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So if function is analytic( ~ holomorphic) in Ω⊂C, then it satisfies C-R equations. And if f satisfies C-R equations and the functions u(x,y) and v(x,y) have first partial derivatives which are continuous, then f is analytic(~holomorphic). Am I right? Then what about relationship between real differentiability and complex differentiability? Does they need C-R equations to distinguish them from each other? I mean that the relationship between those properties in case of complex function are defined by C-R equations .

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up vote 3 down vote accepted

It's not quite clear exactly what the quesion is.

A real differentiable function satsifying Cauchy-Riemann's equations on an open set is holomorphic (or complex differentiable if you prefer), even if it's not a priori $C^1$. If it doesn't satisfy C-R, it's certainly not complex differentiable.

(If you only assume existence of partial derivatives, it's more tricky, look up Looman-Menchoff's theorem.)

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The question is: When is a continuously differentiable vector-valued function $${\bf f}:\quad {\mathbb R}^2\to{\mathbb R}^2\ ,\qquad {\bf z}=(x,y)\ \mapsto\ {\bf w}=\bigl(u(x,y),v(x,y)\bigr)$$ essentially the same as an analytic function $$f:\quad{\mathbb C}\to{\mathbb C}\ ,\qquad z=x+iy\ \mapsto\ w=u(x,y)+i\, v(x,y)\ .$$ Answer: The function ${\bf f}$ has at every point ${\bf z}$ a Jacobian $$\left[\matrix{u_x&u_y\cr v_x&v_y\cr}\right]_{\bf z}\ .$$ The entries of this matrix at a given ${\bf z}$ can be any four numbers, and they encode how ${\bf f}$ changes when moving from ${\bf z}$ to an infinitesimally near point ${\bf z}+{\bf Z}$.

When ${\bf f}$ aspires to be the vectorial expression of an analytical $f$ then the four partial derivatives have to satisfy at each point the conditions $u_x=v_y$, $u_y=-v_x$; in other words, the Jacobian of ${\bf f}$ has to be at all points ${\bf z}$ of the form $$\left[\matrix{A&-B\cr B&A\cr}\right], \qquad A,B\in{\mathbb R}\ .$$ The values $A$, $B$ pertaining to a particular ${\bf z}=(x,y)$ are then related to the complex derivative of $f$ at the point $z=x+iy\ $ via $\ f'(z)=A+iB$.

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