Show using the definition, that f is differentiable at $x_0+i0$ 
Question:
Show using the definition, that f is differentiable at $x_0+i0$ where $$f= x^2+y^2+2xyi$$

My attempt:
I know I must use the definition of differentiability but I cannot see where to start in the case of complex numbers.
Where would I begin with this question?
 A: You want to show that 
$$
\lim_{(x,y) \rightarrow (x_0, 0)} \frac{f(x,y) - f(x_0,0)}{(x,y) - (x_0,0)} = \frac{x^2 + y^2 + 2xy - x_0^2}{(x-x_0,y)}
$$
exists. It will help if you know where you are going, i.e. to know what limit you should be looking for. You can use Cauchy-Riemann equations to do that. Indeed, letting $u(x,y)$ and $v(x,y)$ denote the real and imaginary parts of $f$, we have that
$$
u_x = 2x, u_y = 2y, v_x = 2y, v_y = 2x.
$$
Plugging in the point $(x_0,0)$, we have that $u_x(x_0,0) = 2x_0, u_y(x_0,0) = 0, v_x(x_0,0) = 0,$ and $v_y(x_0,0) = 2x_0.$ 
We see by the Cauchy-Riemann equations that indeed $f$ is complex-differentiable at any point $(x_0,0)$. If you were allowed to solve the problem this way, you would be done. Anyway, now we know what the derivative is: $f'(x_0,0) = 2x_0,$ since in general $f'(z) = u_x + iv_x$.
Now try to show that for any $\epsilon > 0$ there exists $\delta > 0$ such that
$$
\left|\frac{x^2 + y^2 + 2xy - x_0^2}{(x-x_0,y)} - 2x_0 \right| < \epsilon
$$
whenever $|(x,y) - (x_0,0)| < \delta$.
It takes a bit of work, but it does come out. 
