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If $f:\mathbb{C}\to\mathbb{C}$ is any smooth function and $z\in\mathbb{C}$, the derivative $df_z$ of $f$ at $z$ is a $\mathbb{R}$-linear operator from the tangent space $T_{z}\mathbb{C}$ to $\mathbb{C}$ (and $T_{z}\mathbb{C}$ can of course be canonically identified with $\mathbb{C}$ since $\mathbb{C}$ is a vector spacespace). Said differently, $df$ is a complex-valued differential $1$-form on $\mathbb{C}$ (i.e. an element of the space $\Omega^1(\mathbb{C};\mathbb{C})=\Omega^1(\mathbb{C})\otimes_{\mathbb{R}} \mathbb{C}$).

Any $\mathbb{R}$-linear operator $A$ from one complex vector space to another can be written uniquely as $A=B+C$, where $B$ is complex-linear (i.e. $B(iv)=iBv$ for all $v$) and $C$ is complex anti-linear (i.e. $B(iv)=-iBv$). Specifically, let $Bv=\frac{1}{2}(Av-iAiv)$ and $Cv=\frac{1}{2}(Av+iAiv)$.

The operators $\partial$ and $\bar{\partial}$ can then be characterized as follows: for a function $f:\mathbb{C}\to\mathbb{C}$, one has the unique decomposition $$df=\partial f+\bar{\partial} f$$ where the complex-valued $1$-forms $\partial f$ and $\bar{\partial}f$ are such that, at each $z\in\mathbb{C}$, $(\partial f)_z$ is complex-linear and $(\bar{\partial} f)_z$ is complex anti-linear. At least to me that justifies the notation--it makes more sense to have $\partial f$ be the linear part of $df$ and $\bar{\partial} f$ be the antilinear part than the other way around.

As for $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$, note first that as a special case of the above discussion one has functions $z:\mathbb{C}\to\mathbb{C}$ (which is just the identity) and $\bar{z}:\mathbb{C}\to\mathbb{C}$, and therefore complexified one-forms $dz,d\bar{z}\in\Omega^1(\mathbb{C};\mathbb{C})$. (More specifically, since $z=x+iy$ and $\bar{z}=x-iy$, one has $dz=dx+idy$ and $d\bar{z}=dx-idy$, so for $dz$ and $d\bar{z}$ the signs on $i$ are what you wanted them to be.)

At each point of $\mathbb{C}$, $\{dz,d\bar{z}\}$ is a basis (over $\mathbb{C}$) for the complexified cotangent space. The operators $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$ are then naturally defined as the complexified vector fields such that at every point the basis $\{\frac{\partial}{\partial z},\frac{\partial}{\partial\bar{z}}\}$ for the complexified tangent space is the dual basis to the basis $\{dz,d\bar{z}\}$ for the complexified cotangent space. Imposing this dual basis requirement, the formulas $dz=dx+idy$ and $d\bar{z}=dx-idy$ then readily yields yield the standard formulas for $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$, and gives give rise to the pleasant identity, for any smooth $f:\mathbb{C}\to\mathbb{C}$, $$df=\partial f+\bar{\partial} f=\frac{\partial f}{\partial z}dz+\frac{\partial f}{\partial{\bar{z}}}d\bar{z}$$ just like one would get if $z$ and $\bar{z}$ were real coordinates on a real two-manifold.

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If $f:\mathbb{C}\to\mathbb{C}$ is any smooth function and $z\in\mathbb{C}$, the derivative $df_z$ of $f$ at $z$ is a $\mathbb{R}$-linear operator from the tangent space $T_{z}\mathbb{C}$ at $z$ to $\mathbb{C}$ (and $T_{z}\mathbb{C}$ can of course be canonically identified with $\mathbb{C}$ since $\mathbb{C}$ is a vector space. Said differently, $df$ is a complex-valued differential $1$-form on $\mathbb{C}$ (i.e. an element of the space $\Omega^1(\mathbb{C};\mathbb{C})=\Omega^1(\mathbb{C})\otimes_{\mathbb{R}} \mathbb{C}$).

Any $\mathbb{R}$-linear operator $A$ from one complex vector space to another can be written uniquely as $A=B+C$, where $B$ is complex-linear (i.e. $B(iv)=iBv$ for all $v$) and $C$ is complex anti-linear (i.e. $B(iv)=-iBv$). Specifically, let $Bv=\frac{1}{2}(Av-iAiv)$ and $Cv=\frac{1}{2}(Av+iAiv)$.

