# Homework Help: Differentiable functions in complex analysis

1. Nov 7, 2013

### flooey.D

Hello all, I have the following problem from Complex Analysis that I would like for someone to check my understanding on:
1. The problem statement, all variables and given/known data
The problem is to find the derivative if it exists of
$f(z) = \frac{e^{i\theta}}{r^2} = r^{-2}\cos \theta + i r^{-2}\sin \theta$
where I have already changed the complex function f(z) into polar form.

2. Relevant equations
The Cauchy-Riemann conditions in polar coordinates are
$\frac{\partial{u}}{\partial{r}} = \frac{1}{r}\frac{\partial{v}}{\partial{\theta}}$
and
$\frac{\partial{u}}{\partial{\theta}} = -r\frac{\partial{v}}{\partial{r}}$

3. The attempt at a solution
For the first Cauchy-Riemann condition above, I get
$\frac{\partial{u}}{\partial{r}} = \frac{-2}{r^3}\cos \theta$ and
$\frac{1}{r}\frac{\partial{v}}{\partial{\theta}} = \frac{1}{r^3}\cos \theta$
For the second Cauchy-Riemann condition above I get
$\frac{\partial{u}}{\partial{\theta}} = \frac{-1}{r^{2}}\sin \theta$ and
$-r\frac{\partial{v}}{\partial{r}} = \frac{2}{r^2}\sin \theta$

To check my understanding of this, the Cauchy-Reimann conditions are not satisfied "symbolically", but there could be certain values of $r$ or $\theta$ where the Cauchy-Reimann conditions are satisfied, right? However, in this problem, I am not seeing any combination of $r$ or $\theta$ 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?

2. Nov 7, 2013

### brmath

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. Nov 7, 2013

### flooey.D

Well, it wasn't in terms of z. It was $\frac{x+iy}{x^2 + y^2}$ which i thought would work better if I changed it to polar form.

4. Nov 7, 2013

### Dick

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. Nov 7, 2013

### flooey.D

Ahh, I'm seeing that I should have written $x+iy = re^{i\theta}$, 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.

Last edited: Nov 7, 2013
6. Nov 7, 2013

### Dick

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. Nov 7, 2013

### flooey.D

New CR equations are

1st set:
$\frac{\partial{u}}{\partial{r}} = -\frac{1}{r^2}\cos \theta$ and
$\frac{1}{r}\frac{\partial{v}}{\partial{\theta}} = \frac{1}{r^2}\cos \theta$

2nd set:
$\frac{\partial{u}}{\partial{\theta}} = -\frac{1}{r}\sin \theta$ and
$-\frac{r\partial{v}}{\partial{r}} = \frac{1}{r}\sin \theta$

This one I'm having some trouble with:
If I try $\theta = 0$
then the 2nd set of CR equations is satisfied, but the 1st set is not. And vice versa for $\theta = \pi / 2$.
If I try angles like $\frac{\pi}{4}, \frac{3\pi}{4}, \ldots$ 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. Nov 7, 2013

### Dick

Ooops. Actually, I made a sign error. I think you are right.

Last edited: Nov 7, 2013
9. Nov 7, 2013

### D H

Staff Emeritus
That is correct given the corrected function.

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?

10. Nov 7, 2013

### flooey.D

Thank you everyone.

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