Differentiate a function containing a variable and its complex conjugate If I have a function of x:
$$f(x) = x + \frac K{x^*}$$
Where $x$ is a complex number and $x^*$ is its conjugate.
How can I find $f'(x)$ ?
My first thoughts are to rearrange:
$$f(x) = x + \frac K{x-2 Im(x)}$$
But in general I am unsure where to start with this. 
Can anyone help?
 A: I'm not ready to use $x$ for a complex number, so I'll write $z$ instead of $x$ and $\bar z$ instead of $z^*$.  In terms of Wirtinger derivatives we have 
$$\frac{\partial }{\partial z}\left(z+\frac{K}{\bar z}\right) = 1\quad \text{and } \ \frac{\partial }{\partial \bar z}\left(z+\frac{K}{\bar z}\right) = -\frac{K}{\bar z^2} \tag1$$
-- one can operate with these derivatives as if $z$ and $\bar z$ were independent  variables. 
Incidentally, we can get "real" derivatives out of (1) pretty easily, using
$$
\frac{\partial }{\partial x} = \frac{\partial }{\partial z} + \frac{\partial }{\partial \bar z},\qquad \frac{\partial }{\partial y} = i\frac{\partial }{\partial z} - i\frac{\partial }{\partial \bar z} \tag2
$$
Namely, 
$$\frac{\partial }{\partial x}\left(z+\frac{K}{\bar z}\right) = 1-\frac{K}{\bar z^2}\quad \text{and } \ \frac{\partial }{\partial y}\left(z+\frac{K}{\bar z}\right) = i+\frac{iK}{\bar z^2} \tag3$$
A: Let us compute $\frac{\partial f}{\partial x}$. By definition this is
$$\frac{\partial f}{\partial x}(x)=\lim_{y\rightarrow0}\frac{f(x+y)-f(x)}{h}.$$
The most interesting feature of this formula is the quotient in the right hand side, which is a quotient between complex numbers (two-dimensional real vectors).
We get 
\begin{align}
\frac{\partial f}{\partial x}(x)&=\lim_{y\rightarrow0}\frac{f(x+y)-f(x)}{y}\\
&=\lim_{y\rightarrow0}\frac{x+y+K/(x+y)^*-x-K/x^*}{y}\\
&=1+K\cdot\lim_{y\rightarrow0}\left[\frac{-1}{x^*(x^*+y^*)}\cdot\frac{y^*}{y}\right].
\end{align}
As an aside $\lim_{y\rightarrow0}\frac{-1}{x^*(x^*+y^*)}=-1/(x^*)^2\neq0$. Therefore, the limit defining $\frac{\partial f}{\partial x}(x)$ exists if and only if $\lim_{y\rightarrow0}\frac{y^*}{y}$ exists. But this limits doesn't exist. For example, if $y$ approaches $0$ along the reals, then $y^*/y=1$ (because for reals $y^*=y$) but if $y$ approaches $0$ along the pure imaginary numbers $y^*/y=-1$ (because for pure imaginary numbers $y^*=-y$). Notice, this is happening precisely because these are complex numbers. 
