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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,

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    $\begingroup$ What happens if you expand $f(x,y) = (x(x^2 + y^2), y(x^2 + y^2))$ further? E. g. use the fact that $(a,b) = a + ib$ in complex notation. You can see immediately that $f$ is real-entire because it is a polynomial in $x,y$! $\endgroup$ – Christopher A. Wong Apr 10 '14 at 7:46
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    $\begingroup$ Related: math.stackexchange.com/questions/189366/… $\endgroup$ – Micah Apr 10 '14 at 16:33
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(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.

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