Show polynomial function is infinitely differentialble Let for $\alpha=(\alpha_0,...,\alpha_n)\in\mathbb{C}^{n+1}$ the polynomial $p_\alpha :\mathbb{C}\to\mathbb{C}$ be given by $p_\alpha(z)=\sum_{k=0}^n \alpha_kz^k$.

Show that with the identification $\mathbb{C}\cong\mathbb{R}^2$ with $x+iy\cong (x,y)$, the function $p_\alpha$ is element of $C^\infty(\mathbb{R}^2,\mathbb{R}^2)$.

I read such things can be shown using Taylor's theorem but how can I apply it here?
 A: I leave the details to you.
It is not very difficult (and very useful) to prove that:


*

*All constant functions are $C^\infty$.

*If $f:\mathbb C\to \mathbb C$ and $g:\mathbb C\to \mathbb C$ are $C^\infty$ (in the sense given in the OP question), then $fg$  and $f+g$ are also $C^\infty$.

*$f(z)=z$ is $C^\infty$, so $f_k(z)=z^k$ is $C^\infty$. 

*Any polynomial $p$ is $C^\infty$. Clearly, by 0. and 1. (product), every monomial $a_kz^k$ is $C^\infty$. Using $1.$ (sum) and induction, every polynomial, which is a finite sum of monomial, is $C^\infty$.  

A: How about this:
EDIT: I mistakenly thought the series was infinite, and not a polynomial. I thought the question was about showing that any complex-analytic function was $C^{\infty}$ when seen as a function from $\mathbb R^2$ to itself, under the equivalence $x+ iy \rightarrow (x,y)$. My proof below refers to this case.
If $f(z)=u(x,y)+iv(x,y)$ is complex-differentiable, i.e., analytic in a region $R$ , then the partials $u_x, u_y, v_x, v_y$ exist and are continuous. Since a complex-analytic function $f(z)$ is infinitely-differentiable, the same goes for all partials of all orders, i.e., $u_{xx}, u_{xy}, u_{yx}, u_{yy}, u_{xxy},....$  are all infinitely-differentiable .Moreover, $$f'(z)=u_x+iv_x $$. With the identification $x+iy \rightarrow (x,y)$ , we get $$f'(z)=u_x+iv_x =\nabla f(x,y)\rightarrow (u_x,v_x) $$ , where $\nabla$ is the gradient, i.e., the total derivative of $f$. Now, we have from above that both $u_x, v_x$ both exist and are continuous, so that $f(x,y)$ is differentiable. Now we can induct on this result (using the fact from above that all partials exist and are continuous/differentiable, i.e., $C^{\infty}$) to differentiate $f$ any number of times to get the result we want.
Another result you can use that is more abstract and is more of a hand-waving result is that $\mathbb C$ and $\mathbb R^2$ are diffeomorphic as manifolds (actually, using the map in your post), so that every function differentiable in one manifold is (after change of coordinates given by the diffeomorphism) differentiable in its diffeomorphic image.
