Why is f(z) = |z| not holomorphic at (0,0)?

[fauboca]

$$f(z) = |z|$$

By the Cauchy-Riemann equations,

$$u_x = \frac{x}{\sqrt{x^2+y^2}}$$

$$v_y = -v_x = 0$$

$$u_y = \frac{y}{\sqrt{x^2+y^2}}$$

Since the C.R. equations don't work at (0,0), how can show $f(z)$ is not holomorphic at (0,0)?

 

[Dick]

$$f(z) = |z|$$

By the Cauchy-Riemann equations,

$$u_x = \frac{x}{\sqrt{x^2+y^2}}$$

$$v_y = -v_x = 0$$

$$u_y = \frac{y}{\sqrt{x^2+y^2}}$$

Since the C.R. equations don't work at (0,0), how can show $f(z)$ is not holomorphic at (0,0)?

You mean you are claiming the CR equations DO WORK at z=0, right? Actually, they don't. u isn't differentiable with respect to x at (0,0). u(x,0)=|x|.

 

[fauboca]

You mean you are claiming the CR equations DO WORK at z=0, right? Actually, they don't. u isn't differentiable with respect to x at (0,0). u(x,0)=|x|.

I didn't claim they work at z = 0. I asked how to show they don't.

 

[susskind_leon]

A function is complex differentiable if their partial derivatives for u and v exist and they satisfy the C-R-eq. Since the p.d. for u do not exist, f(z) is not complex differentiable (in z=0). This means that f(z) is not holomorphic in z=0.

 

[fauboca]

A function is complex differentiable if their partial derivatives for u and v exist and they satisfy the C-R-eq. Since the p.d. for u do not exist, f(z) is not complex differentiable (in z=0). This means that f(z) is not holomorphic in z=0.

So just take the limit of f(z) approaching from the x and y-axis to show they limits are different. Thus, f(z) is not differentiable at z = 0?

 

[Dick]

So just take the limit of f(z) approaching from the x and y-axis to show they limits are different. Thus, f(z) is not differentiable at z = 0?

That's the way to show f(z) is discontinuous at z=0. It's not. Write down a difference quotient for the x derivative of u(x,y) at z=0. (u(0+h,0)-u(0,0))/h. Show different ways of letting h->0 lead to different limits.

 

[fauboca]

That's the way to show f(z) is discontinuous at z=0. It's not. Write down a difference quotient for the x derivative of u(x,y) at z=0. (u(0+h,0)-u(0,0))/h. Show different ways of letting h->0 lead to different limits.

This

$$\lim_{h\to 0}\frac{\sqrt{(x+h)^2+y^2}-\sqrt{x^2+y^2}}{h}$$

??

 

[Dick]

This

$$\lim_{h\to 0}\frac{\sqrt{(x+h)^2+y^2}-\sqrt{x^2+y^2}}{h}$$

??

Yes. You were wondering what happens at z=0. So put x=0 and y=0.

 

[fauboca]

Yes. You were wondering what happens at z=0. So put x=0 and y=0.

That is what I essentially did but my professor wrote, I need to consider (0,0) separately.

 

[Dick]

That is what I essentially did but my professor wrote, I need to consider (0,0) separately.

We ARE doing (0,0) separately. Away from (0,0) you can use your formulas for u_x and u_y. At (0,0) we are looking at the difference quotient. BTW what do you conclude from that?

 

[fauboca]

We ARE doing (0,0) separately. Away from (0,0) you can use your formulas for u_x and u_y. At (0,0) we are looking at the difference quotient. BTW what do you conclude from that?

The limit is $\pm 1$.

What I was saying is on an assignment that is what my professor wrote even though I showed the $\pm 1$ too.

 

[Dick]

The limit is $\pm 1$.

What I was saying is on an assignment that is what my professor wrote even though I showed the $\pm 1$ too.

Since you can find two different limits that actually means the limit doesn't exist and the function isn't differentiable. Better have a talk with your professor about what is actually required here.