Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
7  
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
1  
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
1  
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
add comment

2 Answers

up vote 5 down vote accepted
+50

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

share|improve this answer
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.