# Complex functions and where they are complex differentiable and holomorphic

How can I show where $f(z)=3x^3 − 6xy^2 + i(3x^2y + 2x^3)$ is complex-differentiable and where it is holomorphic.

I solved the Cauchy-Riemann equations:

$U_x=6x^2-6y^2=V_y=3x^2$

$-U_y=12xy=V_x=6xy+6x^2$

To which I found $f(z)$ is complex differentiable at $(0,0)$ and $\left (\frac{2}{3},\frac{2}{3} \right)$ (which could very well be wrong...)

How do I show where it is holomorphic?

Many thanks!

• For $f$ to be analytic/holomorphic , Cauchy-Riemann must be satisfied in an open set. Otherwise it is only differentiable. Maybe you can use, e.g., Wolfram to double-check your work. – Gary. Aug 17 '15 at 20:16
• Your $U_x$ is wrong. – mrf Aug 17 '15 at 20:25

Let $f(z)=3x^3 − 6xy^2 + i(3x^2y + 2x^3)$ and setting $U(x,y)=3x^3 − 6xy^2$ and $V(x,y)=3x^2y + 2x^3$, then $U_x=9x^2-6y^2$, $U_y=-12xy$; also, $V_x= 6xy+6x^2$, $V_y=3x^2$.

For C.R. eq. to be hold we must have $U_x=V_y$ and $U_y=-V_x$ so that $$9x^2-6y^2= 3x^2.........(1)$$ and $$-12xy=- (6xy+6x^2)..........(2)$$ thus eqs. (1) and (2) simplified respectively to be $$x^2-y^2=0 \Rightarrow x=y..........(3)$$ and $$x^2-xy=x(x-y)=0.......(4)$$ From (4) we have $x=0$ or $x=y$, (which gives the same result in eq (3)).

So that C.R. eqs., are satisfied at $x=y$, where $x,y \in \mathbb{R}$.

Or we can say $f$ satisfies the C.R. eqs. on the set $E=\{{(x,y):y=x,x\in\mathbb{R}}\}$