This is from page $46$ of Visual Complex Analysis by Needham:
This is my interpretation of the situation, which I think to be wrong:
I think my interpretation is wrong for the following reason: Take $L$ to be the vertical line passing through $(2, 0)$. Then $\phi = \dfrac {\pi}{2}$ and $d = 2$. Take $z = (2, 0)$. Then the point $e^{-i \frac {\pi}{2}}z$ lies on the horizontal axis, so its imaginary part is $0 \not = d$.
Another reason why I think it's wrong is that the expression $d$ is supposed to be equal to does not seem constant.
$e^{-i\pi/2}z = -2i$, which is purely imaginary, and things work out nicely. Note that $z$ is not part of the exponent.