# Question regarding analyticity, differentiability and the Cauchy-Riemann Equations ( Complex Analysis )

For non-analyticity, we can show that the Cauchy-Riemann equations don't hold anywhere. But here we can see that the Cauchy-Riemann equations hold when $$y=2x$$. I understand what the book did here but I just want to clarify something: If Cauchy-Riemann equations hold for $$y=2x$$, it doesn't necessarily mean that the function is differentiable. Do they just assume here that the function is differentiable along the line $$y=2x$$ and then talk about how the function won't be differentiable in any neighborhood of any complex number on the line $$y=2x$$ and so it won't be analytic anywhere? ( Since Cauchy-Riemann equations won't hold in the neighborhoods and hence not analytic). I just need to understand if they assumed that the function is differentiable along the line $$y=2x$$ or not. What's happening here? Thank you.

I cannot use "Sufficient Condition for Differentiability", since it is introduced later on in the book.

The problem below was solvable since I could use the "Sufficient Condition for Differentiability":

Show that the function $$f(z) = x^2 - x + y + i(y^2 - 5y - x)$$ is nowhere analytic but differentiable along $$y = x + 2$$. I tried doing this and so far the Cauchy-Riemann equations are only satisfied when we have $$y = x + 2$$. So when we have $$y = x + 2$$, all the demands for "Sufficient Condition for Differentiability" are met and so the function is differentiable along $$y = x + 2$$. However, the function won't be differentiable at every point in the neighborhood of any complex number along $$y = x + 2$$, so it won't be analytic.

The question I have is, do we need differentiability along a curve to talk about differentiability at neighborhoods of points along the curve?

So far I've learnt that:

Analyticity at a point implies differentiability at a point but the converse is not true, Analyticity at every point in a domain D implies differentiability at every point in a domain D and the converse is true, Analytic at a point implies differentiable at a point implies Cauchy-Riemann equations hold at that point but the converses are not true.

Are all of these correct?

I know this is a lengthy question, but please help me understand this and clear my confusions. Thank you!

No, they did not assume that $$f$$ is differentiable at the pints of the form $$x+2ix$$. What happens is that the set $$L=\{x+2ix\mid x\in\Bbb R\}$$ is a line and therefore, for any $$z\in L$$, no disk centered at $$z$$ is contained in $$L$$. And so, for any disk centered at any point of $$L$$ there is a point of that disk that doesn't belong to $$L$$ and so, in particular, there is a point in that disk at which $$f$$ is not differentiable.