A function is real-differentiable iff it has a complex-differentiable extension Is this conjecture true?

A function $f:\Bbb R\to\Bbb R$ is real differentiable at $a$ if and only if there exists a complex-differentiable function $g:A\to\Bbb C$ for some neighborhood of $a\in A\subseteq \Bbb C$ such that $f\Big|_{A\cap \Bbb R}=g\Big|_{\Bbb R\cap A}$.

The reverse implication is trivial; a complex differentiable function $g$ is real differentiable on $A$, and since differentiability is local and $f$ and $g$ agree on $A$, $f$ is real differentiable as well. But the forward implication is probably too strong to be true, although I can't think of any counterexamples. Is it true if $g$ is only required to be complex differentiable at $a$? If we only require $f-g\in o(x-a)$, then the statement can be satisfied by letting $g(z)=f(a)+f'(a)(z-a)$.
I'm interested in finding a characterization of real differentiability in terms of complex differentiability; other suggestions for definitional theorems of this sort are also welcome.

No.  Put $f(x) = x^2{\rm sgn}(x)$.  You are done right there.
