# 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:

$$f(x) = x + \frac K{x-2 Im(x)}$$

Can anyone help?

-
What do you mean with $df/dx$? Since $f$ isn't holomorphic, $f'$ doesn't exist. If you meant the Wirtinger derivative $\partial/\partial x$, you get $\partial f/\partial x = 1$, and $\partial f/\partial x^\ast = -K/(x^\ast)^2$. –  Daniel Fischer Jul 16 '13 at 10:22
Thanks, I need to do some reading! –  atomh33ls Jul 16 '13 at 10:48
Why have you added a bounty to this? As already pointed out, the question doesn't make any sense as it stands. –  Rhys Jul 22 '13 at 15:31
Thanks, I am struggling to understand why it makes no sense at present. Any pointers much appreciated. –  atomh33ls Jul 22 '13 at 16:01
Is $f'$ supposed to be the differential of $f$, when viewed as a function $\mathbf R^2\to\mathbf C$? –  jathd Jul 22 '13 at 22:22

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, $$\frac{\partial }{\partial x}\left(z+\frac{K}{\bar z}\right) = 1-\frac{K}{\bar z^2}\quad \text{and } \ \frac{\partial }{\partial y}\left(z+\frac{K}{\bar z}\right) = i+\frac{iK}{\bar z^2} \tag3$$

-

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.

-
From the second line of your second sequence, it appears that h is a typo for y. Am I right? –  Ross Presser Jul 29 '13 at 5:48