Let $z = x + i y$, and consider the following function:
$$
f(z) = e^{\frac{1}{1 + z^2}} \qquad z \in \mathbb{C}
$$
Note that at $z = \pm i$ the function does not converge.
We see that on $\mathbb{R}$ when $y = 0$, this becomes:
$$
f(x) = e^{\frac{1}{1 + x^2}} \qquad x \in \mathbb{R},
$$
which converges nicely for all values $x \in \mathbb{R}$.
Furthermore, this function is analytic on $\mathbb{R}$, and so there exists some power series of form:
$$
f(x) = \sum_{n = 0}^{\infty} a_n (x – x_0)^n \qquad a_n, x, x_0 \in \mathbb{R}
$$
that converges for all $x \in \mathbb{R}$.
Does this power series naturally extend to the complex plane if we have $x \mapsto z$, i.e.:
$$
f(z) = \sum_{n = 0}^{\infty} b_n (z – z_0)^n \qquad z \in \mathbb{C}, \quad z_0 = x_0 + i0, \quad b_n = a_n + i 0
$$
In this case, would the function shown above thus demonstrate that a complex-valued power series that converges on all of $\mathbb{R}$ not necessarily converge in $\mathbb{C}$?
Best Answer
No. If a power series converges on all $\mathbb{R}$, it(s extension) converges for all complex numbers. One quick way to see why this holds is considering the formula for the radius of convergence and recalling that one proves absolute convergence with it, thus holding for $\mathbb{C}$ as well.