Complex analysis using definition of the derivative 
Question: 
  $f(z) = z + 2iz^2 \operatorname{Im}(z)$
Is the function differentiable at $z = 0$?
Where is $f(z)$ analytic?

Is there any way to do this using the definition of a derivative? 
I tried plugging in $z=x+iy$ and I got to the point $-2x^2y-4xi(y^3)+x-iy+2y^3$ and now I'm stumped. 
There is a way using Cauchy but I'm not too sure. Can anyone give me a head start?
 A: Derivative at $0$:
$$
\lim_{h\to0}\frac{f(h)-f(0)}{h}=
\lim_{h\to0}(1+2ih\mathrm{Im}(h))=1
$$
So, yes, the function is differentiable at $0$. It is differentiable nowhere else, because
$$
f(z+h)=z+h+2i(z+h)^2\mathrm{Im}(z+h)
$$
and so
$$
f(z+h)-f(z)=
h+2i(2zh+h^2)\mathrm{Im}(z)+2i(z+h)^2\mathrm{Im}(h)
$$
Therefore
$$
\frac{f(z+h)-f(z)}{h}=
1+2i(2z+h)\mathrm{Im}(z)+2i\frac{(z+h)^2\mathrm{Im}(h)}{h}
$$
After expanding $(z+h)^2$, it's easy to reduce differentiability  to the existence of
$$
\lim_{h\to0}z^2\frac{\mathrm{Im}(h)}{h}
$$
If $z\ne0$, this limit doesn't exist: if you approach $0$ along the real axis you get $0$; if you approach $0$ along the imaginary axis, you get $z^2$.
A: I am not sure, because of your notation, which  is exactly the function that you mean,
 but  as a rule of thumb: functions that depend on $\overline{z}=Re(z)-Im(z)$ are not derivable. As your function depends in a single place of  on $Im(z)=\frac{z-\overline{z}}{2}$, your function have a dependence  on $\overline{z}$ and therefore it is not derivable.  
