Is the complex tangent space closed under conjugation? Let $D$ be a domain in $\mathbb{C}^n$ with $C^2$ boundary and let $\rho$ be a defining function for $D$.  Let $m \in \partial D$.  The complex tangent space to $\partial D$ at $m$ is $\mathcal{T}_m(\partial D) =  \{w \in \mathbb{C^n}: \sum_{i=1}^n \frac{\partial{\rho}}{\partial{z_i}}(m) w_i = 0\}$.  Is this set closed under complex conjugation; that is, if $w \in \mathcal{T}_m(\partial D)$, then is $\overline{w} \in \mathcal{T}_m(\partial D)$?
$\textbf{Update}:$ Here are some facts about defining functions.
A $C^2$ function $\rho$ is a defining function if $\rho = 0 $ on $\partial D$, $\rho < 0$ in D, $\rho > 0$ in $\bar{D}^c$, and $\nabla{\rho} \neq 0$ on $\partial D$.  
For example, if $D = B^2(0, 2)$ then a defining function is $\rho(z) = |z|^2 - 2$.
It can be shown that the complex tangent space is independent of the choice  of defining function.
 A: Take the following example. I will use $(z,w) \in \mathbb{C}^2$ for my coordinates.  $\rho = z+\bar{z} +i w - i \bar{w}$.  That's a real valued function and defines a plane.   You can check that $(i,1)$ is in your space but $(-i,1)$ is not.  Complex conjugation does not preserve complex vector subspaces.
But the question is deeper, and really the definition that way is wrong.  Or at least it is unnatural.  A tangent vector for a hypersurface in $\mathbb{C}^n$ is not a point in $\mathbb{C}^n$.  Really a tangent vector is an element of the tangent space.  And the tangent space is the space of differentiations.  So for a point $p$, let's think of the tangent vectors of $\mathbb C^n$ as if it were $\mathbb R^{2n}$, ignoring the complex structure for a moment:
$$
T_p {\mathbb C}^n = \operatorname{span}_{\mathbb R} \left\{ \frac{\partial}{\partial x_1}, \ldots, \frac{\partial}{\partial y_n} \right\}
$$
Then, this is a real vector space ($2n$-dimensional).  You can "complexify" this by tensoring with $\mathbb C$, that is, just taking the complex span:
$$
\mathbb C \otimes T_p {\mathbb C}^n = \operatorname{span}_{\mathbb C} \left\{ \frac{\partial}{\partial x_1}, \ldots, \frac{\partial}{\partial y_n} \right\}
$$
which is a complex $2n$-dimensional space, and this is where $\frac{\partial}{\partial z_j}$ and $\frac{\partial}{\partial \bar{z}_j}$ live.
Now define
$$
T^{(0,1)}_p \mathbb C^n = 
\operatorname{span}_{\mathbb C} \left\{ \frac{\partial}{\partial \bar{z}_1}, \ldots, \frac{\partial}{\partial \bar{z}_n} \right\}
$$
and similarly $T^{(1,0)}_p \mathbb C^n$.  Then the vectors that kill a defining function $\rho$ for a real manifold $M$ are the tangent spaces for $M$.  That is, e.g., $v \in T^{(1,0)}_p M \subset T^{(1,0)}_p \mathbb{C}^n$ whenever $v(\rho) = 0$ (remember $v$ is a derivation).
Now $T^{(0,1)}_p M$ and $T^{(1,0)}_p M$ are disjoint subspaces (only intersect at the origin) of $\mathbb C \otimes T_p M$.  If you define the conjugation operator (a real linear operator on $\mathbb C \otimes T_p M$) to take $\frac{\partial}{\partial z_j}$ to $\frac{\partial}{\partial \bar{z}_j}$, you can find that vectors from $T^{(0,1)}_p M$ are the conjugates of vectors from $T^{(1,0)}_p M$.  Those two vector spaces are therefore isomorphic as real vector spaces.
So the issue is that what you want is "unnatural", that is, the (0,1) and (1,0) vector spaces are same dimension, but have different basis, and conjugation which takes one basis to the other is only real-linear and not complex linear.  Your definition is really identifying $T^{(1,0)}_p M$ with a subspace of $\mathbb C^n$, by identifying the standard euclidean basis $e_j$ with $\frac{\partial}{\partial z_j}$.  Then complex conjugation on $\mathbb{C}^n$ does not preserve complex vector subspaces.
