Question: Let $R = \{z \in \mathbb{C} \colon |\Re(z)|, |\Im(z)| < 1\}$. Determine all functions $f\colon R \to \mathbb{C}$ that are holomorphic in $R$ and satisfies $$f'(z) = \overline{f(\overline{z})},\ \text{for all}\ z \in R.$$
My attempt: Claim that the only functions are of the form $A e^z$ where $A \in \mathbb{C}$. Suppose $f$ is such a function, then $f$ is analytic in $R$ and it admits a unique power series representation at $z=0$ that is $f(z) = \sum_{n=0}^\infty a_n z^n$ for all $z \in R$ where $a_n \in \mathbb{C}$ for all $n =0,1,2,\dots$. By the identity we have $\sum_{n=0}^\infty a_{n+1} (n+1) z^n = \sum_{n=0}^\infty \overline{a_n} z^n$ for all $z \in R$. By uniqueness of the power series, $\overline{a_n} = a_{n+1} (n+1)$ for all $n =0,1,2,\dots$, where the radius of convergence is precisely $\infty$. However, I am not sure how to proceed from here.
With $C = a_0 = f(0)$ the recursion formula $a_{n+1} = \overline{a_n}/(n+1)$ gives $a_n = C/n!$ for even $n$, and $a_n = \overline C/n!$ for odd $n$. Therefore $$ \begin{align} f(z) &= C + \frac{\overline C}{1!}z + \frac{C}{2!}z^2 + \frac{\overline C}{3!}z^3 + \frac{C}{4!}z^4 + \frac{\overline C}{5!}z^5 +\cdots \\ &= C\left(1 + \frac{1}{2!}z^2 + \frac{1}{4!}z^4 + \cdots\right) + \overline C\left(z + \frac{1}{3!}z^3 + \frac{1}{5!}z^5 + \cdots\right) \\ &= C \cdot \cosh(z) + \overline C \cdot \sinh(z) \, . \end{align} $$ You can also write it as $$ f(z) = C \cdot \frac{e^z + e^{-z}}{2} + \overline C\cdot \frac{e^z - e^{-z}}{2} = \operatorname{Re}(C) \cdot e^z + i \operatorname{Im}(C)\cdot e^{-z} \, , $$ or $f(z) = a e^z + ibe^{-z}$ with $a, b \in \Bbb R$.
If you are familiar with the basic theory of linear differential equations with constant coefficients then you can also argue as follows: Differentiating $f'(z) = \overline{f(\overline{z})}$ gives $$ f''(z) = \overline{f'(\overline{z})} = f(z) \, , $$ so $f$ is a solution of the ordinary differential equation (ODE) $$ w'' - w = 0 \, . $$
Both $(e^z , e^{-z})$ and $(\cosh(z), \sinh(z))$ are pairs of linearly independent solutions of that ODE, so that $f$ is necessarily a linear combination of those functions.