Power series of $f$ where $f(B(0,R)) \subset \mathbb{R}$

Question

Let $f$ a power series with radius $R>0$ such that $f(B(0,R)) \subset \mathbb{R}$, considering :

$$f(z) = \sum_{n=0}^\infty a_n z^n$$

The question I'm asking myself is : Is $f$ a constant ?

I can find that $\forall n\in \mathbb N, a_n \in \mathbb R$. I don't know if it could help.

I appreciate any feedback !

EDIT : I don't know what are the Cauchy-Riemann equations. So is it possible to have another way ?

A power series defines a holomorphic function. What do you know about holomorphic functions which take only real values? (See https://math.stackexchange.com/q/438089/42969) — Martin R, Nov 27 '22 at 19:03
I am deeply sorry but I don't know what is the Cauchy-Riemann equations. Is this possible without this principle ? — Jotadiolyne Dicci, Nov 27 '22 at 19:14
Suppose $f'(x_0) \neq 0$. Compute the derivative of the function $t \mapsto f(x_0+i t \overline{f'(x_0)})$, it will be imaginary. However, if $f$ only takes real values then this is a contradiction. Hence $f'(x_0) = 0$ everywhere. — copper.hat, Nov 27 '22 at 19:42
@copper.hat I am sorry but why if $f$ only takes real values then it is a contradiction ? — Jotadiolyne Dicci, Nov 27 '22 at 19:50
in case text takes a power series first (Weierstrass) approach: by translation $f(0) = 0$; let $M$ be max modulus of $f\Big(\overline B\big(0,\frac{R}{2}\big)\Big)$... then $g(z):=M^2-f(z)\cdot f(z)$ is analytic and achieves maximum modulus at interior point $0 \in \overline B\big(0,\frac{R}{2}\big)$ hence is constant, so $f(z)^2 =f(0)^2 =0$ i.e. $f(z) =0$ everywhere in the ball — user8675309, Nov 27 '22 at 19:57
@copper.hat I get $t \mapsto i\overline{f'(x_0)}f'(x_0+i t \overline{f'(x_0)})$ — Jotadiolyne Dicci, Nov 27 '22 at 19:59
Take the derivative with respect to $t$ ans set $t=0$, what do you get, is the answer real or imaginary? — copper.hat, Nov 27 '22 at 20:02
However, if $f$ is real everywhere, then what must the above derivative be? — copper.hat, Nov 27 '22 at 20:04
What is a real number less a real number divided by a real number? Is it real or imaginary? And what would the limit of such number be? — copper.hat, Nov 27 '22 at 20:16
You may want to review your complex analysis, the Cauchy Riemann equations are fairly elementary. — copper.hat, Nov 27 '22 at 20:18

score 1 · Accepted Answer · answered Nov 28 '22 at 03:17

The problem can be answered without complex analysis tools. Let $n\ge 1$ be the smallest number for which $a_n\neq 0.$ Then $$f(z)=a_nz^{n}+z^{n+1}g(z)$$ where $$g(z)=\sum_{k=n+1}^\infty a_kz^{k-n-1}$$ The function $g$ is bounded for $|z|\le R/2,$ say $|g(z)|\le M.$ Hence $$\left |{f(z)\over z^n}-a_n\right |\le M|z|,\quad |z|\le {R\over 2}$$
For $z_0=re^{{-\pi i\over 2n}},$ with $0<r\le R/2,$ we get $$|a_n|\le \left |{f(z_0)\over r^n}i-a_n\right |\le Mr$$ Hence $a_n=0,$ a contradiction.

Power series of $f$ where $f(B(0,R)) \subset \mathbb{R}$

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