complex number rotation and negation Given

I have 2 questions.

*

*How do I calculate $e^{-i\nu_j}$ with $\zeta_{j+1}$ and $\zeta_j$

*Is $-i$ on the RHS is really required ? I get the correct value from python when $-i$ is not there. could this be a typo ?

This is related to https://stackoverflow.com/questions/75538966/numpy-eitheta-and-trigonometric-costheta-isintheta-does-not-match
I'am having a difficulty understanding why my numerical results are different from above.
Thanks
REF: "The Complex Variable Boundary Element Method for Potential Flow Problems" -- Miroslav Mokry
 A: First about question 2: your intention.
From page 212 of your REF, Figure 1(b) and the introduction below defines angle $\nu$ to be the angle of normal, not the angle of just $\zeta_{j+1}-\zeta_j$.
In contrast, in your linked stackoverflow question, that OP focuses on an angle $\vartheta$ -- the angle of $\zeta_{j+1}-\zeta_j$. This might explain why there's the $-i$ factor in your question image, if the goal is still to find $\nu$ (instead of $\vartheta$). The $-i$ factor rotates the angle clockwise by $\pi/2$.

Now, as $e^{i\nu_j}$ is defined as
$$e^{i\nu_j} = -i \frac{\zeta_{j+1}-\zeta_j}{\left|\zeta_{j+1}-\zeta_j\right|},$$
the required $e^{-i\nu_j}$ is simply the reciprocal of $e^{i\nu_j}$:
$$\begin{align*}
e^{-i\nu_j} &= \frac{1}{e^{i\nu_j}}\\
&= \cfrac{1}{-i \cfrac{\zeta_{j+1}-\zeta_j}{\left|\zeta_{j+1}-\zeta_j\right|}}\\
&= i \frac{\left|\zeta_{j+1}-\zeta_j\right|}{\zeta_{j+1}-\zeta_j}
\end{align*}$$
Or if you already calculated $e^{i\nu_j} = -i \frac{\zeta_{j+1}-\zeta_j}{\left|\zeta_{j+1}-\zeta_j\right|}$ (e.g. in numpy), then perform the division $1/e^{i\nu_j}$ to get $e^{-i\nu_j}$.
A: You should be able to verify that
$$(1) \quad |e^{i\theta}|=1, \quad \forall \theta \in \cal R,$$
$$(2) \quad e^{-i\theta} = \frac{ 1}{e^{i\theta}}.$$

*

*Assuming the $\zeta$'s are real (are they?), (1) and (2) imply that

$$e^{-i\nu_j}= \cos \nu_j - i \sin \nu_j = i \frac{\zeta_{j+1}-\zeta_j}{|\zeta_{j+1}-\zeta_j|}$$


*Hard to say unless you tell us what $\nu_j$ and $\zeta_j, \zeta_{j+1}$ are.

UPDATE
If $\zeta$'s are complex, as you say, then
$$e^{-i\nu_j}= \cos \nu_j - i \sin \nu_j = i \frac{\overline{\zeta}_{j+1}-\overline{\zeta}_j}{|\zeta_{j+1}-\zeta_j|}, \quad \text{ where } \overline{\zeta} = \text{Re}( \zeta ) - i \,\text{Im}(\zeta)$$
