Holomorphic function help with checking the solution

[Stephen88]

Homework Statement

I'm sorry in advance but this will contain a lot of words to describe the soluton as I"m not good with Latex.

z_=conjugate
Show that f(z) = z^2 * z_ is not holomorphic in C. At which points is it
complex-differentiable?
I think I solved the problem I"m just looking for a second view.(or multiples).

Homework Equations

Cauchy–Riemann equations

The Attempt at a Solution

Ok first I wrote f(z)=z*z*z_=z*|z^2|=(x+yi)(x^2+y^2)
Using the first Cauchy–Riemann equation and calculating the partial diff with respect to x of the real part then calculating the partial diff with respect to y for the imaginary part I get:
2x^2+y^2=3yi+ix^2,this is not holomorphic but is complex diff at x=y=0.
Is this correct?If not what should I do.(if you can provide answer with examples,or at least detailed, that with be great)
Thank you.

 

[Dick]

Your conclusion is right, but the partial derivatives are wrong and there is no 'i' in the Cauchy-Riemann equations. What are u(x,y) and v(x,y)? Try and clean that up.

 

[Stephen88]

...f(z)=x^3 +x*y^2+i(y*x^2+y^3).
u(x,y)=x^3 +x*y^2
v(x,y)=i(y*x^2+y^3)

C-R=>2x^2+y^2=3y^2+x^2...this is equal for x=y=0...
Is there anything else I should do?

 

[Dick]

...f(z)=x^3 +x*y^2+i(y*x^2+y^3).
u(x,y)=x^3 +x*y^2
v(x,y)=i(y*x^2+y^3)

C-R=>2x^2+y^2=3y^2+x^2...this is equal for x=y=0...
Is there anything else I should do?

v(x,y)=y*x^2+u^3. There is no 'i' in v(x,y). There are two C-R equations. I assume you are doing u_x=v_y. If so your x partial derivative of u(x,y) is still wrong.

 

[Stephen88]

sorry about that, typo mistake...3x^2+y^2=3y^2+x^2...for any x=y I assume

 

[Dick]

sorry about that, typo mistake...3x^2+y^2=3y^2+x^2...for any x=y I assume

Yes, any x=y works. x=(-y) also works. But you aren't using the other CR equation.

 

[Stephen88]

I can but if one of the C-R eq is not satisfied then the holomorphic part is done...

 

[Dick]

I can but if one of the C-R eq is not satisfied then the holomorphic part is done...

True. But there are lots of solutions of the CR equation you just wrote down. That doesn't mean there are lots of points where the function has a complex derivative.

 

[Stephen88]

Understood.Thank you