Limit set of a complex difference quotient Let f be $\Bbb R$-differentiable at $z_0$. Let $g$ Be the ratio:
$$g(z)=\frac{f(z)-f(z_0)}{z-z_0}, z\neq z_0$$
Prove the limit set of $g$ is the circle centered at the point $\frac{\partial f}{\partial z}(z_0)$ of radius $|\frac{\partial f}{\partial \bar z}|$.
The limit set is defined as the set of all limit points of $g$ at $z_0$, I.e., of all complex numbers A for which there is a sequence $(z_n)$ not including $z_0$, with $|z_n -z| \to 0$ as $n \to \infty$ such that $|g(z_n)-A| \to 0$. 
My attempt: Since $f$ is $\Bbb R$-differentiable at $z_0=x_0+iy_0$,we know that if we write $f(x+iy)=u(x,y)+iv(x,y)$ as $f(x,y)=(u(x,y),v(x,y))$, then $u(x,y) = u(x_0,y_0)+u_x|x-x_0|+u_y|y-y_0|+o(x-x_0,y-y_0)|$ and $v(x,y) = v(x_0,y_0)+v_x|x-x_0|+v_y|y-y_0|+o(x-x_0,y-y_0)|$.
Then we can re-write $g(z+iy)$ as $$g(x,y) =\frac{(u(x,y),v(x,y))-(u(x_0,y_0),v(x_0,y_0))}{(x,y)-(x_0,y_0)} = \frac{(u(x,y)-u(x_0,y_0),v(x,y)-v(x_0,y_0))}{(x-x_0,y-y_0)} 
$$
So $$\Re(g) = \frac{u(x,y)-u(x_0,y_0)}{x-x_0}=u_x$$
$$\Im(g) = \frac{v(x,y)-v(x_0,y_0)}{y-y_0}$$
Thus, $\lim_{x \to x_0} \Re (g) =u_x$
and $\lim_{x \to x_0} \Im (g) =v_y$
What have I done wrong here? Am I going in the wrong direction? I feel like I don't properly understand how to convert between a complex values function and a $\Bbb{R}^2$ valued function, as when I re-wrote $g$ I had a vector on the denominator, which I'm not sure is defined. 
I understand that the question is asking me to show that this limit set is equal to $$\{x+iy:(x-\Re\frac{\partial f}{\partial z}(z_0))^2+(y-\Im\frac{\partial f}{\partial z}(z_0))^2=|\frac{\partial f}{\partial \bar z}|^2\}$$
$$= \{x+iy:(x-0.5\frac{\partial f}{\partial x}(z_0))^2+(y+0.5\frac{\partial f}{\partial y}(z_0))^2=|\frac{\partial f}{\partial \bar z}|^2\}$$But writing it this way hasn't helped me see what to do. 
Thanks very much in advance.
 A: $\mathbb{R}$-differentiability of $f:\mathbb{C}\supset A\rightarrow\mathbb{C}$ at $z_0$ implies (by definition) that:
$$f(z)=f(z_0)+\frac{\partial f}{\partial x}\Bigg\rvert_{z_0}dx+\frac{\partial f}{\partial y}\Bigg\rvert_{z_0}dy+o(|d z|)\tag{1}$$
where $dz=z-z_0$, $dx=\frac{dz+\overline{dz}}{2}$ and $dy=\frac{dz-\overline{dz}}{2i}$ and $\overline{dz}$ is the complex conjugate of $dz$, that is $\overline{dz}=\begin{bmatrix}1&\phantom-0\\0&-1\end{bmatrix}dz$ (in the sense of $(\mathbb{R}^2,\mathbb{C})$, not in the sense of $(\mathbb{C}^2,\mathbb{C})$)
So $(1)$ can b rewritten as
$$f(z)=f(z_0)+\frac{\partial f}{\partial z}\Bigg\rvert_{z_0}dz+\frac{\partial f}{\partial \bar z}\Bigg\rvert_{z_0}d\bar z+o(|d z|)\tag{2}$$
