I am having a hard time understanding how to derive a cartesian complex number with respect to its polar magnitudes. Particularly, I have struggled understanding the section 3 of this technical note.
In the note we have the complex number $V$:
Voltage $V=|V| \cdot e^{j \theta}$
Voltage unitary vector $E=\frac{V}{|V|}$
Until here, all fine. Now comes the derivatives (which I do not understand)
Voltage derivative with respect to the voltage module: $\frac{\partial V}{\partial |V|} =\frac{V}{|V|}=E $
Voltage derivative with respect to the voltage angle: $\frac{\partial V}{\partial \theta} = j V$
I don't know which rules have been used to obtain those results.
Think of $V$ as a function, namely $V: \mathbb{R}^+_0 \times \mathbb{R} \to \mathbb{C}$ given by $$V(m, a) := me^{ja},$$ where $m$ stands for modulus and $a$ for angle. Morally, this should be equivalent to saying $V = |V| e^{j\theta}$. But let's follow the formal route. We can now define the voltage unitary vector (function) $E: \mathbb{R}^+_0 \times \mathbb{R} \to \mathbb{R}$ as $$E(m,a) := \frac{V(m,a)}{|V(m,a)|} = e^{ja}.$$
The derivative of $V$ wrt. the modulus is simply $\frac{\partial V}{\partial m}(m,a) = e^{ja}$. This equals $E(m,a)$ for all values of $m, a$. If you leave out the function arguments, this is equivalent to $\frac{\partial V}{\partial m} = E$ — exactly what is written in your post.
The derivative of $V$ wrt. the angle is simply $\frac{\partial V}{\partial a}(m,a) = jme^{ja} = jV(m, a)$. Again by leaving out the arguments, that is $\frac{\partial V}{\partial a} = jV$.