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$e^z=e^x cos\,{y}+ie^x sin\,{y}=u+iv$, then $u_x=v_y, u_y=-v_x$ for all $z\in C$. So $e^z$ is differentiable everywhere in $C$.

True?

But we know that if

$f(z)$ satisfies Cauchy Riemann equations at $z_0$ does not imply $f(z)$ is differentiable at $z_0$.

Please help me to get the right conclusion. Please help me to understand "If Cauchy-Riemann equations are satisfied everywhere, then does it mean the function is differentiable for all $z$"

user1942348
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  • What's the question? Yes, the CR equations at one point do not imply differentiability at that point. And yes, the whole point is that (if we assume $f$ is $C^1$) then knowing the C-R equations hold everywhere imply that $f$ is differentiable, – David C. Ullrich Apr 23 '20 at 13:04
  • @DavidC.Ullrich Question was "If Cauchy-Riemann equations are satisfied everywhere, then does it mean the function is differentiable for all z " – user1942348 Apr 23 '20 at 13:06
  • Right. Do you know a reference that introduces the CR equations but does not answer that question? – David C. Ullrich Apr 23 '20 at 13:08
  • @DavidC.Ullrich I have reference that if f(z) satisfies Cauchy Riemann equations at z0 does not imply f(z) is differentiable at z0, then why "If Cauchy-Riemann equations are satisfied everywhere, then it means that the function is differentiable for all z " is correct? – user1942348 Apr 23 '20 at 13:11
  • What degree of regularity are you assuming for $f$? – Moishe Kohan Apr 23 '20 at 13:32

2 Answers2

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Say $f=u(x,y)+iv(x,y)$ as usual. Note that saying $f\in C^1$ means that $u_x, u_y, v_x$ and $v_y$ are continuous.

No, knowing the CR equations at a point does not imply differentiability at a point. But:

Easy Lemma. If $f$ is $C^1$ and satisfies the C-R equations at $z_0$ then $f$ is differentiable at $z_0$.

(Proof: Just a little calculus, as in almost every elementary complex book.)

You ask how the following theorem can be true:

Theorem. If $f$ is continuous and satisfies the CR equations at every point of an open set then it is differentiable there.

It's hard to see how to answer "how can this be?"; it's a theorem, with a proof. Of course the theorem is trivial if we assume $f\in C^1$, but it's true, although far from trivial, as stated. That's the Looman-Menschoff Theorem.

quid
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David C. Ullrich
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If u and v are also harmonic for all z, then f(z) will indeed be holomorphic over C, which means it will be differentiable over C. However if u or v are not harmonic then it’s not necessarily true that f(z) will be differentiable on all of C, even though they may satisfy the Cauchy-Riemann Equations.

Ximenez
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    What matters is just continuity of the partials of $u$ and $v$ (which of course does follow if they're harmonic). – David C. Ullrich Apr 23 '20 at 13:21
  • Exactly, if they are continuous and have continuous first order partial derivatives and satisfy the CR equations for all of C, then f will be differentiable everywhere. – Ximenez Apr 23 '20 at 13:23