# Prove that real-values function $f$ on the interval $(x_0 - r , x_0 + r)$, is analytic at $x_0$

Let F be a holomorphic function defined in an open disk $$D(x_0, r)$$ where $$x_0 ∈ R$$. Define a function $$f$$ in the interval $$(x_0 − r, x_0 + r)$$ by $$f(x) = F(x$$) for all $$x ∈ (x-0 − r, x_0 + r)$$. Suppose that $$f$$ is a real-valued function on the interval $$(x_0 − r, x_0 + r)$$. Prove that f is analytic at $$x_0$$.

I know that since $$F$$ is holomorphic in the open disk, $$F$$ has the power series representation that is also analytic in the open disk.

How do I use that to prove the real-valued function is analytic at $$x_0$$?

Let $$f(z)= \sum_{n=0}^{\infty}a_n(z-x_0)^n$$ be the power series representation in $$D(x_0,r)$$. Then
$$f(x)= \sum_{n=0}^{\infty}a_n(z-x_0)^n$$ for all $$x \in (x_0 − r, x_0 + r).$$
Now use $$f(x) \in \mathbb R$$ for $$x \in (x_0 − r, x_0 + r)$$ , to obtain that all $$a_n$$ are real.