Does this formula derived from the Cauchy-Riemann differential equations have a name? I was trying to prove the Cauchy-Riemann differential equations and I found something interesting while up in the derivation. What I will present below is somehow my reversed proof for them, that showed the following formula: For a complex differentiable  function $f(x+yi) = u(x,y) + v(x,y)i$ and any vector (or maybe unit vector) $\langle a, b\rangle$, we have $$D_{\langle a, b\rangle}u + iD_{\langle a, b \rangle}v = (a+bi)f'(x+yi)$$
1) Is this true? Are my proofs below correct?
2) Does this formula have a name. Are there any references for it?
Proof 1 (this is the first part of my proof for the Cauchy-Riemann differential equations that led to the formula above):
Starting by looking at the fact that a function $f : G \subseteq \mathbb{C} \to \mathbb{C}$, with $u = \mathop{\rm Re} f$ and $v = \mathop{\rm Im} f$ is (complex-)differentiable at $z \in G$ if the limit $\lim_{h\to 0} \frac{f(z+h)-f(z)}{h}$ exists, and that means that the limit is equal for all ways $h$ could possibly approach 0, as $h$ is a number on the complex plane.
Then, I took the limit definition of $f'(z)$ and I let
$$ h = k(a+bi), \hspace{0.5cm} z = x+yi,\hspace{0.5cm} k,a,b \in \mathbb{R} \\ k = \frac{h}{a+bi}, \ \lim_{h\to 0} k = 0$$
Under this substitution, the limit becomes
$$ f'(z) = \lim_{k\to 0} \frac{f((x+ka) + (y+kb)i) - f(x+yi)}{k(a+bi)} $$
And this limit must be equal for any $a, b \in \mathbb{R}$. Further
$$ \begin{align}f'(z) &= \lim_{k\to 0} \frac{u(x+ka, y+kb) + v(x+ka, y+kb)i - u(x,y) - v(x,y)i}{k(a+bi)} \\ &= \lim_{k\to 0} \frac{u(x+ka, y+kb) - u(x,y)}{k(a+bi)} + i\lim_{k\to k} \frac{v(x+ka, y+kb) - v(x,y)}{k(a+bi)} \\ &= \frac{1}{a+bi}\left(\lim_{k\to 0} \frac{u(x+ka,y+kb) - u(x,y)}{k} + i\lim_{k\to 0} \frac{v(x+ka, y+kb)-v(x,y)}{k}\right) \\ &= \frac{1}{a+bi}\left(D_{\langle a, b \rangle} u(x,y) + iD_{\langle a, b \rangle}v(x,y)\right) \end{align}$$
This leads to the formula above.
Proof 2 (This is the reversed rest of the proof I done):
For a complex-differentiable function, the Cauchy-Riemann differential equations tell that $u_x = v_y$ and $f'(z) = \frac{\partial f}{\partial x} = u_x + v_xi$. So for a vector $\langle a, b\rangle$ we have:
$$ (a+bi)f'(z) = (a+bi)(u_x + v_xi) = au_x - bv_x + (av_x + bu_x)i = au_x + bu_y + (av_x + bv_y)i = \langle a, b\rangle \cdot \nabla u + i\langle a, b\rangle \cdot \nabla v = D_{\langle a, b\rangle}u + iD_{\langle a, b\rangle}v$$
 A: This is true, your proofs seem ok, and fiddling around with limits gets you the right answer, however I have provided a slightly more intuitive proof that you may find insightful.
Proving the Cauchy-Riemann equations by considering rotations, will quickly lead to the general case you found in one go!
If you are familiar with linear algebra, we can re-write multiplication of complex numbers by a multiplication of a matrix with basis $\ [1,i] $.
$$\ (u+iv)(x+iy)=ux-vy + (uy+vx) $$
we muts have the matrix of the following form,
$$\begin{bmatrix}u & -v \\v & u \end{bmatrix}*\begin{bmatrix}  x \\ y \end{bmatrix} =\begin{bmatrix}ux -vy\\vx+uy \end{bmatrix}$$
Hence multiplication will work. Similarly if we consider the derivative with respects to basis $\ [1,i] $, 
$$\ D_{[1,i]}= \begin{bmatrix}u_{x} & u_{y} \\v_{x} & v_{y} \end{bmatrix} $$
By uniqueness of multiplication, this gives us the Cauchy-Riemann equations. To explore what happens with another basis, we simply multiply by the change of basis matrix. In our case, we change to basis $\ [(a+bi), i(a+bi)] $. 
Note that this is also a multiplication, hence the change of base matrix simply represents the multiplication matrix of $\ a+bi $ which is
$$\ A= \begin{bmatrix}a & -b \\b & a \end{bmatrix} $$
So,
$$\ D_{[(a+bi),i(a+bi)]}=D_{[1,i]}*A $$
which, since all of the above are in the same multiplicative group, is isomorphic to the complex numbers and hence represents precisely what you stated above.
(For a bit more explanation of the last step, consider right multiplying by any point in the complex plane, wrt to the $\ [1,i] $ basis)
Since all we are doing is a change of basis, I don't think it has a special name. I would say these are just identities/alternatives of the Cauchy Riemann equations. Since Cauchy Riemann is the simplest of these forms it works when ever we work with $\ z=x+iy   $ etc. and so we don't ever really require this general result.
