0
$\begingroup$

I am reading a particular paper and in it, they have the following derivation for some complex function $x$,

$$\frac{\partial \, x^* x}{\partial \, q} = 2 x \frac{\partial \, x^*}{\partial \, q}$$

where $q$ is just some parameter that $x$ depends on.

I don't understand how this works. Is this is a general property of how complex conjugates and derivatives work? Does the function have to have a special property to be true? For what class of functions should this work?


In trying to understand this, I got as far as,

$$\frac{\partial \, x^* x}{\partial \, q} = x \frac{\partial \, x^*}{\partial \, q} + x^* \frac{\partial \, x}{\partial \, q}$$

using the product rule of derivatives, but I don't see how,

$$x \frac{\partial \, x^*}{\partial \, q} = x^* \frac{\partial \, x}{\partial \, q}$$

to get the result above.

$\endgroup$
  • $\begingroup$ You're correct. It is not true that $\bar z=z$ unless $\text{Im}(z)=0$. $\endgroup$ – Mark Viola Mar 24 '17 at 22:07
1
$\begingroup$

Write $x= a+ib$, with $a,b \in \mathbb R$. Then $x^*x = a^2+b^2$ and

$$\frac{\partial(x^*x)}{\partial q} = 2a \frac{\partial a}{\partial q}+ 2b \frac{\partial b}{\partial q}. $$

On the other hand,

$$2x \frac{\partial x^*}{\partial q} = 2(a+ib) \left(\frac{\partial a}{\partial q}- i \frac{\partial b}{\partial q}\right) = 2a\frac{\partial a}{\partial q}+2b \frac{\partial b}{\partial q} -2ai \frac{\partial b}{\partial q}+2ib \frac{\partial a}{\partial q}.$$

So your equality holds if and only if

$$a\frac{\partial b}{\partial q}=b\frac{\partial a}{\partial q}. $$

$\endgroup$
  • $\begingroup$ Is this a well-known class of complex functions? $\endgroup$ – XYZT Mar 24 '17 at 22:46
  • $\begingroup$ Not as far as I know, but I'm not an expert in complex analysis. $\endgroup$ – Stefano Mar 24 '17 at 22:52
0
$\begingroup$

Just write out the components: $x = a + i b$ and $x^* = a - i b$ and perform all the derivatives, then gather your terms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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