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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!

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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.

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  • $\begingroup$ Yes, I understand that. We have points where f is not differentiable and hence f is not analytic anywhere. So we don't need differentiability along y = x + 2 to discuss differentiability at neighborhoods of points along y = x + 2? $\endgroup$ Commented Dec 8, 2020 at 15:23
  • $\begingroup$ Yes, that is correct. $\endgroup$ Commented Dec 8, 2020 at 15:24
  • $\begingroup$ Could you explain why? $\endgroup$ Commented Dec 8, 2020 at 15:26
  • $\begingroup$ Isn't that what my answer is about? What more is needed? $\endgroup$ Commented Dec 8, 2020 at 15:29
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    $\begingroup$ Yes, all of them are correct. $\endgroup$ Commented Dec 8, 2020 at 15:39

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