Analytic function that does not depend on $\bar z$ 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.
 A: 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
$$
\frac{\partial f}{\partial \overline{z}} = 0
\quad\Longleftrightarrow\quad
\left\{ 
\begin{array}{}
\cfrac{\partial u}{\partial x}=\cfrac{\partial v}{\partial y} \\ 
\cfrac{\partial u}{\partial y}=-\cfrac{\partial v}{\partial x}
\end{array}
\right.
$$
