Show $\frac{df}{dz}=\frac{\hat{r}\bullet\nabla f}{e^{j\phi}}$ . Where $z$ is a complex number and $f$ is differentiable at z. Show $\frac{df}{dz}=\frac{\hat{r}\bullet\nabla f}{e^{j\phi}}$ . 
Where $z$ is a complex number and f is differentiable at $z$.
The $\bullet$ denotes the dot(inner) product.
$\nabla$ is the gradient.
$j=\sqrt{-1}$.
$\hat{r}=cos(\phi)\hat{x}+sin(\phi)\hat{y}$is an arbitrary unit vector in xy plane.
$\phi$ is an arbitrary angle associated with the direction of $\hat{r}$.
 A: By definition $$\frac{df}{dz}=\lim_{\Delta z\to 0} \frac{f(z+\Delta z)-f(z)}{\Delta z}$$
Let $z=x+jy$ where $x$ and $y$ are real and imaginary parts of $z$.
Let $\Delta z=\Delta\rho e^{(j\phi)}$ and $f=u(x,y)+jv(x,y)$ wher u and v are real functions of $x$ and $y$. So we have
$$\frac{df}{dz}=\lim_{\Delta\rho\to 0} \frac{u(x+\Delta \rho \cos(\phi),y+\Delta \rho \sin(\phi))-u(x,y)}{\Delta \rho e^{(j\phi)}}+j \frac{v(x+\Delta \rho \cos(\phi),y+\Delta \rho \sin(\phi))-v(x,y)}{\Delta \rho e^{(j\phi)}}$$
the above expression can be written as the directional derivative of u and v in the direction of $\hat{r}=\cos(\phi) \hat{x}+\sin(\phi) \hat{y}$, $\mathbf {D}_\mathbf{\hat{r}}$, as below
$$\frac{df}{dz}=\frac{\mathbf {D}_\mathbf{\hat{r}}u+j \mathbf{D}_\mathbf{\hat{r}}v}{e^{(j\phi)}}$$
since $\mathbf {D}_\mathbf{\hat{r}}u=\hat{r}\bullet\nabla u$ we get
$$\frac{df}{dz}=\frac{\hat{r}\bullet\nabla u+j \hat{r}\bullet\nabla v}{e^{(j\phi)}}$$
or finally:
$$\frac{df}{dz}=\frac{\hat{r}\bullet\nabla f}{e^{j\phi}}$$
