Complex Analysis: Holomorphic Functions I am currently learning about holomorphic functions, and right after the statement about the Cauchy-Riemann equations, there is a proposition that states, "If f is holomorphic at $z_0$, then $ \frac{\partial f}{\partial {\bar{z}}} = 0"$.  For the explanation it says "Taking real and imaginary parts, it is easy to see that the Cauchy-Riemann Equations are equivalent to $ \frac{\partial f}{\partial {\bar{z}}}(z_0) = 0".$ 
However, I don't understand that explanation at all and it seems like the $\bar{z}$ came out of left field. Can someone perhaps expand on the explanation and the significance of the statement?  Thanks in advance!

Generally, $x = (z+\bar{z})/2$ and $y = (z-\bar{z})/(2i)$. Thus, any function of $x,y$ can be recast as a function of $z$ and $\bar{z}$. Also, differentials can be cast in the same light: for $f = f(x,y)$ or $f=f(z, \bar{z})$ we have
$$ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy  = 
\frac{\partial f}{\partial z} dz + \frac{\partial f}{\partial \bar{z}} d\bar{z}  $$
Setting $dx = (dz+d\bar{z})/2$ and $dy = (dz-d\bar{z})/(2i)$. Plug these into the above to derive:
$$ df = \frac{\partial f}{\partial x} (dz+d\bar{z})/2 + \frac{\partial f}{\partial y} (dz-d\bar{z})/(2i)  = 
\frac{\partial f}{\partial z} dz + \frac{\partial f}{\partial \bar{z}} d\bar{z}  $$
from which we obtain:
$$ df = \frac{1}{2}\left( \frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y} \right) dz +  \frac{1}{2}\left( \frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y} \right) d\bar{z} = \frac{\partial f}{\partial z} dz + \frac{\partial f}{\partial \bar{z}} d\bar{z} $$
hence,
$$\frac{\partial f}{\partial z} = \frac{1}{2}\left( \frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y} \right)   \ \ \& \ \ 
\frac{\partial f}{\partial \bar{z}} = \frac{1}{2}\left( \frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y} \right).  $$
Let $f=u+iv$ and notice $u_x=v_y$ and $v_x=-u_y$ is equivalent to $\frac{\partial f}{\partial \bar{z}} =0$. (work it out, it isn't long)
