# $f'(z)$ is imaginary on the complex axis

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

$$f'(iy) = \lim_{h \to 0} \frac{f(iy+h) - f(iy)}{h}$$ 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

$$f'(iy) = \lim_{r \to 0} \frac{f(i(y+r)) - f(iy)}{ir} = \frac{1}{i} \lim_{h \to 0} \frac{f(i(y+r)) - f(y)}{r}$$ and $f(iy)$ and $f(i(y+r))$ are real by assumption. Hence the derivative is imaginary.

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First, when you pull the $i$ out in front of the limit, it is $\frac{1}{i}$. Second, $y$ is not imaginary. You state that $y\in\mathbb{R}$. – Joe Johnson 126 Nov 8 '12 at 11:09

I don't know of a standard symbol for the imaginary line. Perhaps $i \mathbb{R}$ (I think I've seen $e^{i\theta}\mathbb{R}$ used for a line at angle $\theta$).

Your proof is along the right lines, but the details are wrong; you may have just made some mistakes in typing it out. Let $y$ and $h$ be real, and consider $$f'(iy) = \lim_{h\to 0} \frac{f(iy+ih) - f(iy)}{ih} ~.$$

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