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

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In summary: 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.
  • #1
fauboca
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[tex]f(z) = |z|[/tex]

By the Cauchy-Riemann equations,

[tex]u_x = \frac{x}{\sqrt{x^2+y^2}}[/tex]

[tex]v_y = -v_x = 0[/tex]

[tex]u_y = \frac{y}{\sqrt{x^2+y^2}}[/tex]

Since the C.R. equations don't work at (0,0), how can show [itex]f(z)[/itex] is not holomorphic at (0,0)?
 
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  • #2
fauboca said:
[tex]f(z) = |z|[/tex]

By the Cauchy-Riemann equations,

[tex]u_x = \frac{x}{\sqrt{x^2+y^2}}[/tex]

[tex]v_y = -v_x = 0[/tex]

[tex]u_y = \frac{y}{\sqrt{x^2+y^2}}[/tex]

Since the C.R. equations don't work at (0,0), how can show [itex]f(z)[/itex] 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|.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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?
 
  • #6
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.
 
  • #7
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

[tex]\lim_{h\to 0}\frac{\sqrt{(x+h)^2+y^2}-\sqrt{x^2+y^2}}{h}[/tex]

??
 
  • #8
fauboca said:
This

[tex]\lim_{h\to 0}\frac{\sqrt{(x+h)^2+y^2}-\sqrt{x^2+y^2}}{h}[/tex]

??

Yes. You were wondering what happens at z=0. So put x=0 and y=0.
 
  • #9
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 [itex]\pm 1[/itex].

What I was saying is on an assignment that is what my professor wrote even though I showed the [itex]\pm 1[/itex] too.
 
  • #12
fauboca said:
The limit is [itex]\pm 1[/itex].

What I was saying is on an assignment that is what my professor wrote even though I showed the [itex]\pm 1[/itex] 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.
 

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

What does it mean for a function to be holomorphic at (0,0)?

Being holomorphic at a point means that the function is differentiable at that point and in a neighborhood around it. In other words, the function is smooth and has no sharp turns or corners at that point.

How can I determine if a function is holomorphic at (0,0)?

To determine if a function is holomorphic at (0,0), you can use the Cauchy-Riemann equations. These equations relate the partial derivatives of a function in terms of its real and imaginary parts. If the partial derivatives satisfy these equations at (0,0), then the function is holomorphic at that point.

What is the significance of holomorphicity in complex analysis?

Holomorphic functions play a crucial role in complex analysis because they are the most well-behaved functions in the complex plane. They have many useful properties, such as being analytic and having a power series representation, which make them easier to work with in mathematical calculations and applications.

Can a function be holomorphic at (0,0) but not in its entire domain?

Yes, a function can be holomorphic at (0,0) but not in its entire domain. This means that the function may be differentiable at (0,0) but not in other parts of its domain. For example, a function with a singularity at (0,0) cannot be holomorphic in its entire domain.

How can I use the concept of holomorphicity to solve problems in complex analysis?

Holomorphicity is a useful concept in complex analysis because it allows us to simplify complex functions and make them easier to work with. By understanding the properties of holomorphic functions, we can use techniques such as Cauchy's integral theorem and the Cauchy integral formula to solve problems involving contour integrals and complex functions.

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