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Analytic vs differentiable

Hi,

I need some help with the question attached.

It's my understanding that an analytic function is differentiable at every point in the domain.

What does "nowhere analytic" mean. Is the premise behind this question that the function is differentiable on the coordinate (does coordinate mean real?) axes and not differentiable on the imaginary axes?

OR, is it that, there is some point where the function is differentiable, but it is not an entire function?

i found the C-R Equations.

du/dx = 3x^2 + 3y^3 -3 , dv/dy = 3y^2 - 2 + 3x^2

du/dy = 6xy , dv/dx = 6xy

du/dx = dv/dy

and du/dy = - dv/dx where x = 0 or y = 0

and thus C-R equations are not satisfied on an open disk, but only on the line x=0 or y=0

I think this is wrong.

Any help appreciated, especially if you point out the thought process required here.

Thanks.

Re: Analytic vs differentiable

Quote:

Originally Posted by

**99.95** Hi,

I need some help with the question attached.

It's my understanding that an analytic function is differentiable at every point in the domain.

What does "nowhere analytic" mean. Is the premise behind this question that the function is differentiable on the coordinate (does coordinate mean real?) axes and not differentiable on the imaginary axes?

The point is that a function that is differentiable at every point on an **open** set (the "open disc" that you refer to) is analytic on that set. And, no "coordinate" does not mean "real". You used the phrase "coordinates axes" which is plural! Both real and imaginary axes are "coordinate axes".

Quote:

OR, is it that, there is some point where the function is differentiable, but it is not an entire function?

i found the C-R Equations.

du/dx = 3x^2 + 3y^3 -3 , dv/dy = 3y^2 - 2 + 3x^2

du/dy = 6xy , dv/dx = 6xy

du/dx = dv/dy

and du/dy = - dv/dx where x = 0 or y = 0

and thus C-R equations are not satisfied on an open disk, but only on the line x=0 or y=0

I think this is wrong.

Any help appreciated, especially if you point out the thought process required here.

Thanks.

Re: Analytic vs differentiable

I think you have your answer in front of you already. Your thought process is right. The C-R eqns. are satisfied only if the point is on either of the coordinate axes which means that the function is differentiable only on those points which are on one of the coordinate axes. However the function is not differentiable in any open disc in the complex plane. (Try to pick one!) So it not analytic at any point in the complex plane.