1. Complex Directional Derivative.

I have the following definition of the complex directional derivative :
$D_{w_1} f(z) = lim_{t \rightarrow 0} \frac{f(z+tw_1)-f(z)}{t}$ where $|w_1| = 1$.
$D_{w_1} f(z) = lim_{t \rightarrow 0} \frac{u(x+tx_1,y+ty_1)-u(x,y)}{t}+i\frac{v(x+tx_1,y+ty_1)-v(x,y)}{t}$
From the definition of real directional derivative
$D_{w_1} f(z) = x_1\frac{du}{dx}+y_1\frac{du}{dy}+i(x_1\frac{dv}{d x}+y_1\frac{dv}{dy})$
But now if $w_1 = i$, $x_1 = 0$, $y_1 = 1$ and $D_{w_1} f(z) = \frac{du}{dy}+i \frac{dv}{dy}$ which does not agree with Cauchy Riemann equations.
Could someone tell me where I am mistaken please?

2. The directional derivative does not need to "agree with" the Cauchy–Riemann equations. In fact, it has nothing to do with them. For the directional derivative to exist, the function does not even have to be analytic. For example, the complex conjugate function $f(z) = \bar{z}$ is not analytic and so does not satisfy the C–R equations, but it has directional derivatives in every direction.

3. Thanks I think I understand why.
So is $

D_{w_1} f(z) = x_1\frac{du}{dx}+y_1\frac{du}{dy}+i(x_1\frac{dv}{d x}+y_1\frac{dv}{dy})
$
So is $