0

What is derivative of $z^*z$ : I find two contradictory results

we have $f(z)=z^*z=x^2+y^2=u+iv$

so we have $(du/dx)+i(dv/dx)=2x$.

But according to formula 30 and 31 of http://www1.spms.ntu.edu.sg/~ydchong/teaching/06_complex_derivatives.pdf

we could compute in another way : we have $(dv/dy)-i(du/dy)=0-i2y=-2iy$, which is not equal to previous result.

Are the formulas (30) and (31) of this reference wrong ?

Is there a good reference for computing a complex derivative ?

J. W. Tanner
  • 60,406
Mathieu Krisztian
  • 240
  • 1
    consider your function may not be complex differentiable; cf. this question – J. W. Tanner Dec 30 '20 at 13:10
  • @J. W. Tanner This is very interesting. It seems that indeed, from Cauchy-Riemann equations, it would no tbe differentiable. But I don't understand how it could not be differentiable since it appears in the Higgs potential : $\mu \phi^* \phi+...$, and we could differentiate the Higgs potential in order to obtain mass, so I don't understand... – Mathieu Krisztian Dec 30 '20 at 13:14
  • 1
    It is definitely not complex differentiable. On the other hand, it is a differentiable function of two real variables. – GEdgar Dec 30 '20 at 13:18
  • Thanks a lot. Your information is extremely important. – Mathieu Krisztian Dec 30 '20 at 14:00

1 Answers1

3

While partial derivatives of complex functions are straightforward, the total derivative is trouble. Proceeding in a calculus-$101$ analogy, we write $$\frac{dw\left(z,\bar{z}\right)}{dz} = \lim_{\Delta z \rightarrow 0} \frac{\partial w}{\partial z} + \frac{\partial w}{\partial \bar{z}}\frac{\Delta\bar{z}}{\Delta z} \:,$$ where if $\Delta z=\left|\Delta z\right| e^{i\theta}$, then the ratio $\Delta\bar{z}/\Delta z$ becomes $e^{-2i\theta}$, which can have any phase $\theta$ as $\Delta z\rightarrow 0$. The best thing we can do about the total derivative is to restrict $w$ to have no explicit $\bar{z}$-dependence, eliminating the second term altogether. We therefore take the following two equations as criteria of the total derivative: \begin{equation}\label{totalderiv}\frac{dw}{dz} = \frac{\partial w}{\partial z} \hspace{2.54cm} \frac{\partial w}{\partial\bar{z}}=0\end{equation}

The Cauchy-Riemann conditions follow right from this. I think the PDF you linked made the same point. Not sure if this helps overall.

Daniele Tampieri
  • 9,727