0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Gary. Aug 17 '15 at 20:16
  • $\begingroup$ Your $U_x$ is wrong. $\endgroup$ – mrf Aug 17 '15 at 20:25
0
$\begingroup$

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 $E=\{{(x,y):y=x,x\in\mathbb{R}}\}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.