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