Analytic function must be free from $\bar{z}.$ How can i say that analytic function must be free from $\bar{z}?$ I verified it for many examples like $\bar{z}, z^2+\bar{z} $ etc and found that the results seems to be true according to me . I am thinking  like as if any complex function contains $\bar{z}$  term in any form(obviously in nontrivial form  unlike $\bar{z}-\bar{z}$) then that terms is never analytic and hence whole function is also never analytic. Please suggest . Thanks .
 A: This is an interesting question, which is usually glossed over when doing complex calculus in $z$ and $\bar z$ terms. It is also not so clear what the actual claim is. E.g., the function $q(z):={\rm conjugate}(\bar z)$ is analytic $\ldots$ 
I'm going to argue in the following setup: Let
$$\Phi:\quad\Omega\to{\mathbb C},\qquad(z,\zeta)\to\Phi(z,\zeta)$$
be defined on the  domain $\Omega\subset{\mathbb C}^2$. Assume that $\Phi$ is (i) a $C^1$ function of its four real variables $(x, y, \xi,\eta)$, and is (ii)   separately analytic in each of the two complex variables $z$ and $\zeta$, meaning that the partial functions
$$z\mapsto\Phi(z,\zeta_0),\qquad {\rm resp.,}\qquad \zeta\mapsto\Phi(z_0,\zeta)$$
are analytic  functions of $z$, resp., of $\zeta$, as long as $(z,\zeta_0)\in\Omega$, resp., $(z_0,\zeta)\in\Omega$. Such a $\Phi$ could be given as a series, or as an "analytic expression" in terms of the variables $z$ and $\zeta$.
Given such a $\Phi$ we consider the function
$$f(z):=\Phi(z,\bar z)\qquad(z\in\Omega')\ ,\tag{1}$$
whose domain is the open set $\Omega':=\bigl\{z\in{\mathbb C}\bigm| (z,\bar z)\in\Omega\bigr\}\subset{\mathbb C}$. We now pose the following question:  
What are necessary and sufficient conditions on $\Phi$ making  $f$  an analytic function on $\Omega'$?
Claim. The function $f$ is analytic on $\Omega'$ iff $$\Phi_{.2}(z,\bar z)=0\qquad\forall\>z\in\Omega'\ .$$
Here $\Phi_{.2}$ denotes the partial derivative of $\Phi$ with respect to its second complex variable.
Proof. Fix a $z\in\Omega'$, and let $h$ be a complex increment variable. Then
$$\eqalign{f(z+h)-f(z)&=\Phi(z+h,\bar z+\bar h)-\Phi(z,\bar z)\cr  &=\Phi_{.1}(z,\bar z)\>h+\Phi_{.2}(z,\bar z)\>\bar h\ +o\bigl(|h|\bigr)\qquad(h\to0\in{\mathbb C})\ .\cr}\tag{2}$$
Here we just have used the assumptions on $\Phi$ and the definition of derivative. Now the main part of the right hand side of $(2)$ is a complex linear function of $h$, i.e., of the form $h\mapsto \lambda h$ for some $\lambda\in{\mathbb C}$, iff  $\Phi_{.2}(z,\bar z)=0$.$\quad\square$
When we are applying this principle  there will be given a function $f$ defined in the form $(1)$. By abuse of notation one then does not explicitly  introduce an auxiliary variable $\zeta$. Instead one just considers the $\bar z$ appearing in $(1)$ as an "independent complex variable" and formally differentiates the expression $\Phi(z,\bar z)$ with respect to this variable. In this way one arrives at the handy condition
$${\partial\Phi(z,\bar z)\over\partial\bar z}=0\qquad\forall z\in\Omega'\ .$$
