Analyzing the Holomorphicity of f(z) at (0,0)

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Homework Help Overview

The discussion revolves around the holomorphicity of the function f(z) = |z| at the point (0,0). Participants are exploring the implications of the Cauchy-Riemann equations and the conditions for complex differentiability.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the applicability of the Cauchy-Riemann equations at (0,0) and question the differentiability of the function. There are attempts to analyze limits approaching (0,0) from different directions and the implications of these limits on differentiability.

Discussion Status

The discussion is active, with participants providing insights into the nature of the limits and the behavior of the function at (0,0). Some guidance is offered regarding the use of difference quotients and the necessity of considering (0,0) separately. Multiple interpretations of the problem are being explored.

Contextual Notes

There is mention of a professor's feedback regarding the need to consider (0,0) separately, which adds a layer of complexity to the discussion. Participants are navigating the nuances of the definitions and conditions for holomorphicity.

fauboca
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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)?
 
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fauboca said:
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|.
 
Dick said:
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.
 
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.
 
susskind_leon said:
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?
 
fauboca said:
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.
 
Dick said:
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}

??
 
fauboca said:
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.
 
Dick said:
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.
 
  • #10
fauboca said:
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?
 
  • #11
Dick said:
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.
 
  • #12
fauboca said:
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.
 

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