Differentiable functions in complex analysis

In summary, Complex Analysis homework statement states that the derivative of a complex function cannot be found at the origin.
  • #1
flooey.D
6
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Hello all, I have the following problem from Complex Analysis that I would like for someone to check my understanding on:

Homework Statement


The problem is to find the derivative if it exists of
[itex] f(z) = \frac{e^{i\theta}}{r^2} = r^{-2}\cos \theta + i r^{-2}\sin \theta [/itex]
where I have already changed the complex function f(z) into polar form.



Homework Equations


The Cauchy-Riemann conditions in polar coordinates are
[itex] \frac{\partial{u}}{\partial{r}} = \frac{1}{r}\frac{\partial{v}}{\partial{\theta}} [/itex]
and
[itex] \frac{\partial{u}}{\partial{\theta}} = -r\frac{\partial{v}}{\partial{r}} [/itex]



The Attempt at a Solution


For the first Cauchy-Riemann condition above, I get
[itex] \frac{\partial{u}}{\partial{r}} = \frac{-2}{r^3}\cos \theta [/itex] and
[itex] \frac{1}{r}\frac{\partial{v}}{\partial{\theta}} = \frac{1}{r^3}\cos \theta [/itex]
For the second Cauchy-Riemann condition above I get
[itex] \frac{\partial{u}}{\partial{\theta}} = \frac{-1}{r^{2}}\sin \theta [/itex] and
[itex] -r\frac{\partial{v}}{\partial{r}} = \frac{2}{r^2}\sin \theta [/itex]

To check my understanding of this, the Cauchy-Reimann conditions are not satisfied "symbolically", but there could be certain values of [itex] r [/itex] or [itex] \theta [/itex] where the Cauchy-Reimann conditions are satisfied, right? However, in this problem, I am not seeing any combination of [itex] r [/itex] or [itex] \theta [/itex] where the Cauchy-Riemann conditions could be satisfied, and so this function is not differentiable anywhere. Is this the right understanding of complex functions being differentiable or not?
 
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  • #2
Usually we differentiate with respect to z. If the function has a derivative with respect to z it is analytic throughout some neighborhood of z.

If you break the function back down either into x,y or r ##\theta##, getting it all back together is a major hassle.

So what was the original function of z?
 
  • #3
Well, it wasn't in terms of z. It was [itex] \frac{x+iy}{x^2 + y^2} [/itex] which i thought would work better if I changed it to polar form.
 
  • #4
flooey.D said:
Well, it wasn't in terms of z. It was [itex] \frac{x+iy}{x^2 + y^2} [/itex] which i thought would work better if I changed it to polar form.

Yes, it is easier in polar. But you don't have the right polar form. x+iy isn't the same as ##e^{i \theta}##.
 
  • #5
Dick said:
Yes, it is easier in polar. But you don't have the right polar form. x+iy isn't the same as ##e^{i \theta}##.

Ahh, I'm seeing that I should have written [itex] x+iy = re^{i\theta} [/itex], is this what you are hinting at?

Re-working through it all I get down to similar expressions for the C-R conditions above with 2 changing to 1, and the powers of r decreasing by 1. My conclusion above is still the same.
 
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  • #6
flooey.D said:
Ahh, I'm seeing that I should have written [itex] x+iy = re^{i\theta} [/itex], is this what you are hinting at?

Re-working through it all I get down to similar expressions for the C-R conditions above with 2 changing to 1, and the powers of r decreasing by 1. My conclusion above is still the same.

Yes, it's not differentiable everywhere, but there are some values of r and θ where it is differentiable. Can you find them? And could you spell out what you got for the new CR equations?
 
  • #7
Dick said:
And could you spell out what you got for the new CR equations?
New CR equations are

1st set:
[itex]\frac{\partial{u}}{\partial{r}} = -\frac{1}{r^2}\cos \theta [/itex] and
[itex]\frac{1}{r}\frac{\partial{v}}{\partial{\theta}} = \frac{1}{r^2}\cos \theta [/itex]

2nd set:
[itex]\frac{\partial{u}}{\partial{\theta}} = -\frac{1}{r}\sin \theta [/itex] and
[itex]-\frac{r\partial{v}}{\partial{r}} = \frac{1}{r}\sin \theta [/itex]

Dick said:
there are some values of r and [itex]\theta[/itex] where it is differentiable. Can you find them?
This one I'm having some trouble with:
If I try [itex] \theta = 0 [/itex]
then the 2nd set of CR equations is satisfied, but the 1st set is not. And vice versa for [itex]\theta = \pi / 2 [/itex].
If I try angles like [itex] \frac{\pi}{4}, \frac{3\pi}{4}, \ldots [/itex] I get one negative and one positive number for each set of CR equations.
Any r I choose except r=0, I get one negative and one positive number for each set of CR equations.

