Question about Analytic Functions

Question

Let $f \colon U \subset \mathbb{C} \longrightarrow \mathbb{C}$, in which $U$ is an open set of $\mathbb{C}$ such that $U \supset \mathbb{R}$, be an analytic function such that $f(\mathbb{R}) \subset \mathbb{R}$ and $\lambda i \in U$, for $\lambda = \frac{1\,-\,e}{ 1\,+\,e}$.
Furthermore, suppose that the function $g \colon \mathbb{R} \longrightarrow \mathbb{R}$ such that $g(t) = f(t)$, for all $t \in \mathbb{R}$, satisfies $g'(t) = \frac{1}{1\,+\,t^2}$ and $g(0) = \pi/4$. I want to compute $f(\lambda i)$.
So, what I've thought is, since the restriction $g$ of $f$ to $\mathbb{R}$ satisfies $g(t) = \arctan(t) + k$, $k \in \mathbb{R}$, one has $g(t) = \arctan(t) + \pi/4$ because $g(0) = \pi/4$.
Now, since $f$ is analytic in $U \supset \mathbb{R}$, the Identity Principle tells us that $f(z) = g(z)$ for all $z \in U$, because $\mathbb{R} \cong \mathbb{R} \times \left\{0\right\}$ has limit points in $U$ (in fact, $\mathbb{R} \times \left\{0\right\}$ is closed in $U$), so that $f(\lambda i) = \arctan(\lambda i) + \pi/4$. By applying the identity $\arctan(z) = \frac{i}{2} \log \left(\frac{i\,-\,z}{i\,+\,z}\right)$, one has $\arctan(\lambda i) = \frac{i}{2} \log \left(\frac{1\,-\,\lambda}{1\,+\,\lambda}\right) = \frac{i}{2} \log \left(1/e\right)$, so that $f(\lambda i) = - \, i/2 + \pi/4$.
Is this reasoning correct? I came across this question in a problem list and found it interesting, so I've tried to solved it.
Any help is appreciated.