Find all points where f(z) is differentiable.

Question

I have $f(z)=2x+ixy^2$. I want to see if what I've done is correct or not.

I've found the Cauchy-Riemann equations: $2=2xy$ and $0=-y^2$. Obviously the function is differentiable for all $y=0$. Does that mean that the function is differentiable at $(1/y, 0)$ for all y in the real numbers?

Answer 1

The system that you got was$$\left\{\begin{array}{l}2=2xy\\0=-2y^2\end{array}\right.$$The second equation has only one solution: $y=0$. But if $y=0$, then the first equality doesn't hold.

The conclusion is that $f$ is differentiable nowhere.

Answer 2

The idea is that a function f is "complex differentiable" if both equations hold. The second equation you have implies that y must be 0.

If you plug that into the first equation you get that 2=0, which is absurd.

Therefore, f is differentiable nowhere (empty set)

Answer 3

Actually, the function is not differentiable anywhere since your first equality becomes invalid for $y = 0$.