# Determing where function is differentiable (Complex Analysis)

• Oct 4th 2007, 07:01 PM
scothoward
Determing where function is differentiable (Complex Analysis)
Determine where the function f has a derivative, as a function of a complex variable:

f(x +iy) = 1/(x+i3y)

I know the cauchy-riemann is not satisfied, so does that simply mean the function is not differentiable anywhere?

• Oct 4th 2007, 08:14 PM
Jhevon
Quote:

Originally Posted by scothoward
Determine where the function f has a derivative, as a function of a complex variable:

f(x +iy) = 1/(x+i3y)

I know the cauchy-riemann is not satisfied, so does that simply mean the function is not differentiable anywhere?

it depends. what points did you try? it seems to me it is differentiable at points (x,y) where x = 3y or x = -3y
• Oct 4th 2007, 08:17 PM
scothoward
Alright, so doing the partials and equating I get

x = 3y and
xy = 3xy

So, subbing first equation into the second we get,
3y^2=9y^2, so y=0. If y=0 then x=0. So differentiable only at 0+0i.

But, because 0 + i0 is undefined in the original function, would this be differentiable nowhere?

Am I approaching this right?
• Oct 4th 2007, 08:29 PM
Jhevon
Quote:

Originally Posted by scothoward
Alright, so doing the partials and equating I get

x = 3y and
xy = 3xy

So, subbing first equation into the second we get,
3y^2=9y^2, so y=0. If y=0 then x=0. So differentiable only at 0+0i.

But, because 0 + i0 is undefined in the original function, would this be differentiable nowhere?

Am I approaching this right?

yup, that's it. just was making sure you checked the possibility
• Oct 4th 2007, 08:30 PM
scothoward
Great..

so just to clarify, in general there are cases where cauchy-riemann are not satisfied, however derivatives do exisit at some points? (In Other Words, if Cauchy-Riemann is not satisfied, it does not neccessarily mean the function isnt differentiable everywhere?)
• Oct 4th 2007, 08:47 PM
Jhevon
Quote:

Originally Posted by scothoward
Great..

so just to clarify, in general there are cases where cauchy-riemann are not satisfied, however derivatives do exisit at some points? (In Other Words, if Cauchy-Riemann is not satisfied, it does not neccessarily mean the function isnt differentiable everywhere?)

correct (if you're saying what i think you're saying), and also, the equations can be satisfied, but the function is still not differentiable. The Cauchy-Riemann is a necessary but not sufficient condition for differentiability (i don't like how you phrased what you said though, the equations must be satisfied for differentiability, but the formulas don't have to look the same, that's what i'm saying, you just have to check what points work, if any)
• Oct 5th 2007, 03:50 AM
Plato
It is not clear to me from the discussion that the function was ever put in proper u+iv form.
The function must be rewritten as $\frac{1}{{x + 3yi}} = \frac{x}{{x^2 + 9y^2 }} + i\frac{{ - 3y}}{{x^2 + 9y^2 }}$.

So $u(x,y) = \frac{x}{{x^2 + 9y^2 }}\;\& \;v(x,y) = \frac{{ - 3y}}{{x^2 + 9y^2 }}$.

That may have been done, but it is clear. Was it done?
• Oct 5th 2007, 03:52 AM
Jhevon
Quote:

Originally Posted by Plato
It is not clear to me from the discussion that the function was ever put in proper u+iv form.
The function must be rewritten as $\frac{1}{{x + 3yi}} = \frac{x}{{x^2 + 9y^2 }} + i\frac{{ - 3y}}{{x^2 + 9y^2 }}$.

So $u(x,y) = \frac{x}{{x^2 + 9y^2 }}\;\& \;v(x,y) = \frac{{ - 3y}}{{x^2 + 9y^2 }}$.

That may have been done, but it is clear. Was it done?

i did it, and i believe the poster did as well, since he said this:

Quote:

3y^2=9y^2, so y=0. If y=0 then x=0. So differentiable only at 0+0i.