would say it is rather the opposite : if it is differentiable then it does not depend on the path.

There are examples where the function is differentiable along every direction but it is not differentiable.
 
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  • #7
jk22 said:
If you center around 0 you get independent of the angle but it is not differentiable (|r|)

I would say it is rather the opposite : if it is differentiable then it does not depend on the path.

There are examples where the function is differentiable along every direction but it is not differentiable.
I'm not sure what you mean there. IIRC, a function is differentiable at a point if the derivative exists at that point and is continuous at that point. A derivative exists at a point if the limit, from the definition of a derivative, exists. A limit exists iff all one-sided limits exist and are the same value. So a polar form (in 2D case anyways) would consider all paths and, if the limit wrt to the radius exists and is independent of the angle, then the function is differentiable at that point, given that it is also continuous.

EDIT: Granted, your statement isn't wrong from a logic standpoint. Of course differentiability implies path independence but it also implies continuity. I suppose I left out the latter in my previous post since I thought it was already understood.
 
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  • #8
Since |z| is real, [itex]\frac{\partial \lvert z \rvert}{\partial y} = 0 [/itex]. Cauchy-Riemann implies that [itex] \frac{\partial \lvert z \rvert}{\partial x} [/itex] must also be zero, which again means that for |z| to be analytic, it must be a constant. And since |z| obviously is not a constant...
 
  • #9
[tex]\lim_{\Delta z\to 0}\frac{|z_0+\Delta z|-|z_0|}{\Delta z}[/tex]

By triangle inequality,

[tex]|z_0+\Delta z|-|z_0| \leq |\Delta z|[/tex]

Using special case where,

[tex]|z_0+\Delta z|-|z_0| = |\Delta z|[/tex]

Gives,

[tex]\lim_{\Delta z\to 0}\frac{|\Delta z|}{\Delta z} [/tex]

When approaching zero on positive real axis, this limit is equal to one. When approaching zero on negative real axis, this limit is equal to -1.

Rigorous?

Chad
 
  • #10
inversquare said:
Gives,

limΔz→0|Δzz

Writing [itex] z=re^{i\phi}[/itex], [itex] \vert z \vert = r [/itex]. Thus [itex] \frac{\vert z \vert}{z} = e^{-i\phi} [/itex]...
 
  • #11
Actually, it is differentiable at z=0 but nowhere analytic , because there is no open set where C-R is satisfied.
 
  • #12
Svein said:
Since |z| is real, [itex]\frac{\partial \lvert z \rvert}{\partial y} = 0 [/itex].
This is true on the real axis only, or for the imaginary part of the expression (I guess you mean that).
WWGD said:
Actually, it is differentiable at z=0
What is its derivative? Not even the restriction to the real values is differentiable there.
 
  • #13
mfb said:
This is true on the real axis only, or for the imaginary part of the expression (I guess you mean that).

What is its derivative? Not even the restriction to the real values is differentiable there.

But it satisfies C-R there, doesn't that imply differentiability?
 
  • #14
WWGD said:
But it satisfies C-R there, doesn't that imply differentiability?
No it does not, the derivatives of the real part are not even defined.
 
  • #15
Ah, right we must have that the partial derivatives exist and are continuous. EDIT I realized
I was for some reason thinking about f(z)=z^ , i.e., f(x+iy)=x-iy for some reason instead.
 
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  • #16
mfb said:
This is true on the real axis only, or for the imaginary part of the expression (I guess you mean that).

OK. I apologize. The full text should be: In order to use the Cauchy-Riemann test, you write [itex] f(z)=u(z) + iv(z)[/itex], with u and v both real. When [itex]f(z)=\vert z \vert [/itex], [itex]v=0 [/itex] for all z, therefore [itex]\frac{\partial v}{\partial y} = 0 [/itex] and [itex]\frac{\partial v}{\partial x} = 0 [/itex]. If [itex]\vert z \vert [/itex] were analytic, Cauchy-Riemann would force [itex]\frac{\partial u}{\partial y} = 0 [/itex] and [itex]\frac{\partial u}{\partial x} = 0 [/itex], which would imply that [itex]\vert z \vert [/itex] were a constant.
 
  • #17
A correction of my last post, #15. I thought the function being used was ## f(z)=|z|^2## which is actually differentiable at ##z=0##, since the partials exist therein -- they are 2x and 2y respectively --and are continuous. Then ##f(z)## is differentiable at ##0## with derivative ## f'(z)=u_x(0,0)+iv_x(0,0)=0## but it is nowhere-analytic since , e.g., C-R is not satisfied in any open set.
 

1. What is the definition of an analytic function?

An analytic function is a complex-valued function that is differentiable at every point in its domain. This means that at every point, the limit of the function's difference quotient (also known as the derivative) exists.

2. How can we prove that f(z) = |z| is not analytic?

We can prove that f(z) = |z| is not analytic by showing that it fails to satisfy the Cauchy-Riemann equations, which are necessary conditions for a function to be analytic. In this case, the Cauchy-Riemann equations are not satisfied because the partial derivatives of f(z) with respect to x and y do not exist at the point z = 0.

3. Can you provide an example of f(z) = |z| not being analytic?

Yes, at the point z = 0, the function f(z) = |z| fails to be analytic because the partial derivatives of f(z) with respect to x and y do not exist. Therefore, the limit of the difference quotient does not exist at this point, and the function is not differentiable.

4. What is the geometric interpretation of an analytic function?

The geometric interpretation of an analytic function is that it represents a function that is smooth and continuous, without any sharp corners or edges. This means that the function's graph can be drawn without lifting the pencil from the paper, and the function's value changes smoothly and continuously as the input changes.

5. Why is it important to know if a function is analytic or not?

It is important to know if a function is analytic or not because analytic functions have many useful properties, such as being able to be represented by power series and having a well-defined derivative at every point. These properties make analytic functions easier to work with and allow for the use of powerful mathematical tools, such as Cauchy's integral theorem and the Cauchy integral formula.

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