Complex analysis analytic functions

Question

I need to tell where $f(z)$ is analytic.$f(z)=(2x+y-x^2y)+i(3+2y-xy^2)$. I found that the cauchy riemann eqns for $u_x=v_y$ but when comparing $-u_y$ and $v_x$ I get $-1+x^2=-y^2$

What does this tell me about the functions analytic behavior?

Thanks

Answer 1

A function $f(z)=u(x,y)+iv(x,y)$ is analytic a region $\Omega \subset \mathbb{C}$ if and only if the Cauchy-Riemann equations are satisfied for all $z \in \Omega$. In your case, you have $u_x=v_y$, but

$$\ \ \ \ -u_y=v_x \ \ \ \ \text{ if and only if } \ \ \ \ x^2+y^2=1$$

That is, the Cauchy-Riemann equations are satisfied only on the unit circle in $\mathbb{C}$. Therefore, there exists no region where $f$ is analytic. This means, for example, that $f$ is complex differentiable only on the unit circle, but does not have a power series expansion in any region of the complex plane.