1. ## Complex Directional Derivative.

I have the following definition of the complex directional derivative :
$\displaystyle D_{w_1} f(z) = lim_{t \rightarrow 0} \frac{f(z+tw_1)-f(z)}{t}$ where $\displaystyle |w_1| = 1$.
$\displaystyle D_{w_1} f(z) = lim_{t \rightarrow 0} \frac{u(x+tx_1,y+ty_1)-u(x,y)}{t}+i\frac{v(x+tx_1,y+ty_1)-v(x,y)}{t}$
From the definition of real directional derivative
$\displaystyle D_{w_1} f(z) = x_1\frac{du}{dx}+y_1\frac{du}{dy}+i(x_1\frac{dv}{d x}+y_1\frac{dv}{dy})$
But now if $\displaystyle w_1 = i$, $\displaystyle x_1 = 0$, $\displaystyle y_1 = 1$ and $\displaystyle D_{w_1} f(z) = \frac{du}{dy}+i \frac{dv}{dy}$ 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 $\displaystyle f(z) = \bar{z}$ 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 $\displaystyle D_{w_1} f(z) = x_1\frac{du}{dx}+y_1\frac{du}{dy}+i(x_1\frac{dv}{d x}+y_1\frac{dv}{dy})$ correct, if I may ask?

4. Originally Posted by vincisonfire
Thanks I think I understand why.
So is $\displaystyle D_{w_1} f(z) = x_1\frac{du}{dx}+y_1\frac{du}{dy}+i(x_1\frac{dv}{d x}+y_1\frac{dv}{dy})$ correct, if I may ask?
Yes (the derivatives should be partials, of course).

5. Yes I didn't take the time to put the partial symbols.