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.

Question

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 $\hat{r}$.

Answer 1

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: $$\frac{df}{dz}=\frac{\hat{r}\bullet\nabla f}{e^{j\phi}}$$