tangent vector on a complex manifold Let $x_1,x_2, \ldots, x_n$ be local coordinates on a manifolds $M$. One can interpret $\frac{\partial}{\partial x_i}(p)$ as a tangent vetor to a curve with constant $x_j$ (where $j \neq i$).
What is the interpretation of $\frac{\partial}{\partial z_i}(p)$ on a complex manifold in local coordinates?
I'll be glad for any references.
 A: I feel like the standard texts are a little brief explaining this. I was confused about it for a really long time. This is going to be a fairly long answer.
First up, lets start with a smooth manifold $M$ with dimension $n$. Let $ \mathscr{S}$ be the sheaf of smooth functions on $M$. More concretely, $ \mathscr{S}(U)$ is the $ \mathbb{R}$-algebra of smooth functions $U \to \mathbb{R}$. We are going to define another sheaf on $M$. Let $ \mathscr{C}$ be the sheaf of complex valued smooth functions on $M$. More concretely, $ \mathscr{C}(U)$ is the $ \mathbb{C}$-algebra of smooth functions $ U \to \mathbb{C}$, where by smooth, I mean that each component function is smooth. It is a standard result from differential geometry that vector fields on $ U \subseteq M$ are the same as $ \mathbb{R}$-linear maps $ D : \mathscr{S}(U) \to \mathscr{S}(U)$ satisfying $D(fg) = f D(g) + g D(f) $. These are called $ \mathbb{R}$-linear derivations.
It is natural to ask the following question: Are the $ \mathbb{C}$-linear derivations $ \mathscr{C}(U) \to \mathscr{C}(U)$ smooth sections of some vector bundle over $U$? The answer is yes! Here is the idea (to make this rigorous quite a lot more work is required. Inpaticular, at some point you need to muck around with partitions of unity). Firstly, if $ D: \mathscr{S}(U) \to \mathscr{S}(U)$ is an $ \mathbb{R}$-linear derivation, then $D(\alpha + i \beta) = D \alpha + i D \beta$ defines a $\mathbb{C}$-linear derivation $ \mathscr{C}(U) \to \mathscr{C}(U)$. Now assume that $U$ is small, $x_1, \dots, x_n$ are coordinates on $U$ and $ D : \mathscr{C}(U) \to \mathscr{C}(U)$ is a $ \mathbb{C}$-derivation. Taylor's theorem tells us that 
$$ D (\alpha + \beta i) = \sum c_j \frac{\partial (\alpha + \beta i)}{\partial x_j} \quad c_j \in \mathscr{C}(U). $$
This is exactly a smooth section of
$$ TM \otimes_{\mathbb R} \mathbb{C} = \coprod_{p \in M} T_p \otimes_{\mathbb R} \mathbb{C} $$
Before we start talking about complex manifolds, notice that we still have the non degenerate pairing
$$ TM \otimes_{\mathbb{R}} \mathbb{C} \times TM^* \otimes_{\mathbb{R}} \mathbb{C} \to \mathscr{C} $$
defined by $( \partial / \partial x_i , dx_j) \mapsto \delta_{ij}$. This allows us to define the differential of $ f \in \mathscr{C}(U) $ by $ (\partial / \partial x_i, df) = \partial f / \partial x_i $. 
OK, Now let $X$ be a complex manifold with complex dimension $n$. Let $z_j = x_j + i y_j$  be holomorphic coordinates on $U \subseteq X$. Then both $z_j $ and $ \overline{z_j}$ are in $ \mathscr{C}(U)$. Inpaticular, we have that $ dz_j = d x_j + i d y_j $ and $ d \overline{z_j} = dx_j - i dy_j$. From the equations $ 2 dx_j = dz_j + d \overline{z_j}$ and $ 2i dy_j = dz_j - d \overline{z_j} $, it follows that the sections of $ TM^* \otimes_{\mathbb{R}} \mathbb{C}$ over $U$ are given by 
$$ \sum_j f_j dz_j + g_j d \overline{z_j} \quad f_j,g_j \in \mathscr{C}(U) $$ Under the non-degenerate pairing, the dual basis to $ dz_j, d \overline{z_j}$ is given by 
$$ \frac{\partial}{\partial z_j} = 1/2 \left( \frac{\partial}{\partial x_j} - i \frac{\partial}{\partial y_j} \right) $$
$$ \frac{\partial}{\partial \overline{z_j}} = 1/2 \left( \frac{\partial}{\partial x_j} + i \frac{\partial}{\partial y_j} \right) $$
This is how I think about $ \partial / \partial z_j$.
A: DBr's answer is very good.  Here is another way to think about it:
In a smooth manifold, as the OP says, one way to think of $\dfrac{\partial}{\partial {x_i}}$ at $p$ is as taking the derivative along the curve $x_j =$ constant for $j \neq i$.  In other words, we get a map $\gamma$ from an interval $I$ around $0$ 
to the manifold $M$, taking $0$ to $p$, and taking $t$ to $(0,\ldots, t, \ldots, 0)$ (assuming coords. are chosen so that $p$ is at the origin), where $t$ appears in the $i$th place and all other coords. are $0$.  We can compose a function $f$
with $\gamma$, and then differentiate $f\circ \gamma$ w.r.t. $t$.  This gives the value of $\dfrac{\partial f}{\partial x_i}$ at $p$.
Similarly, on a complex manifold with local coords. $z_i$, we may define a map $\gamma$ in exactly the same way, but now taking a n.h. $U$ of $0$ in the complex plane to $M$ (so $t$ is now a complex variable rather than a real one).
If $f$ is a holomorphic function on $M$, then the composite $f\circ \gamma$
is a holomorphic function on $x$, and we may differentiate it w.r.t. $t$.  This gives
$\dfrac{\partial f}{\partial z_i}$ at $p$.
