$\frac{d}{dz}$ and $\frac{d}{d\overline{z}}$ for $|z-a|^p$ I wish to find $\frac{d}{dz}$ and $\frac{d}{d\overline{z}}$ of $f(z)=|z|$ and $|z-a|^p$, $-\infty < p < \infty$.
$\frac{d}{dz} = \frac{1}{2} \left ( \frac{du}{dx} + \frac{dv}{dy}\right ) + \frac{1}{2}\left ( \frac{dv}{dx} - \frac{du}{dy}\right )$
$\frac{d}{d\overline{z}} = \frac{1}{2} \left ( \frac{du}{dx} - \frac{dv}{dy}\right ) + \frac{i}{2}\left ( \frac{dv}{dx} + \frac{du}{dy}\right )$
How does this work for $|z-a|^{p}$?
 A: HINT:
Write $|z|=(z\bar z)^{1/2}$ and $|z-a|^p=(z-a)^{p/2}(\bar z-\bar a)^{p/2}$
Then, take partial derivatives with respect to $z$ and $\bar z$.
SPOLIER ALEERT Scroll over the highlighted area to reveal the answer

We have the transformation $(x,y)\to (z,\bar z)$ as given by $$z=x+iy$$and $$\bar z=x-iy$$and the inverse transformation $(z,\bar z) \to (x,y)$ as given by $$x=\frac12 (z+\bar z)$$and $$y=\frac{1}{2i}(z-\bar z)$$Then, from the chain-rule, we can write $$\frac{\partial f}{\partial z}=\frac12\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+\frac i2\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right) \tag 1$$along with $$\frac{\partial f}{\partial \bar z}=\frac12\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+\frac i2\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\tag 2$$We note that if $f$ is analytic, $\frac{\partial f}{\partial \bar z}=0$ and thus, $(2)$ yields the Cauchy-Riemann equations.  Now, if $f=|z-a|^p$, we have $f=(z-a)^{p/2}(\bar z-\bar a)^{p/2}$ and $$\frac{\partial f}{\partial z}=\frac p2(z-a)^{p/2-1}(\bar z-\bar a)^{p/2}=\frac p2 (\bar z-\bar a)|z-a|^{p-2}$$Alternatively, we can use $(1)$ to calculate $\frac{\partial f}{\partial z}$.  Noting that the $v=0$, we have $$\begin{align}\frac{\partial f}{\partial z}&=\frac12 \frac{\partial u}{\partial x}-\frac i2 \frac{\partial u}{\partial y}\\\\&=\frac p2 (x-a_r) |z-a|^{p-2}-i\frac12 (y-a_i)|z-a|^{p-2}\\\\&=\frac p2 (\bar z-\bar a)|z-a|^{p-2} \end{align}$$as expected!  A parallel development is straightforward for evaluating $\frac{\partial f}{\partial \bar z}$ and is left as an exercise for the reader.

