$F$ is linear so is equal to its derivative at every point I am reading Stein's Complex analysis. It currently discusses the difference between real and complex derivatives and uses the example $f(z) = \bar{z}.$ It then states that this corresponds to the map $F: (x, y) \mapsto (x, -y)$ which is differentiable in the real sense. I am confused by the next statement: "$F$ is linear and is therefore equal to its derivative at every point." What does that mean? I can sort of see it if they mean the derivative of linear functions are equal everywhere...Is that what they are getting at? 
 A: The derivative of a function $F \colon \mathbb{R}^m \to \mathbb{R}^n$ at a point $p \in \mathbb{R}^m$ is the linear map $DF \colon \mathbb{R}^m \to \mathbb{R}^n$ given by the matrix of partial derivatives at $p$:
$$DF(p) = \begin{pmatrix}
\frac{\partial F^1}{\partial x^1}(p) & \cdots & \frac{\partial F^1}{\partial x^m}(p) \\
\vdots & & \vdots \\
\frac{\partial F^n}{\partial x^1}(p) & \cdots & \frac{\partial F^n}{\partial x^m}(p) \end{pmatrix}$$
Notice that the entries of this matrix are functions: they depend on the choice of $p$.
Suppose now that $F \colon \mathbb{R}^m \to \mathbb{R}^n$ is a linear map, let's say
$$F(x_1, \ldots, x_m) = \begin{pmatrix}
a_{11} & \cdots & a_{1m} \\
\vdots & & \vdots \\
a_{n1} & \cdots & a_{nm} \end{pmatrix} \begin{pmatrix} x_1 \\ \vdots \\ x_m \end{pmatrix} = \begin{pmatrix} a_{11}x_1 +  \cdots + a_{1m}x_m \\ \vdots \\ a_{n1}x_1 + \cdots + a_{nm}x_m \end{pmatrix}.$$
Its derivative $DF(x_1, \ldots, x_m)$ at $(x_1, \ldots, x_m) \in \mathbb{R}^m$ will be the matrix of partial derivatives at $(x_1, \ldots, x_m)$.  Calculate for yourself: $DF(x_1, \ldots, x_m)$ is exactly the constant matrix
$$DF(x_1, \ldots, x_m) = \begin{pmatrix}
a_{11} & \cdots & a_{1m} \\
\vdots & & \vdots \\
a_{n1} & \cdots & a_{nm} \end{pmatrix}$$
That is, all the entries are constants and the matrix we get is the one we started with.
In your situation, the map $F \colon \mathbb{R}^2 \to \mathbb{R}^2$ is the linear function
$$F(x,y) = (x, -y) = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.$$
The derivative of $F$ is then the constant matrix
$$DF(x,y) = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$
