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I want to show that the following function is not analytic anywhere. $$f(z)=x^2+iy^3$$

Now I don't really understand the Cauchy-Riemann equations, but it seems we take:

$$u(x,y)=x^2,v(x,y)=y^3$$ as we normally would, and take the partial derivatives:

$$u_x =2x, u_v = 0$$ $$v_x = 0, v_y = 3y^2$$

And we want:

$$u_x=v_y,u_y = -v_x$$

For necessary condition for being analytic at some point.

$$2x=3y^2,0=0$$

So we are potentially(not necessarily) analytic and any point such that $2x=3y^2$ I believe.

Now, how then I have ruled out almost every point, how do I rule out these remaining points?

Skies burn
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1 Answers1

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Note the precise statement of the theorem. http://www.math.columbia.edu/~rf/complex2.pdf Go to the theorem on page 8. The CR equations must hold on an OPEN subset. A single point or line is not that.

  • Since for any open ball around a point on that line, you will find points that don't satisfy the CR equations, the function is not analytic. @SkiesBurn –  Apr 20 '15 at 05:40
  • Why is$f(z)=x^2-y^2 + 2xyi$ analytic?

    It has $u_x=2x=v_y$ and $u_y=-2y=-v_x$ and $f'(z)=2z$

    How do I deduce it is an open subset?

     – Skies burn Apr 20 '15 at 05:44
  • $u_x=v_y$ and $u_y=-v_x$ for all $x,y$. Certainly all of $\mathbb{C}$ is open.@SkiesBurn –  Apr 20 '15 at 05:47
  • Wait I think I misunderstood something.

    Do the Cauchy Riemann equations require that $u_x=v_y$ and $u_y=-v_x$ in general or do I set these, and find which points it isn't true?

     – Skies burn Apr 20 '15 at 05:50
  • @SkiesBurn read the link I sent you. If the CR equations hold on an open subset of C, then the function is analytic on that subset. –  Apr 20 '15 at 05:52
  • Oh I see now thank you. We have satisfaction at $(0,0)$ which makes it not open and hence it is nowhere analytic. – Skies burn Apr 20 '15 at 05:54
  • But the function is complex differential at ${z = x+iy \mid 2x = 3y^2 }$, right? – Leo Mar 10 '16 at 08:18