Show a limit doesn't exists, complex case

In summary, the student is trying to solve a problem involving limits and derivatives. However, he is confused about how to do so. He is unsure of how to show that a limit doesn't exist in complex analysis.
  • #1
moxy
41
0

Homework Statement



[itex]f(z) = |z|[/itex]
I'm looking to show that [itex]f'(z)[/itex] does not exist for any [itex]z \in ℂ[/itex].

Homework Equations



[itex]f'(z) = \lim_{z_0 → 0}{\frac{f(z) - f(z_0)}{z - z_0}} [/itex]

[itex]z = x + iy = Re(z) + i Im(z)[/itex]

[itex]|z| = \sqrt{x^2 + y^2}[/itex]

The Attempt at a Solution


Clearly I just have to show that [itex]\lim_{z_0 → 0}{\frac{f(z) - f(z_0)}{z - z_0}}[/itex] does not exist. However, I'm confused about how to do this. I'm unsure of how to show that a limit doesn't exist in complex analysis.

I tried to take the limit in two cases, when [itex]Re(z) = Re(z_0)[/itex] and then again when [itex]Im(z) = Im(z_0)[/itex]. Is this the correct approach?
 
Physics news on Phys.org
  • #2
Have you done the Cauchy-Riemann equations?
 
  • #3
I was only trying to prove it with limits because it's an exercise in my complex analysis book that comes before the Cauchy-Riemann equations are introduced.

[itex]f(z) = U(x,y) + iV(x,y)[/itex]

[itex]f(z) = |z| = |x + iy| = \sqrt{x^2 + y^2}[/itex]

[itex]U(x,y) = \sqrt{x^2 + y^2}[/itex]
[itex]V(x,y) = 0[/itex]

[itex]U_x = 0.5(x^2 + y^2)^{-1/2}(2x) = x(x^2 + y^2)^{-1/2}[/itex]
[itex]U_y = y(x^2 + y^2)^{-1/2}[/itex]
[itex]V_x = 0[/itex]
[itex]V_y = 0[/itex]

[itex]U_x ≠ V_y[/itex]
[itex]U_y ≠ -V_x[/itex]However, does the derivative exist when

[itex]x(x^2 + y^2)^{-1/2} = 0[/itex]
[itex]y(x^2 + y^2)^{-1/2} = 0[/itex]

i.e. when z = 0?
 
Last edited:
  • #4
moxy said:
I was only trying to prove it with limits because it's an exercise in my complex analysis book that comes before the Cauchy-Riemann equations are introduced.

[itex]f(z) = U(x,y) + iV(x,y)[/itex]

[itex]f(z) = |z| = |x + iy| = \sqrt{x^2 + y^2}[/itex]

[itex]U(x,y) = \sqrt{x^2 + y^2}[/itex]
[itex]V(x,y) = 0[/itex]

[itex]U_x = 0.5(x^2 + y^2)(2x) = x(x^2 + y^2)[/itex]
[itex]U_y = y(x^2 + y^2)[/itex]
[itex]V_x = 0[/itex]
[itex]V_y = 0[/itex]

[itex]U_x ≠ V_y[/itex]
[itex]U_y ≠ -V_x[/itex]


However, does the derivative exist when

[itex]x(x^2 + y^2) = 0[/itex]
[itex]y(x^2 + y^2) = 0[/itex]

i.e. when z = 0?

U_x=x/sqrt(x^2+y^2). That's not what you wrote. And the derivatives don't exist at (x,y)=(0,0) either, for the same reason the derivative of |x| with respect to x doesn't exist at x=0.
 
  • #5
Yeah, I forgot the (-1/2) exponents when I typed it out.

Okay, so z = 0 is a "potential" solution after I check the C-R equations, but then when I check it case by case, I see that the derivative doesn't exist when z = 0. Is this a correct?
 
  • #6
moxy said:
Yeah, I forgot the (-1/2) exponents when I typed it out.

Okay, so z = 0 is a "potential" solution after I check the C-R equations, but then when I check it case by case, I see that the derivative doesn't exist when z = 0. Is this a correct?

Right. U_x and U_y at z=0 have the form 0/0. That doesn't tell you much. But if you look at the details, they don't exist at z=0 either.
 
  • #7
Thank you! I'm still confused about limits in the complex plane and my original approach to this problem, but I'll save that for another time and another problem :) I really appreciate your help.
 

Related to Show a limit doesn't exists, complex case

1. What is a limit?

A limit is a fundamental concept in calculus that describes the behavior of a mathematical function as its input approaches a certain value. It represents the value that the function "approaches" as the input gets closer and closer to the specified value.

2. How do you determine if a limit exists?

A limit exists if the function approaches a single, finite value as the input approaches the specified value from both the left and right sides. If the function approaches different values from the left and right sides, the limit does not exist.

3. What is a complex case for showing a limit doesn't exist?

A complex case refers to a situation where the input and output of a function are complex numbers, rather than just real numbers. This adds an extra level of complexity in determining whether a limit exists or not.

4. What are some methods for showing a limit doesn't exist in a complex case?

One method is to use the epsilon-delta definition of a limit, where we can find two different paths for approaching the specified value that lead to different limit values. Another method is to use polar coordinates and show that the limit value depends on the angle of approach.

5. Why is it important to show a limit doesn't exist in a complex case?

In calculus, we use limits to understand the behavior of functions and make predictions about their values. If a limit doesn't exist, it means that the function is not behaving in a predictable way and we cannot make accurate predictions. This is particularly important in complex cases, where the behavior of the function can be more unpredictable.

Similar threads

  • Topology and Analysis
Replies
14
Views
615
  • Calculus and Beyond Homework Help
Replies
8
Views
135
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
545
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Topology and Analysis
Replies
2
Views
783
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
Back
Top