The operators $\partial$ and $\bar{\partial}$ can then be characterized as follows: for a function $f:\mathbb{C}\to\mathbb{C}$, one has the unique decomposition $$df=\partial f+\bar{\partial} f$$ where the complex-valued $1$-forms $\partial f$ and $\bar{\partial}f$ are such that, at each $z\in\mathbb{C}$, $(\partial f)_z$ is complex-linear and $(\bar{\partial f})_z$ (\bar{\partial} f)_z$is complex anti-linear. At least to me that justifies the notation--it makes more sense to have$\partial f$be the linear part of$df$and$\bar{\partial} f$be the antilinear part than the other way around. As for$\frac{\partial}{\partial z}$and$\frac{\partial}{\partial \bar{z}}$, note first that as a special case of the above discussion one has functions$z:\mathbb{C}\to\mathbb{C}$(which is just the identity) and$\bar{z}:\mathbb{C}\to\mathbb{C}$, and therefore complexified one-forms$dz,d\bar{z}\in\Omega^1(\mathbb{C};\mathbb{C})$. (More specifically, since$z=x+iy$and$\bar{z}=x-iy$, one has$dz=dx+idy$and$d\bar{z}=dx-idy$, so for$dz$and$d\bar{z}$the signs on$i$are what you wanted them to be.) At each point of$\mathbb{C}$,${dz,d\bar{z}}$\{dz,d\bar{z}\}$ is a basis (over $\mathbb{C}$) for the complexified cotangent space. The operators $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$ are then naturally defined as the complexified vector fields such that at every point the basis ${\frac{\partial}{\partial z},\frac{\partial}{\partial\bar{z}}}$ \{\frac{\partial}{\partial z},\frac{\partial}{\partial\bar{z}}\}$for the complexified tangent space is the dual basis to the basis${dz,d\bar{z}}$\{dz,d\bar{z}\}$ for the complexified cotangent space. Imposing this dual basis requirement, the formulas $dz=dx+idy$ and $d\bar{z}=dx-idy$ then readily yields the standard formulas for $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$, and gives rise to the pleasant identity, for any smooth $f:\mathbb{C}\to\mathbb{C}$, $$df=\partial f+\bar{\partial} f=\frac{\partial f}{\partial z}dz+\frac{\partial f}{\partial{\bar{z}}}d\bar{z}$$ just like one would get if $z$ and $\bar{z}$ were real coordinates on a real two-manifold.

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If $f:\mathbb{C}\to\mathbb{C}$ is any smooth function and $z\in\mathbb{C}$, the derivative $df_z$ of $f$ at $z$ is a $\mathbb{R}$-linear operator from the tangent space $T_{z}\mathbb{C}$ at $z$ to $\mathbb{C}$ (and $T_{z}\mathbb{C}$ can of course be canonically identified with $\mathbb{C}$ since $\mathbb{C}$ is a vector space. Said differently, $df$ is a complex-valued differential $1$-form on $\mathbb{C}$ (i.e. an element of the space $\Omega^1(\mathbb{C};\mathbb{C})=\Omega^1(\mathbb{C})\otimes_{\mathbb{R}} \mathbb{C}$).

Any $\mathbb{R}$-linear operator $A$ from one complex vector space to another can be written uniquely as $A=B+C$, where $B$ is complex-linear (i.e. $B(iv)=iBv$ for all $v$) and $C$ is complex anti-linear (i.e. $B(iv)=-iBv$).

The operators $\partial$ and $\bar{\partial}$ can then be characterized as follows: for a function $f:\mathbb{C}\to\mathbb{C}$, one has the unique decomposition $$df=\partial f+\bar{\partial} f$$ where the complex-valued $1$-forms $\partial f$ and $\bar{\partial}f$ are such that, at each $z\in\mathbb{C}$, $(\partial f)_z$ is complex-linear and $(\bar{\partial f})_z$ is complex anti-linear. At least to me that justifies the notation--it makes more sense to have $\partial f$ be the linear part of $df$ and $\bar{\partial} f$ be the antilinear part than the other way around.

As for $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$, note first that as a special case of the above discussion one has functions $z:\mathbb{C}\to\mathbb{C}$ (which is just the identity) and $\bar{z}:\mathbb{C}\to\mathbb{C}$, and therefore complexified one-forms $dz,d\bar{z}\in\Omega^1(\mathbb{C};\mathbb{C})$. (More specifically, since $z=x+iy$ and $\bar{z}=x-iy$, one has $dz=dx+idy$ and $d\bar{z}=dx-idy$, so for $dz$ and $d\bar{z}$ the signs on $i$ are what you wanted them to be.)

At each point of $\mathbb{C}$, ${dz,d\bar{z}}$ is a basis (over $\mathbb{C}$) for the complexified cotangent space. The operators $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$ are then naturally defined as the complexified vector fields such that at every point the basis ${\frac{\partial}{\partial z},\frac{\partial}{\partial\bar{z}}}$ for the complexified tangent space is the dual basis to the basis ${dz,d\bar{z}}$ for the complexified cotangent space. Imposing this dual basis requirement, the formulas $dz=dx+idy$ and $d\bar{z}=dx-idy$ then readily yields the standard formulas for $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$, and gives rise to the pleasant identity, for any smooth $f:\mathbb{C}\to\mathbb{C}$, $$df=\partial f+\bar{\partial} f=\frac{\partial f}{\partial z}dz+\frac{\partial f}{\partial{\bar{z}}}d\bar{z}$$ just like one would get if $z$ and $\bar{z}$ were real coordinates on a real two-manifold.