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Let $f:\mathbb{C}\to\mathbb{C}$, and let $u,v:\mathbb{R}^2\to\mathbb{R}$ by $u(x,y)=\Re f(x+iy)$, $v(x,y)=\Im f(x+iy)$. Is it true that $f$ is complex-differentiable at $z_0$ if and only if $u$ and $v$ are differentiable at $(\Re z_0,\Im z_0)$ and $u_x=v_y$ and $u_y=-v_x$ at $(\Re z_0,\Im z_0)$?

I think I proved that the answer is yes. However, I also know that $f$ is complex-differentiable at $z_0$ if and only if $u_x=v_y$ and $u_y=-v_x$ and $u$ and $v$ have continuous first partials at $z_0$. (See Churchill and Brown (7th ed.), the theorem in Section 21 and the corollary in Section 48.) This implies that for any $u,v$ satisfying $u_x=v_y$ and $u_y=-v_x$ at a point, $u$ and $v$ are differentiable at that point if and only if $u$ and $v$ have continuous first partials at that point. It seems like this statement is definitely false, though I can't come up with any counterexamples at the moment. ...So what is going on?

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    $\begingroup$ wait so (CR + cont partial)-(iff)-(CR + real diff)-(iff)-(complex-diff), but (cont partial)-(only if but not if)-(real diff)? $\endgroup$
    – BCLC
    Apr 7, 2021 at 7:00

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The answer is yes, $f$ is complex-differentiable if and only if its real and imaginary parts are real-differentiable and satisfy the Cauchy-Riemann equations. Notice I said they are real differentiable, not that their partial derivatives exist. Actually, there are functions where both partial derivatives exist but the function is not differentiable. It is a theorem that if that function has continuous partial derivatives then it is real differentiable. This is explains the statement of Churchill and Brown, they choose to say "CR + continuous partials" rather than "CR + real differentiable".

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    $\begingroup$ wait so (CR + cont partial)-(iff)-(CR + real diff)-(iff)-(complex-diff), but (cont partial)-(only if but not if)-(real diff)? $\endgroup$
    – BCLC
    Apr 7, 2021 at 7:00

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