$\frac{\partial f}{\partial \bar z}\equiv 0$ as a definition for a Holomorphic function? In (Krantz, 2008; pg14) and in other places I have seen, the definition of a Holomorphic function is one for which:
$$\frac{\partial f}{\partial \bar z}\equiv 0$$
rather then the standard definition of $f$ been complex differentiable. My question is are these two definitions equivalent, and how does the former take account of singularities where the function is not analytic?
 A: There are at least three equivalent ways to define complex differentiability. Let $f\colon \mathbb C\to\mathbb C$ be a real differentiable function (i.e., differentiable as a function from $\mathbb R^2$ to $\mathbb R^2$). Then $f$ is complex differentiable if any of the three equivalent statements holds:
$(i)$ Cauchy-Riemann equations hold, i.e. if $f(x+iy) = u(x,y)+iv(x,y)$, then
\begin{align}
\dfrac{\partial u}{\partial x} &= \dfrac{\partial v}{\partial y},\\
\dfrac{\partial u}{\partial y} &= -\dfrac{\partial v}{\partial x}.
\end{align}
$(ii)$ $f(z)$ is independent of $\bar z$, i.e. $\dfrac{\partial f}{\partial\bar z} = 0$.
$(iii)$ Differential of $f$ acts as complex multiplication, i.e. $Df(z) = \lambda z$, $\lambda\in\mathbb C$.
I'm guessing that Cauchy-Riemann equations are "the familiar definition", so let us reduce $(ii)$ and $(iii)$ to $(i)$.

First of all, when we write $f(x+iy) = u(x,y)+iv(x,y)$, we are thinking of $x$ and $y$ as coordinate functions for which we have corresponding directional derivatives $\frac{\partial}{\partial x}$ and  $\frac{\partial}{\partial y}.$ But, we could also look at another coordinates, namely $z = x + iy$ and $\bar z = x - iy$. We would like to represent $\frac{\partial}{\partial \bar z}$ in terms of $\frac{\partial}{\partial x}$  and $\frac{\partial}{\partial y}$, so first we calculate:
\begin{alignat}{2}
\frac{\partial z}{\partial x} &=&\ 1,\ \frac{\partial z}{\partial y} &=&\ i,\\
\frac{\partial \bar z}{\partial x} &=&\ 1,\ \frac{\partial \bar z}{\partial y} &=&-i,
\end{alignat}
which gives us 
\begin{align}
\frac{\partial}{\partial x} &= \frac{\partial}{\partial z} + \frac{\partial}{\partial \bar z},\\
\frac{\partial}{\partial y} &= i\frac{\partial}{\partial z} - i\frac{\partial}{\partial \bar z},
\end{align}
and inverting the system matrix gives us
\begin{align}
\frac{\partial}{\partial z} &= \frac 12\left(\frac{\partial}{\partial x} -i \frac{\partial}{\partial y}\right),\\
\frac{\partial}{\partial \bar z} &= \frac 12\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right),
\end{align}
and finally
$$2\frac{\partial f}{\partial \bar z} = \left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right)(u+iv) = \left(\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}\right) + i\left(\frac{\partial u}{\partial y} + \frac{\partial v }{\partial x}\right)$$
and you can easily see that $(i)$ and $(ii)$ are equivalent.

Personally, $(iii)$ is my favorite way of looking at complex differentiation and to understand completely its relation to $(i)$, one should look at $\mathbb C$ as $2$-dimensional real algebra.
For now, we have differential of $f$ as real vector function, its matrix representation is the Jacobian
$$J_f=\begin{pmatrix}
\dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y}\\
\dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y}
\end{pmatrix}.$$
Let us examine what does it mean for linear operator on $\mathbb R^2$ to act as complex multiplication. Take some linear operator $A$. We would like that $Az = (a+ib) z$, so and let us see how it acts on basis $\{1,i\}$. $A(1)$ should of course be equal to $a + ib$, while $A(i) = (a+ib)i = -b + ia$. We see that $A$ is represented by matrix
$$\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}.$$ It is immediate that not only is this necessary, but sufficient as well, since
$$\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}
\begin{pmatrix}
x\\
y
\end{pmatrix}=
\begin{pmatrix}
ax-by\\
bx+ay
\end{pmatrix},$$ precisely what we expect of complex multiplication. If you inspect Jacobian again, you will see that $(ii)$ is equivalent to Cauchy-Riemann equations.
What is happening here is that we have faithful representation of real algebra $\mathbb C$
$$\pi\colon \mathbb C\to M_2(\mathbb R),\\
\pi(a+ib) = \begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}.$$ Moreover, this representation gives $\mathbb R^2$ structure of $\mathbb C$-module isomorphic to $\mathbb C$ with isomorphism $\varphi(a+ib) = (a,b)$. Indeed, $\varphi$ is obviously bijective and  $\varphi(\lambda z) = \pi(\lambda)\varphi(z)$.
To sum it up, with real differential we want differential to be $\mathbb R$-linear, while for complex differential we want it to be $\mathbb C$- linear. Since we understand how real differentiation works, we exploit $\mathbb C$-linear isomorphism of $\mathbb C$ and $\mathbb R^2$ to write $\mathbb C$-linearity of complex differential in those terms.
A: $f$ is complex differentiable at $a$ iff $$f(a+z) = f(a)+ cz + o(|z|)$$ for some $c \in \mathbb{C}$, thus we can say $ f'(a)= c$. 
If it is only real-differentiable then $cz$ is replaced by $M (x,y)$ for some matrix $M \in \mathbb{R}^{2 \times 2}$, equivalently by $c z+\overline{d} \overline{z}$, thus we can say $$\frac{\partial f}{\partial z}(a) = c ,\quad \frac{\partial f}{\partial \bar z}(a) = \overline{d}, \qquad f(a+z) = f(a)+ cz +\overline{d} \overline{z}+ o(|z|)$$
