A general proof of $f\left(\bar{z}\right)=\overline{f\left(z\right)}$ As a school student I have seen a striking property of functions .
$$f\left(\bar{z}\right)=\overline{f\left(z\right)}$$
Where $z$ is a complex number and $\bar{z}$ it's complex conjugate. For eg: $z=x+iy$ then $\bar{z}=x-iy$ where $x,y \in \mathbb{R}$
How do we in generally prove the result ?
I have consulted some books on complex numbers and haven't yet seen a general proof, google search also didn't help, with this specific query.
All places I have consulted just mentions this just like a physical law in nature (no questions regarding it's validity).But I believe only when we have a formal proof we could use it more power, and presently I don't believe that all functions obey this property .
If all functions do not obey the mentioned property(I have never came across a function that doesn't obey the result, but I agree that doesn't mean there doesn't exist one) , is there a specific name for such functions and how do we identify it without actually proving(I mean, proof not by showing that both the LHS and RHS are equal)
Proofs, references and any Google search tag's will be deeply appreciated 
 A: If $f$ is analytic on some non-empty, connected, open set $D$ that is 'symmetric' in the sense that $\overline{D} = D$, then we have $f(\mathbb{R} \cap D) \subset \mathbb{R}$ iff $f(\overline{z}) = \overline{f(z)}$ for all $z \in D$.
First note that $D$ must intersect $\mathbb{R}$, since if $z \in D$, then $\overline{z} \in D$, and $D$ is path connected since it is open and connected.
($\Rightarrow$): Let $z_0 \in \mathbb{R} \cap D$. Then $f$ has a power series in some neighborhood of $z_0$, say $f(z) = \sum a_n (z-z_0)^n$. By restricting $z$ to the reals in the neighborhood,  we have $f(z) \in \mathbb{R}$. We see (by differentiating and evaluating the result at $z_0$) that $a_n \in \mathbb{R}$ for all $n$, and so we have $f(\overline{z}) = \overline{f(z)}$ in the neighborhood of $z_0$.
A little work shows that the function $\phi(z) = f(z)-\overline{f(\overline{z})}$ is analytic on $D$, and the above shows that $\phi(z) = 0$ on an interval of the real line. Since $D$ is connected, it follows that $\phi(z) = 0$ on all of $D$.
($\Leftarrow$): If $z \in \mathbb{R}$, then $\overline{z} \in \mathbb{R}$, and so $f(z) = f( \overline{z} )$. It follows that $f(\mathbb{R} \cap D) \subset \mathbb{R}$.
A: As mentioned in the comments, the result is not always true. But it is true for an interesting class of functions, namely those which can be written as a power series with real coefficients. Suppose $f(z) = \sum_{n=0}^{\infty} a_n z^n$ and $a_n$ are all real. Then 
$$ \overline{f(z)} = \overline{ \sum_{n=0}^{\infty} a_n z^n}= \sum_{n=0}^{\infty} \overline{a_n z^n} = \sum_{n=0}^{\infty} \overline{a_n} \overline{z^n} =\sum_{n=0}^{\infty} a_n \overline{z}^n = f(\overline{z}). $$
To do those steps we used $\overline{x+y} = \overline{x} + \overline{y}, \overline{xy} = \overline{x}\overline{y}, \overline{z^n}=\overline{z}^n$ and that $\overline{x}=x$ if $x$ is real. 
Included in this class are polynomials with real coefficients (and this fact is why polynomials with real coefficients have roots that come in complex conjugate pairs) and Sine/Cosine (whose power series have real coefficients).
Note to others: I've intentionally left out convergence issues, as I think it would hinder the OP more than it helps here. 
A: I don't think such function has any single word that describe them. If one were to speak about them without formulas I would call them "the functions that commute with complex conjugation".
Here's one example of an otherwise fairly nice function that doesn't satisfy this property: The function $f(z)=\frac{2z}{z^2+1}$ is holomorphic on the domain $$D=\mathbb C\setminus\bigl(\{x+i\mid x\ge 0\}\cup\{x-i\mid x\le 0\}\bigr).$$
Because $D$ is open and simply connected, $f$ has an antiderivative $F$ on $D$; by choosing an appropriate constant of integration we can select $F$ such that $F(0)=0$. Then $F(x)$ maps reals to reals -- in fact $F(x)=\log(x^2+1)$ for real $x$ -- and $F$ is holomorphic throughout its domain. But $F(a-bi)$ differs from $\overline{F(a+bi)}$ by $\pm 2\pi i$ whenever $|b|>1$.
