I'm having trouble with an algebraic operation in a proof, which I will copy here:
Specifically, I do not see the connection between steps 4.8 and 4.9.
As best as I can tell, the equations in 4.8 were multiplied by $\frac1{\Delta z}$ and $\frac i{\Delta z}$ respectively, then added together, to obtain 4.9. When I try to reproduce the operation, however, I get something familiar but off. And I still do not know how the inequality popped up.
Can someone idiot-proof the algebraic operations used to obtain 4.9 from 4.8?
OK, so I've figured out the algebra, at least. For simplicity of notation below, I used $\Delta x=dx$ and $\frac{\partial u}{\partial x}=u_x$.
We start by multiplying the 4.8 equations by $1/dz$ and $i/dz$ respectively, then add them, to obtain:
$$\frac{du+idv}{dz}=\frac1{dz}\left(u_xdx+iv_xdx+u_ydy+iv_ydy\right)+\frac{dx}{dz}(\epsilon_1+i\epsilon_2)+\frac{dy}{dz}(\delta_1+i\delta_2)$$
Then we note $u_xdx+iv_xdx+u_ydy+iv_ydy=u_xdx+iv_xdx-v_xdy+iu_xdy$ by the Cauchy-Riemann equations. Which we turn into a product $(dx+idy)(u_x+iv_x)=dz(u_x+iv_x)$.
Finally we obtain
$$\frac{du+idv}{dz}=\frac1{dz}dz(u_x+iv_x)+\cdots$$
And then
$$\frac{du+idv}{dz}-(u_x+iv_x)=\frac{dx}{dz}(\epsilon_1+i\epsilon_2)+\frac{dy}{dz}(\delta_1+i\delta_2)$$
Noting that
$$\left|\frac{du+idv}{dz}-(u_x+iv_x)\right|=\left|\frac{dx}{dz}(\epsilon_1+i\epsilon_2)+\frac{dy}{dz}(\delta_1+i\delta_2)\right|$$
and
$$\left|\frac{dx}{dz}(\epsilon_1+i\epsilon_2)+\frac{dy}{dz}(\delta_1+i\delta_2)\right|\leq\left|\frac{dx}{dz}(\epsilon_1+i\epsilon_2)\right|+\left|\frac{dy}{dz}(\delta_1+i\delta_2)\right|$$
which is the inequality I was looking for.