How is the complex differential form $d\overline{z}$ defined? Consider a two-dimensional real smooth manifold $M$. For simplicity, take $M=\mathbb{R}^2$. We can talk about the $1$-forms, $dx$ and $dy$; the $2$-form $dx\wedge dy$, where $x$ and $y$ are the coordinate maps. Fix a point $p\in M$. The differential $(dx)_p$ is an element of the dual of the (real) tangent space $T_pM$. 
For a one-dimensional complex analytic manifold $N$, say $\mathbb{C}$, we can talk about the complex $1$-forms, $dz$, $d\overline{z}$. (I know nothing but the names of these two objects in the complex case.) Naively, $(dz)_q$ is an element in the dual of the "(complex) tangent space" $T_qN$ where $z$ is the coordinate map. But there is only one complex dimension. How is $d\overline{z}$ really defined? 
 A: For $M=\mathbb{C}=\{x+\sqrt{-1}y:x,y\in\mathbb{R}\}$, 
$dz=dx+\sqrt{-1}dy$ and $d\overline{z}$ is the conjugate of $dz$, i.e. 
$$d\overline{z}=dx-\sqrt{-1}dy.$$
To think about this, remember that for $M=\mathbb{R}^2$, $1$-forms $dx$ and $dy$ are dual to $\displaystyle\frac{\partial}{\partial x}$ and $\displaystyle\frac{\partial}{\partial y}$ in the sense that 
$$dx(\frac{\partial}{\partial x})=dy(\frac{\partial}{\partial y})=1\mbox{ and }
dx(\frac{\partial}{\partial y})=dy(\frac{\partial}{\partial x})=0.$$
With these in mind, $dz$ and $d\overline{z}$ in $\mathbb{C}$ are dual to 
$\displaystyle\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-\sqrt{-1}\frac{\partial}{\partial y}\right)$ and $\displaystyle\frac{\partial}{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial}{\partial x}+\sqrt{-1}\frac{\partial}{\partial y}\right)$ in the sense that 
$$dz(\frac{\partial}{\partial z})=d\overline{z}(\frac{\partial}{\partial \overline{z}})=1\mbox{ and }
dz(\frac{\partial}{\partial \overline{z}})=d\overline{z}(\frac{\partial}{\partial z})=0.$$
For example, to check $\displaystyle dz(\frac{\partial}{\partial z})=1$, we compute
$$dz(\frac{\partial}{\partial z})=(dx+\sqrt{-1}dy)\left(\frac{1}{2}\left(\frac{\partial}{\partial x}-\sqrt{-1}\frac{\partial}{\partial y}\right)\right)\\
=\frac{1}{2}\left[dx\left(\frac{\partial}{\partial x}\right)
+\sqrt{-1}dy\left(\frac{\partial}{\partial x}\right)
-\sqrt{-1}dx\left(\frac{\partial}{\partial y}\right)
+dy\left(\frac{\partial}{\partial y}\right)\right]
=1.$$
The other can be checked similarly.
