show the equality $f(\bar{z})=\overline{f(z)}$ for entire function $f(z)$

Question

i am struggling with this proof. Do you have some ideas?

Let $f(z)$ be entire function.

Show that, if $f$ maps $\mathbb{R}$ into $\mathbb{R}$, then equality $f(\bar{z})=\overline{f(z)}$ holds for all $z \in \mathbb{C}.$

Thanks for help.

Answer 1

If $f$ maps reals to reals, the coefficients of the power series expansion are real.

Then $\overline{f(z)} = \sum_n \overline{a_n} \overline{z^n} = \sum_n a_n \overline{z}^n = f(\overline{z})$.

To see why the $a_n$ are real, note that $a_0$ must be real since $f(0) = a_0$.

Suppose $s$ is real, then since $f'(s) = \lim_{t \to 0, t \text{ real}} {f(s+t)-f(s) \over t}$ we see that $f'(s)$ must be real, and hence $f'$ maps reals to reals, and so $a_1$ is real. It follows that all the $a_n$ are real.

Answer 2

Hint: $f^{(k)}(0)$ exists for any $k\geq 0$ and differentiability in the complex sense implies differentiability in the real sense, so $$ f^{(k)}(0)=\lim_{h\to 0^+}\frac{1}{h^k}\sum_{j=0}^{k}\binom{k}{j}(-1)^j f(jh) \tag{1}$$ where every term in the sum appearing in the RHS is real. That implies $f^{(k)}(0)\in\mathbb{R}$ for any $k\geq 0$, hence $$ f(z) = \sum_{n\geq 0}c_n z^n = \sum_{n\geq 0}\frac{f^{(k)}(0)}{k!}\,z^n \tag{2}$$ trivially fulfills $f(\overline{z})=\overline{f(z)}$.