1
$\begingroup$

While doing math exercises, I came across the following expression:

$$ \dfrac{d z^*}{dz}, $$ where $\cdot^*$ indicates the conjugate operator and $z \in \mathbb{C}$. The initial guess was to solve it by doing the following: $$ \dfrac{d z^*}{dz} = \left(\dfrac{d z}{dz}\right)^* = 1^* = 1, $$ but I am insecure over the first equality. Furthermore, I've found on internet (without demonstration) that $$ \dfrac{d z^*}{dz} = 0. $$

Is that right?

$\endgroup$
3
  • 2
    $\begingroup$ Cf. math.stackexchange.com/q/3200743/357390 $\endgroup$ Commented Sep 23, 2022 at 15:48
  • $\begingroup$ @ParclyTaxel That link does not answer this question. $\endgroup$
    – Mark Viola
    Commented Sep 23, 2022 at 16:41
  • 1
    $\begingroup$ One really has to be careful about notation here. If $\frac{d}{dx}$ means the complex derivative then that derivative doesn't exist. But we could also be talking about the Wirtinger derivative which is a different operation and equal to $0$ in this case. $\endgroup$ Commented Sep 23, 2022 at 18:21

2 Answers 2

2
$\begingroup$

It depends on what you mean by $\frac{d}{d z}\bar z$. If you want to mean the complex derivative of $\bar z$, that is if

$$ \frac{d}{d z}f(z):=f'(z):=\lim_{w\to 0}\frac{f(z+w)-f(z)}{w} $$

then $\frac{d}{d z}\bar z$ doesn't exists, as can be checked by the given definition. However its possible that you wanted to mean $\frac{\partial}{\partial z}\bar z$, then in this case you will have that

$$ \begin{align*} \frac{\partial}{\partial z}\bar z&=\frac1{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)(x-iy)\\ &=\frac1{2}\left(\frac{\partial}{\partial x}x-i\frac{\partial}{\partial x}y-i\frac{\partial}{\partial y}x-\frac{\partial}{\partial y}y\right)\\ &=\frac1{2}\left(1-0-0-1\right)=0 \end{align*} $$

as $\frac{\partial}{\partial z}:=\frac1{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)$.

To complete a bit more, there is a theorem than says that, if $f$ is holomorphic at $z$ then $f'(z)=\frac{\partial}{\partial z}f(z)$.

$\endgroup$
2
  • 1
    $\begingroup$ @GEdgar is correct. The derivative of the complex conjugate fails to exist since $\lim_{\Delta z\to 0}\frac{\overline{( z+\Delta z)}-\overline z}{\Delta z}=\lim_{\Delta z\to 0}\frac{\overline{\Delta z}}{\Delta z}$ fails to exist. But if we view $f(z, \bar z)=\bar z$ as a function of the two independent variables, $z$ and $\bar z$, then $\frac{\partial f}{\partial z}=0$. $\endgroup$
    – Mark Viola
    Commented Sep 23, 2022 at 16:39
  • $\begingroup$ @MarkViola you and GEdgar were right, I just mixed the notations $\frac{d}{d z}$ and $\frac{\partial}{\partial z}$ in my head. I've rewritten my answer $\endgroup$
    – Masacroso
    Commented Sep 24, 2022 at 9:22
0
$\begingroup$

Note that $* : \mathbb{C} \to \mathbb{C}$ with $*(x, y) = (x, -y)$. The derivative is $$D*(x, y) = \begin{pmatrix}1 & 0 \\ 0 & -1 \\ \end{pmatrix}.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .