$U$ is the square with verices in $0, 1, i, i+1$. I though immediately using the identity principle. The thing is what I know about it is that, if U is an open and connected set, holomorph in U such that exists $z_n, f, g$ such that $f(z_n)=g(z_n)$ and $\lim_\limits{n\to\infty}z_n\in U$ then $f=g$.
The problem I have here, is that the simple $z_n=1/n$ is not un $U$, but in $\bar{U}$. Can I use it anyway?
Thanks
After chatting with different people, I've reached two ansers. In both it can be easily seen that $f(z)=e^z+cos(z)$ is a solution. Let $\bar{f}$ be another function that satisfies the condition:
1) Using the Schwarz reflection principle, I can extend both functions $f$ and $\bar{f}$ to the square with vertices in $0, 1, -i, 1-i$ and consider the function $g(z)=f(z)-\bar{f}(z)$. I call the new domain as $V$. Then, I can define $z_n=0.1+\frac{1}{n+1}$. This sequence is contained in the interior of $V$, and converges in $0.1$ that is in the interior of $V$. Then, the Identity Principle says the new function is $0$ for every $z$ in the domain. Then $g \equiv 0$. Then $f \equiv \bar{f}$
2) I define the function $g(z)=i\cdot(z-\frac{1}{2}-\frac{i}{2}) + \frac{1}{2} + \frac{i}{2}$. It's not hard to prove that this rotates the square in $\frac{\pi}{2}$. I also define the functions $h_0(z)=f(z)-\bar{f}(z)$, $h_1(z)=h(g(z))$, $h_2(z)=h(g^2(z))$, $h_3(z)=h(g^3(z))$. I define $h(z)=h_0(z)\cdot h_1(z) \cdot h_2(z) \cdot h_3(z)$. Then, $z\in\partial U \Rightarrow h(z)=0$. Then, using the maximum module principle, $h\equiv 0$. Then, since $h$ is product of holomorph functions, it must be that one of them is the zero function. But if one of them is, the other ones are too, because I can obtain either of them from the others composing with $g$ enough times. Then $h_0 \equiv 0$. Then $f \equiv \bar{f}$