Analyticity of a function depending on $z$ and $\bar{z}$ Say $z \in \mathbb{C}$ and $\bar{z}$ the complex conjugate (i.e. with $\bar{z} z = \left|z \right|^2$).
Can a function of $z$ and $\bar{z}$ be analytical?
Example: $f(z,\bar{z}) = Az^3 + B \bar{z} z$
I thought no, because the partial derivatives will depend on the direction in the complex plane (i.e. the phase of the line along which you take the derivative limit).
Thanks!
 A: One of the many equivalent definitions for a function to be holomorphic is $\displaystyle \frac{\partial f}{\partial \bar{z}} = 0$
$\displaystyle \frac{\partial f}{\partial \bar{z}} = 0$ is equivalent to Cauchy Riemann equations as shown below.
$$x = \frac{z+\bar{z}}{2} \text{ and } y = \frac{z-\bar{z}}{2i}$$
$$\frac{\partial f}{\partial \bar{z}} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \bar{z}} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right)$$
So if $f = u(x,y) + i v(x,y)$, where $u,v: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$, then
$$\frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}$$
$$\frac{\partial f}{\partial y} = \frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y}$$
Hence,$$\frac{\partial f}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right) + \frac{i}{2} \left( \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \right)$$
Hence, you find that $$\left( \frac{\partial f}{\partial \bar{z}} = 0 \right) \iff \left(\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \text{ and } \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} \right)$$
