TITLE: Whats the differences between the real-entire functions on $\mathbb R^{2}$ and complex entire functions on $\mathbb C$?
QUESTION [5 upvotes]: We note, as set of points, $\mathbb R^{2}= \mathbb C.$
A complex valued function $F,$ defined on  an open set  $E$ in the plane $\mathbb R^{2}$, is said to be real-analytic in $E$ if to every point $(s_{0}, t_{0})$ in there corresponds an expansion with complex coefficients
$$F(s, t)= \sum_{n,m=0}^{\infty} a_{nm}(s-s_{0})^{m} (t-t_{0})^{n},$$ 
which converges absolutely for all $(s,t)$ in some neighbourhood of $(s_{0}, t_{0}).$
If $F$ is defined in the whole plane $\mathbb R^{2}$ by a series
$$F(s, t)= \sum_{n,m=0}^{\infty} a_{nm}s^{m} t^{n},$$
which converges absolutely for every $(s,t),$ the we call $F$ real-entire.
We say $F:\mathbb C \to \mathbb C$ is complex-entire if $F$ is differentiable on whole $\mathbb C.$ 

My Questions: (1)  What are conceptual differences between real-entire and complex-entire  functions defined on $\mathbb R^{2}= \mathbb C.$ (2) Suppose $f:\mathbb C \to \mathbb C$ is differentiable on whole $\mathbb C$. Is it true that $f:\mathbb R^{2}\to \mathbb C$ is real entire ? (3) Suppose that $f:\mathbb R^{2}\to \mathbb C$ is real entire. Is it true that $f:\mathbb C \to \mathbb C$ is differentiable on $\mathbb C.$ ? 

Trivial attempt: (a)If we consider, $f:\mathbb C \to \mathbb C$ such that $f(z)=z|z|^{2},$ 
and put $f=u+iv$, where $u,v:\mathbb R^{2}\to \mathbb R$ and $z=x+iy$; then $u(x,y)= x(x^{2}+y^{2})$ and $v(x,y)= y(x^{2}+y^{2})$; and $f$ satisfies Cauchy- Riemann equations iff $xy=0$; and therefore, $f$ can not be differentiable  on $\mathbb C$ (Please correct me if  I have done some thing wrong); now if we look at $f:\mathbb R^{2}\to \mathbb C$ such that $f(x,y)= (x(x^{2}+ y^{2}), y(x^{2}+y^{2}))$ ; Is $f$  is real-entire ? (I don't know how  to proceed here ) (b) Take, $f:\mathbb R^{2}=\mathbb C \to \mathbb C$ such that $f(z)= |z|z= (x\sqrt{x^{2}+y^{2}}, y\sqrt{x^{2}+y^{2}})$; what can we say about $f$ ? 
Thanks,

REPLY [5 votes]: (1) The conceptual difference is huge. A series
$$F(s,t):=\sum_{m,\>n}a_{mn} s^m t^n\tag{1}$$
has ${1\over2}(r+1)(r+2)$ "free" coefficients $a_{mn}$ up to total degree $r$, whereas a series
$$f(z):=\sum_{k=1}^\infty c_k z^k=\sum_{k=1}^\infty c_k (s+it)^k\tag{2}$$
has only $r+1$ "free" coefficients $a_k$ up to total degree $r$. 
The essential difference is the following: An $F$ as in $(1)$ is a function of the two completely independent variables $s$ and $t$, whereas the $f$ in $(2)$ is a function of (resp., a Taylor series in terms of) the single "artificial" variable $z:=s+it$.
(2) Yes. Just write the right side of $(2)$ as a double sum.
(3) A function $F$ as in $(1)$ is complex differentiable only if each homogeneous part
$$\sum_{m+n=r} a_{mn} s^m t^n$$ can be written in the form
$$c_r(x+it)^r$$
for a suitable $c_r\in{\mathbb C}$. Another way of describing this is the following: Substitute in $(1)$
$$x:={z+\bar z\over 2},\quad y:={z-\bar z\over 2i}\ .$$
When the resulting formal series in powers of $z$ and $\bar z$ contains no $\bar z$ whatsoever, then the $F$ we started with is complex differentiable.