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!
 A: 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)
A: In a different direction from the answer of James S. Cook (and perhaps more than you wanted to know).

Functions Differentiable by $\boldsymbol{z}$
The total derivative of $f=u+iv$ in the direction of $h+ik$ is given by
$$
\begin{align}
\left(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\right)(u+iv)
&=h\frac{\partial u}{\partial x}+ih\frac{\partial v}{\partial x}
+k\frac{\partial u}{\partial y}+ik\frac{\partial v}{\partial y}\\
&=(h+ik)\frac{\mathrm{d}f}{\mathrm{d}z}\tag{1}
\end{align}
$$
To exist, $\frac{\mathrm{d}f}{\mathrm{d}z}$ must not depend on what $h+ik$ is. That is,
$$
\overbrace{\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}}
^{\frac{\mathrm{d}f}{\mathrm{d}z}\text{ when }k=0}
=\overbrace{-i\frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}}^{\frac{\mathrm{d}f}{\mathrm{d}z}\text{ when }h=0}\tag{2}
$$
The real and imaginary parts of $(2)$ give the Cauchy-Riemann equations
$$
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\quad\text{and}\quad\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}\tag{3}
$$
Applying $(3)$ to $(1)$ yields
$$
\begin{align}
\frac{\mathrm{d}f}{\mathrm{d}z}
&=\frac1{h+ik}\left(\color{#C00000}{h\frac{\partial u}{\partial x}}\color{#00A000}{+ih\frac{\partial v}{\partial x}
+k\frac{\partial u}{\partial y}}\color{#C00000}{+ik\frac{\partial v}{\partial y}}\right)\\
&=\frac1{h+ik}\left(\color{#C00000}{(h+ik)\frac{\partial u}{\partial x}}\color{#00A000}{-i(h+ik)\frac{\partial u}{\partial y}}\right)\\
&=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)u\tag{4}\\
&=\frac1{h+ik}\left(\color{#C00000}{(h+ik)\frac{\partial v}{\partial y}}\color{#00A000}{+i(h+ik)\frac{\partial v}{\partial x}}\right)\\
&=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)iv\tag{5}\\
&=\frac12\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)f\tag{6}
\end{align}
$$
where $(6)$ is the average of $(4)$ and $(5)$.

Functions Differentiable by $\boldsymbol{\bar{z}}$
The total derivative of $f=u+iv$ in the direction of $h+ik$ is given by
$$
\begin{align}
\left(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\right)(u+iv)
&=h\frac{\partial u}{\partial x}+ih\frac{\partial v}{\partial x}
+k\frac{\partial u}{\partial y}+ik\frac{\partial v}{\partial y}\\
&=(h-ik)\frac{\mathrm{d}f}{\mathrm{d}\bar{z}}\tag{7}
\end{align}
$$
To exist, $\frac{\mathrm{d}f}{\mathrm{d}\bar{z}}$ must not depend on what $h+ik$ is. That is,
$$
\overbrace{\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}}
^{\frac{\mathrm{d}f}{\mathrm{d}\bar{z}}\text{ when }k=0}
=\overbrace{i\frac{\partial u}{\partial y}-\frac{\partial v}{\partial y}}^{\frac{\mathrm{d}f}{\mathrm{d}\bar{z}}\text{ when }h=0}\tag{8}
$$
The real and imaginary parts of $(8)$ give the conjugate Cauchy-Riemann equations
$$
\frac{\partial u}{\partial x}=-\frac{\partial v}{\partial y}\quad\text{and}\quad\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}\tag{9}
$$
Applying $(9)$ to $(7)$ yields
$$
\begin{align}
\frac{\mathrm{d}f}{\mathrm{d}\bar{z}}
&=\frac1{h-ik}\left(\color{#C00000}{h\frac{\partial u}{\partial x}}\color{#00A000}{+ih\frac{\partial v}{\partial x}
+k\frac{\partial u}{\partial y}}\color{#C00000}{+ik\frac{\partial v}{\partial y}}\right)\\
&=\frac1{h-ik}\left(\color{#C00000}{(h-ik)\frac{\partial u}{\partial x}}\color{#00A000}{+i(h-ik)\frac{\partial u}{\partial y}}\right)\\
&=\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)u\tag{10}\\
&=\frac1{h-ik}\left(\color{#C00000}{-(h-ik)\frac{\partial v}{\partial y}}\color{#00A000}{+i(h-ik)\frac{\partial v}{\partial x}}\right)\\
&=\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)iv\tag{11}\\
&=\frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)f\tag{12}
\end{align}
$$
where $(12)$ is the average of $(10)$ and $(11)$.

Partials with Respect to $\boldsymbol{z}$ and $\boldsymbol{\bar{z}}$
Note that $(3)$ says that for a function differentiable by $z$ we have
$$
\frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)f=0\tag{13}
$$
while $(9)$ says that for a function differentiable by $\bar{z}$ we have
$$
\frac12\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)f=0\tag{14}
$$
Therefore, we call
$$
\frac{\partial}{\partial z}=\frac12\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\tag{15}
$$
and
$$
\frac{\partial}{\partial\bar{z}}=\frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)\tag{16}
$$
since, when $f$ is differentiable by $z$,
$$
\frac{\partial f}{\partial z}=\frac{\mathrm{d}f}{\mathrm{d}z}\quad\text{and}\quad\frac{\partial f}{\partial\bar{z}}=0\tag{17}
$$
and, when $f$ is differentiable by $\bar{z}$,
$$
\frac{\partial f}{\partial\bar{z}}=\frac{\mathrm{d}f}{\mathrm{d}\bar{z}}\quad\text{and}\quad\frac{\partial f}{\partial z}=0\tag{18}
$$
Using $(15)$ and $(16)$ and expanding $\mathrm{d}z=\mathrm{d}x+i\,\mathrm{d}y$ and $\mathrm{d}\bar{z}=\mathrm{d}x-i\,\mathrm{d}y$, we get
$$
\mathrm{d}f=\frac{\partial f}{\partial z}\mathrm{d}z+\frac{\partial f}{\partial\bar{z}}\mathrm{d}\bar{z}\tag{19}
$$

Essentially, $(17)$ answers your question, the rest hopefully adds context.
