What is the value of $\dfrac{d z^*}{dz}$?

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?

• Sep 23, 2022 at 15:48
• @ParclyTaxel That link does not answer this question. Sep 23, 2022 at 16:41
• 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. Sep 23, 2022 at 18:21

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 $$f'(z)=\frac{\partial}{\partial z}f(z)$$.

• @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 $\frac{\partial f}{\partial z}=0$. Sep 23, 2022 at 16:39
• @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 Sep 24, 2022 at 9:22

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