where $\frac{\partial f}{\partial z}\Big\rvert_{z_0}=\frac{1}{2}\left(\frac{\partial f}{\partial x}\Big\rvert_{z_0}-i\frac{\partial f}{\partial y}\Big\rvert_{z_0}\right)$ and $\frac{\partial f}{\partial \bar z}\Big\rvert_{z_0}=\frac{1}{2}\left(\frac{\partial f}{\partial x}\Big\rvert_{z_0}+i\frac{\partial f}{\partial y}\Big\rvert_{z_0}\right)$
Dividing both sides of $(2)$ by $dz\ne0$ yields
$$g(z)=\frac{f(z)-f(z_0)}{dz}=\frac{\partial f}{\partial z}\Bigg\rvert_{z_0}+\frac{\partial f}{\partial \bar z}\Bigg\rvert_{z_0}\frac{\overline{dz}}{dz}+\frac{o(|d z|)}{dz}\tag{3}$$
Being $\frac{\overline{dz}}{dz}=e^{-2i\arg dz}$ and $o(\vert dz\vert)$ a function such that
\begin{equation}
o(\vert dz\vert)=\alpha(dz)\vert dz\vert\\
\lim_{dz\rightarrow0}\alpha(dz)=\vert \alpha(0)\vert=0
\end{equation}
it results that
$$\lim_{n\rightarrow \infty}\frac{f(z_n)-f(z_0)}{z_n-z_0}=\frac{\partial f}{\partial z}\Bigg\vert_{z_0}+\frac{\partial f}{\partial \bar z}\Bigg\vert_{z_0}e^{-2i\lim_{z_n\rightarrow \infty}\arg (z_n-z_0)}$$
that is what you were searching for.
More rapidly: a $\mathbb{R}$-linear operaton on $\mathbb{R}^2$ is determined
by its values on two linearly independent vectors. If $a$ and $b$ are its
values at $(1,0)$ and $(0,1)$ respectively, then its value at z=(x,y)=x+iy is
given by
$$\zeta=ax+by$$
and so
$$\zeta=a\frac{z+\bar z}{2}+b\frac{z-\bar z}{2i}=\frac{1}{2}(a-ib)z+\frac{1}{2}(a+ib)\bar z=\frac{1}{2}(a-ib)z+\frac{1}{2}(a+ib)ze^{-2i\arg{z}}$$
This operator is also $\mathbb{C}$-linear operator only when $a+ib=0$.
Now $f$ is $\mathbb{R}$-differentiable when the principal component of asymptotic expansion of its displacement ($\Delta f$) w.r.t. the displacement of the indepentent variable ($dz$)is a $\mathbb{R}$-linear operator, that is
$$df(dz)=\left(\frac{1}{2}(a-ib)+\frac{1}{2}(a+ib)e^{-2i\arg{dz}}\right)dz$$
From here what you were searching for follows.
A: WLOG $z_0=0.$ Let $\partial f/\partial x(0) = a, \partial f/\partial y(0) = b.$ Because $f$ is real differentiable at $0,$ we have
$$\tag 1 f(z) - f(0) = ax +by + r(z),$$
where $r(z)/|z| \to 0.$ Now $x= (z+\bar z)/2, y = (z-\bar z)/(2i).$ Thus the right side of $(1)$ equals
$$\left (\frac{a}{2}+\frac{b}{2i}\right)z + \left (\frac{a}{2}-\frac{b}{2i}\right)\bar z + r(z).$$
Dividing by $z$ then gives
$$\tag 2 \frac{f(z)-f(0)}{z} = \left (\frac{a}{2}+\frac{b}{2i}\right)+ \left (\frac{a}{2}-\frac{b}{2i}\right)\frac{\bar z}{z} + \frac{r(z)}{z}.$$
The first summand on the right of $(2)$ equals $(a/2 + b/(2i))$ for any $z\ne 0.$ The second summand has constant modulus equal to $|a/2 - b/(2i)|$ for any $z\ne 0.$ And again, $r(z)/z\to 0.$ Put this all together and identify the terms in parentheses as $\partial f/\partial z(0)$ and $\partial f/\partial \bar z(0)$ to get the desired conclusion.
