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In my undergraduate complex analysis textbook, it claims that Cauchy Riemann equations is not a sufficient condition for the existence of derivative. Intuitively, I do not understand why this is true, as if you satisfied $$u_x=v_y$$ $$u_y=-v_x$$

it implies that both $$f_x \;\; and\;\; f_y$$ exist and therefore $f$ should be differentiable

Any hint would be much appreciated

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marked as duplicate by mrf, Jyrki Lahtonen, dfeuer, user67258, Julian Kuelshammer Sep 17 '13 at 6:06

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    $\begingroup$ There are functions where $f_x$ and $f_y$ both exist, but $f$ itself is not continuous and so it is not differentiable. To get differentiability, you need the partials to be continuous. See here: calculus.subwiki.org/wiki/… $\endgroup$ – Ryan Sullivant Sep 17 '13 at 4:32
  • $\begingroup$ if $f$ is differentiable in respect to $x$ and $y$ doesn't it imply continuous of $f$? $\endgroup$ – xbd Sep 17 '13 at 4:34
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    $\begingroup$ Nope, if $f$ itself is differentiable, then $f$ is continuous, but being differentiable in higher dimensions is different than having partial derivatives. $\endgroup$ – Ryan Sullivant Sep 17 '13 at 4:35
  • $\begingroup$ @xbd: Take $f$ to be a function that is one everywhere except on the $x,y$ axes where it takes the value zero. Clearly not continuous, but the partials exist. $\endgroup$ – copper.hat Sep 17 '13 at 5:40
  • $\begingroup$ @xbd a sufficient condition for $f$ to be differentiable is the existence and continuity of all partial derivatives (not continuity of $f$ itself). $\endgroup$ – achille hui Sep 17 '13 at 5:48
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Take $f(x+iy) = \sqrt{|x||y|}$. The Cauchy Riemann equations are satisfied at $z=0$, but $f$ is not differentiable (in $\mathbb{C}$) at $z=0$.

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