Am reading chapter 2 in Sarason's Complex Function Theory book and am really confused about the material.
Definition. A complex function $f:G\to\mathbb{C}$ defined on an open set $G\subset\mathbb{C}$ is said to be differentiable at $z_0\in G$ if there exist a $c\in \mathbb{C}$ such that
$$ \lim_{z\to z_0} \frac{R(z)}{z-z_0}=0$$
where $R(z)=f(z)-f(z_0)-c (z-z_0)$.
Definition. A function $u:G\to\mathbb{R}$ defined on an open set $G\subset\mathbb{R}^2$ is said to be differentiable at $(x_0,y_0)\in G$ if there exist $a,b\in \mathbb{R}$ such that
$$ \lim_{(x,y)\to (x_0,y_0)} \frac{R(x,y)}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0$$
where $R(x,y)=u(x,y)-u(x_0,y_0)-a(x-x_0)-b(y-y_0)$.
So far so good. But now on p.15 Sarason writes "to facilitate a comparison with complex differentiation, we restate the preceding definition in complex notation" and then gives
Definition. A function $u:G\to\mathbb{R}$ defined on an open set $G\subset\mathbb{C}$ is said to be differentiable at $z_0=x_0+i y_0 \in G$ if there exist $a,b\in \mathbb{R}$ such that
$$ \lim_{z\to z_0} \frac{R(z)}{z-z_0}=0$$
where $R(z)=u(z)-u(z_0)-a(x-x_0)-b(y-y_0)$.
Question 1: Where is the absolute value in the denominator?
Next, Sarason writes
Let $f:G\to\mathbb{C}$ be defined on an open set $G\subset\mathbb{C}$ and let $u$ and $v$ denotes its real and imaginary parts, so that $f=u+i v$. Given a point $z_0=x_0+iy_0 \in G$ and a complex number $c=a+bi$ we can write
$$R(z)=f(z)-f(z_0)-c (z-z_0)= \big[ u(z)-u(z_0)-a(x-x_0)+b(y-y_0) \big]+ i \big[ v(z)-v(z_0)-b(x-x_0)-a(y-y_0) \big] = R_1(z)+i R_2(z)$$
Then comes the statement
"Clearly $\lim_{z\to z_0} \frac{R(z)}{z-z_0}=0$ if and only if $\lim_{z\to z_0} \frac{R_1(z)}{z-z_0}=0$ and $\lim_{z\to z_0} \frac{R_2(z)}{z-z_0}=0$ ".
Finally Sarason concludes "$f$ is differentiable (in the complex sense) at $z_0$ if and only if $u$ and $v$ are differentiable (in the real sense) at $z_0$".
Question 2: Why is the iff statement true? If $\lim_{x \to 0} f(x)+g(x) = 0$ we don't necessarily have $\lim_{x \to 0} f(x)=0$ and $\lim_{x \to 0} g(x)=0$ so I don't know how to obtain the "only if" part. The "if" part is ok, but in any case the definition of real differentiability of $u$ is
$$ \lim_{z\to z_0} \frac{R_1(z)}{\lvert z-z_0 \rvert}=0$$ and not
$$ \lim_{z\to z_0} \frac{R_1(z)}{z-z_0}=0$$
And I don't see why these two should be equivalent.
Thanks a lot for your help.