In my analysis 2 book the proof goes like this:
If a complex function $f = P(x,y) + iQ(x,y)$ is differentiable at a point $z$, then
$$ \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z} = \alpha \in \mathbb{C} $$
where $\Delta z = \Delta x + i\Delta y$ doesn't depend on the path that it takes to reach zero. That means that it doesnt depend on the argument, and because
$$ tg \ arg \Delta z = \frac{\Delta y}{\Delta x} $$
that means that the limit does not depend on the quotient $\frac{\Delta y}{\Delta x}$.
Now the part that I'm having trouble with:
Now when we express the derivative using differentials we have \begin{align} \alpha = f'(z) &= \frac{df}{dz} = \frac{d(P(x,y) + iQ(x,y))}{d(x + iy)} = \frac{dP + idQ}{dx + idy} = \\ &= \frac{P_xdx + P_ydy + i(Q_xdx + iQ_ydy)}{dx + idy} = \\ &= \frac{(P_x + iQ_x)dx + (P_y + iQ_y)dy}{dx + idy} \tag{*} \end{align} Independence of the way $\Delta z$ goes to zero implies that the last fraction (*) does not depend on the quotient $\frac{dy}{dx}$. That means that for all $\frac{dy}{dx} = t \in \mathbb{R}$, $ (P_x + iQ_x) + (P_y + iQ_y)t = \alpha + i\alpha t$, which is equivalent to $P_x + iQ_x = \alpha $ and $P_y + iQ_x = i\alpha$, i.e. $$ P_x = Q_y, P_y = -Q_x $$
What bothers me is the use of $\frac{dy}{dx}$ as a simple quotient, when it denotes a derivative, and also the mixing of differentials of multivalued functions such as $P$ and $Q$ and the differential of $x$ and $y$ which are independent variables. I don't understand when can I treat derivatives in Leibniz notation as simple quotients and move them around like in this proof.
