composition of non analytic functions can be analytic?

I was wondering if the following scenarios are possible: 1) composition of an analytic function with a non-analytic continuous function being analytic (except for trivial cases) 2) composition of two non-analytic functions being analytic.

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For the second one, consider composing $\bar z$ with itself. – Potato Dec 9 '12 at 3:52

1) Let $U$ and $V$ be open subsets of the complex plane and let $f\colon U \to \mathbb{C}$ be analytic and $g\colon U \to V$ be real-differentiable. Suppose that $f\circ g$ is analytic in $U$ and that $f'(z)\ne 0$ at every point $z \in U$. Claim: $g$ is analytic in $U$.
Proof. By Cauchy-Riemann equations, for every $w\in U$ the Jacobian matrix $Df(w)$ is a scalar multiple of a rotation matrix:$$Df(w)=\alpha R_\theta.$$ The coefficient $\alpha$ does not vanish because of the assumption on $f'(z)$. Since $f\circ g$ is analytic, the same must happen for the Jacobian matrix $D(f\circ g)(z)$: $$Df(g(z))Dg(z)=\beta R_\phi.$$ So $$Dg(z)=\frac{\beta}{\alpha}R_{\phi-\theta}.$$ Hence, again by Cauchy-Riemann equations, $g$ is analytic.
This does not answer your question, since you requested less regularity on $g$, but at least hints at the fact that the phenomenon is unlikely.
2) The composition of the non-analytic mapping $$z\mapsto \frac{1}{\overline{z}}$$ with itself is analytic. (This is essentially the same example as Potato.)