The divergence of holomorphic vector field Consider the holomorphic vector field. Why the divergence of this field is zero? We can write $f(z)=u(x,y)+iv(x,y)$ and the divergence is not equal to $\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}$? Or if this field are holomorphic, then we should consider only the fields of type $(u,-v)$?
 A: If $f(z) = u + iv$ is holomorphic, it satisfies the Cauchy-Riemann equations; 
$u_x = v_y, \tag 1$
$u_y = - v_x; \tag 2$
thus the divergence of the $2$-dimensional vector field $(u, v)$ is
$\nabla \cdot (u, v) = u_x + v_y = u_x + u_x = 2u_x \ne 0 \tag 3$
in general. For example, taking
$f(z) = z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy, \tag 4$
we have
$u = x^2 - y^2, \tag 5$
$v = 2xy, \tag 6$
$u_x = 2x, \tag 7$
$v_y = 2x, \tag 8$
$\nabla \cdot (u, v) = u_x + v_y = 2x + 2x = 4x \ne 0. \tag 9$
We thus see that vector fields $(u, v)$ associated to holomorphic functions $f(z) = u(x, y) + i v(x, y)$ are not in general divergence free; indeed if 
$u_x + v_y = \nabla \cdot (u, v) = 0, \tag{10}$
then
$u_x = -v_y, \tag{11}$
but by (1) we obtain
$u_x = - u_x \Longrightarrow u_x = 0 = v_y; \tag{12}$
it follows then that $u$ is a function of $y$ alone, and $v$ is a function of $x$ alone; the same is then true or $u_y$ and $v_x$; hence by (2), we must have $u_y$ and $v_x$ constant:
$u_y = - v_x = c \in \Bbb R; \tag{13}$
it then follows that $u$ and $v$ must take the form(s)
$u = cy + u_0, \; v = -cx + v_0; \tag{14}$
hence
$f(z) = u + iv = cy + u_0 - icx + iv_0 = c(y - ix) + (u_0 + iv_0)$
$= -ic(x + iy) + u_0 + iv_0 = -icz + \beta, \tag{15}$
where $\beta \in \Bbb C$ is an arbitrary complex constant.  Holomorphic functions of the form (15) are the only possible ones satisfying (10).
On the other hand, if $f(z) = u + iv$ is holomorphic, then the vector field $(u, -v)$ is in fact divergence free:
$\nabla \cdot (u, -v) = u_x + (-v_y) = u_x - v_y = 0 \tag{16}$
by (1).
It is perhaps worth noting that similar concerns apply to the curl operator $\nabla \times$; we have
$\nabla \times (u, v) = v_x - u_y = -u_y - u_y = -2 u_y \ne 0; \tag{17}$
indeed, in the case $f(z) = z^2$,
$\nabla \times (x^2 - y^2, 2xy) = 2y - (-2y) = 4y \ne 0; \tag{18}$
on the other hand
$\nabla \times (u, - v) = -v_x - u_y = -v_x - (-v_x) = -v_x + v_x = 0. \tag{19}$
If $(u, v)$ is curl free,
$v_x - u_y = \nabla \times (u, v) = 0, \tag{20}$
so
$v_x = u_y = -v_x \tag{21}$
by (2); thus
$v_x = 0 = u_y; \tag{22}$
in this case $v$ depends only on $y$ and $u$ only on $x$; now by (1),
$u_x = v_y = c, \tag{23}$
$c$ constant, and
$u = cx + u_0, \tag{24}$
$c = cy + c_0, \tag{25}$
$u + iv = cx + u_0 + i(cy + v_0) = c(x + iy) + (u_0 + iv_0) = cz + \beta, \tag{26}$
$\beta \in \Bbb C$ again an arbitrary complex constant.  
