Given $$f(z)=u(x,y)+iv(x,y)$$ which is analytic over a region $D$, I have to prove that it does not depend on $\bar z$ (the conjugate of $z$).
I've already tried replacing $x$ and $y$ by $\cfrac{z+\bar z}{2}$ and $\cfrac{z-\bar z}{2i}$.
What I get is a function $f$ which depends on $z$ and $\bar z$
I'm trying to relate the Cauchy-Riemann equations $$ \left\{ \begin{array}{} \cfrac{\partial u}{\partial x}=\cfrac{\partial v}{\partial y} \\ \cfrac{\partial u}{\partial y}=-\cfrac{\partial v}{\partial x} \end{array} \right. $$ to $f(z)$ but I'm having some trouble.
Two-dimensional change of variables: $$ x = \frac{z+\overline{z}}{2},\qquad y=\frac{z-\overline{z}}{2i} $$ Then $$ \frac{\partial x}{\partial z} = \frac{1}{2},\qquad \frac{\partial x}{\partial \overline{z}} = \frac{1}{2}, \\ \frac{\partial y}{\partial z} = \frac{1}{2i},\qquad \frac{\partial y}{\partial \overline{z}} = \frac{-1}{2i}, $$ So, with the chain rule for partial derivatives: $$ \frac{\partial u}{\partial \overline{z}} = \frac{\partial u}{\partial x}\;\frac{\partial x}{\partial \overline{z}} +\frac{\partial u}{\partial y}\;\frac{\partial y}{\partial \overline{z}} =\frac{1}{2}\frac{\partial u}{\partial x}-\frac{1}{2i}\frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial \overline{z}} = \frac{\partial v}{\partial x}\;\frac{\partial x}{\partial \overline{z}} +\frac{\partial v}{\partial y}\;\frac{\partial y}{\partial \overline{z}} =\frac{1}{2}\frac{\partial v}{\partial x}-\frac{1}{2i}\frac{\partial v}{\partial y} $$ Then, with $f = u+iv$ we get $$ \frac{\partial f}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial u}{\partial x} +i\frac{\partial v}{\partial x}\right) +\frac{i}{2}\left(\frac{\partial u}{\partial y} +i\frac{\partial v}{\partial y}\right) \\= \frac{1}{2}\left(\frac{\partial u}{\partial x} -\frac{\partial v}{\partial y}\right) + \frac{i}{2}\left(\frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}\right) $$ Now if $u,v$ are real-valued, we get $$ \frac{\partial f}{\partial \overline{z}} = 0 \quad\Longleftrightarrow\quad \left\{ \begin{array}{} \cfrac{\partial u}{\partial x}=\cfrac{\partial v}{\partial y} \\ \cfrac{\partial u}{\partial y}=-\cfrac{\partial v}{\partial x} \end{array} \right. $$