A question from Bak-Newman. Suppose $f(z)$ is real-valued and differentiable for all imaginary points $z$. Show that $f'(z)$ is imaginary at all imaginary points $z$.
First, is there a standard symbol for the set of imaginary points $\{z \in \mathbb{C}: z = iy, y \in \mathbb{R}\}$?
Second, is the following proof legit? Edited to address some sloppiness on my part (thanks to comments). Let $y \in \mathbb{R}$ be arbitrary. Then $iy$ is an arbitrary point on the complex axis. By the definition of the derivative we have
\begin{equation} f'(iy) = \lim_{h \to 0} \frac{f(iy+h) - f(iy)}{h} \end{equation} where $h \in \mathbb{C}$. However, if the derivative exists it is the same no matter the manner in which $h \to 0$. In particular if we restrict $h$ to be purely imaginary the above equality still holds. In fact, letting $h = ir$, $r \in \mathbb{R}$ we get
\begin{equation} f'(iy) = \lim_{r \to 0} \frac{f(i(y+r)) - f(iy)}{ir} = \frac{1}{i} \lim_{h \to 0} \frac{f(i(y+r)) - f(y)}{r} \end{equation} and $f(iy)$ and $f(i(y+r))$ are real by assumption. Hence the derivative is imaginary.