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

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

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

Correct?

Any idea how to prove it using the limit approach?

Chad
 
Or rather, any idea how to prove it is differentiable nowhere?
 
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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.
 
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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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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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Svein

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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...
 
[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
 

WWGD

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Actually, it is differentiable at z=0 but nowhere analytic , because there is no open set where C-R is satisfied.
 
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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).
Actually, it is differentiable at z=0
What is its derivative? Not even the restriction to the real values is differentiable there.
 

WWGD

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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?
 
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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.
 

WWGD

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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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Svein

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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.
 

WWGD

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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.
 

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