Are my CR equations right?
 
  • #8
flooey.D said:
New CR equations are

1st set:
[itex]\frac{\partial{u}}{\partial{r}} = -\frac{1}{r^2}\cos \theta [/itex] and
[itex]\frac{1}{r}\frac{\partial{v}}{\partial{\theta}} = \frac{1}{r^2}\cos \theta [/itex]

2nd set:
[itex]\frac{\partial{u}}{\partial{\theta}} = -\frac{1}{r}\sin \theta [/itex] and
[itex]-\frac{r\partial{v}}{\partial{r}} = \frac{1}{r}\sin \theta [/itex]This one I'm having some trouble with:
If I try [itex] \theta = 0 [/itex]
then the 2nd set of CR equations is satisfied, but the 1st set is not. And vice versa for [itex]\theta = \pi / 2 [/itex].
If I try angles like [itex] \frac{\pi}{4}, \frac{3\pi}{4}, \ldots [/itex] I get one negative and one positive number for each set of CR equations.
Any r I choose except r=0, I get one negative and one positive number for each set of CR equations.

Are my CR equations right?

Ooops. Actually, I made a sign error. I think you are right.
 
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  • #9
flooey.D said:
New CR equations are

1st set:
[itex]\frac{\partial{u}}{\partial{r}} = -\frac{1}{r^2}\cos \theta [/itex] and
[itex]\frac{1}{r}\frac{\partial{v}}{\partial{\theta}} = \frac{1}{r^2}\cos \theta [/itex]

2nd set:
[itex]\frac{\partial{u}}{\partial{\theta}} = -\frac{1}{r}\sin \theta [/itex] and
[itex]-\frac{r\partial{v}}{\partial{r}} = \frac{1}{r}\sin \theta [/itex]
That is correct given the corrected function.

flooey.D said:
Well, it wasn't in terms of z. It was [itex] \frac{x+iy}{x^2 + y^2} [/itex] which i thought would work better if I changed it to polar form.
Even easier is to use the fact that x2+y2=zz*, which means f(z)=z/(zz*)=1/z*. What does that tell you about differentiability?
 
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  • #10
Thank you everyone.

Even easier is to use the fact that x2+y2=zz*, which means f(z)=z/(zz*)=1/z*. What does that tell you about differentiability?

I wish I would have thought of this in the first place; thanks for the extra tip.
 

1. What is a differentiable function in complex analysis?

A differentiable function in complex analysis is a function that has a derivative at every point in its domain. This means that the function is smooth and can be approximated by a linear function at each point.

2. How does complex differentiability differ from real differentiability?

Complex differentiability is a stronger condition than real differentiability. In complex analysis, a function must not only have a derivative at a point, but that derivative must also exist in a neighborhood around that point. In contrast, a real differentiable function only needs to have a derivative at a point.

3. What is the Cauchy-Riemann equations and how are they related to differentiability in complex analysis?

The Cauchy-Riemann equations are a set of necessary and sufficient conditions for a function to be differentiable in complex analysis. These equations relate the partial derivatives of a function with respect to its real and imaginary components. If a function satisfies the Cauchy-Riemann equations, it is differentiable in the complex sense.

4. Can a function be differentiable in the complex sense but not in the real sense?

Yes, it is possible for a function to be differentiable in the complex sense but not in the real sense. This means that the function satisfies the Cauchy-Riemann equations and is smooth in the complex plane, but it may not have a derivative in the real plane.

5. What are some applications of differentiable functions in complex analysis?

Differentiable functions in complex analysis have a wide range of applications in physics, engineering, and mathematics. They are used to study and model physical phenomena such as fluid flow, electric fields, and wave propagation. They are also essential in the study of complex geometry, number theory, and dynamical systems.

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