# How does $\frac{\partial x^* x}{\partial q}$ simplify to $2 x \frac{\partial x^*}{\partial q}$?

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.

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

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

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

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

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