Showing that $|Im[e^{-i \phi z}]| = d$? The question is taken from Visual Complex Analysis by Needham:



My problem is that my work seems to show that $d$ is the real part, not the imaginary part.



I know our new point is supposed to have angle $\theta - \phi$ and not $\phi - \theta$, but it's okay since the absolute value of the imaginary part should stay the same.
How can I justify switching around the new axes so that I get the correct answer? The reason I think I must label the axes this way is because $\phi - \theta$ is supposed to be in quadrant $1$.
 A: I don't quite see why you say your image suggests that the real part of $e^{-i\phi} z$ is $d$. It looks like your picture pretty much clinches the proof. Consider this updated picture I made:

As you probably know, if you multiply $z$ by $e^{-i\phi}$ you wind up rotating the point $z$ clockwise by $\phi$ radians. Instead of just imagining that you're doing it to $z$, imagine that you're doing it to the entire triangle $ABz$. If you do that, the segment $AB$ aligns with the imaginary axis and the segment $Bz$ becomes parallel to the real axis. From there it should be clear that the length of $AB$ is the (absolute value of the) imaginary part of $e^{-i\phi} z$.
A: I would first start with a hint: Consider the case of $\phi = 0$. The equation becomes $d = \left|\operatorname{Im}[z]\right|$, which should be intuitively true. Check what your argument looks like for this case.
My geometric argument would be as follows (spoilers):

 The operation $\exp(-i\phi)z$ inverts the rotation of the line $L$: It rotates $z$ by $-\phi$ around the origin. Let us call the result $z'$. If we assume that $L$ had been turned along with it, then we have $L'$. For $L'$, its angle is $\phi' = 0$, but its offset $d' = d$ is unchanged. Hence, if we find $d'$ of $L'$, we have found $d$ of $L$. Since $\phi'=0$, we have the "hint" case at the beginning of my answer, where $d' = \left|\operatorname{Im}[z']\right| = d$. 

Now I do believe that you understand why to take the absolute value, but for the sake of completeness, note that

 the distance is undirected - it can only be positive.

I would also like to point you to the mistake in your own solution; but to be honest, I don't really understand it. The first and second picture appear geometrically valid and all angles are indicated correctly (as far as I see), but I'm not really sure if this a promising approach, and you lost me in the third picture. I don't get what's happening there at all. But that may just be me. Anyway, maybe you need to go a bit into detail about your ideas and reasoning, if you would also specifically like to know why your solution is wrong.
