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

#### inversquare

$$z\in\mathbb C$$

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

$$lim_{Δz\rightarrow 0} \frac{f(z+Δz)-f(z)}{Δz}$$

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

$$u(x,y) = x^2 + y^2$$ and $$v(x,y) = 0$$

Then,

$$u_x ≠ v_y$$ and $$u_y ≠ - v_x$$

?

Thanks!

Last edited:
Related Topology and Analysis News on Phys.org

#### mfb

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

#### inversquare

$$u(x) = \sqrt{x^2+y^2}$$

Correct?

Any idea how to prove it using the limit approach?

#### inversquare

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

#### da_nang

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.

#### jk22

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.

Last edited:

#### da_nang

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.

Last edited:

#### Svein

Since |z| is real, $\frac{\partial \lvert z \rvert}{\partial y} = 0$. Cauchy-Riemann implies that $\frac{\partial \lvert z \rvert}{\partial x}$ 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...

#### inversquare

$$\lim_{\Delta z\to 0}\frac{|z_0+\Delta z|-|z_0|}{\Delta z}$$

By triangle inequality,

$$|z_0+\Delta z|-|z_0| \leq |\Delta z|$$

Using special case where,

$$|z_0+\Delta z|-|z_0| = |\Delta z|$$

Gives,

$$\lim_{\Delta z\to 0}\frac{|\Delta z|}{\Delta z}$$

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?

#### Svein

Gives,

limΔz→0|Δzz
Writing $z=re^{i\phi}$, $\vert z \vert = r$. Thus $\frac{\vert z \vert}{z} = e^{-i\phi}$....

#### WWGD

Gold Member
Actually, it is differentiable at z=0 but nowhere analytic , because there is no open set where C-R is satisfied.

#### mfb

Mentor
Since |z| is real, $\frac{\partial \lvert z \rvert}{\partial y} = 0$.
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

Gold Member
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?

#### mfb

Mentor
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

Gold Member
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.

Last edited:

#### Svein

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 $f(z)=u(z) + iv(z)$, with u and v both real. When $f(z)=\vert z \vert$, $v=0$ for all z, therefore $\frac{\partial v}{\partial y} = 0$ and $\frac{\partial v}{\partial x} = 0$. If $\vert z \vert$ were analytic, Cauchy-Riemann would force $\frac{\partial u}{\partial y} = 0$ and $\frac{\partial u}{\partial x} = 0$, which would imply that $\vert z \vert$ were a constant.

#### WWGD

Gold Member
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.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving