Find Where $f(z)=x^2+iy^2$ Differentiable And Analytic

Question

Let $f(z)=x^2+iy^2$ find where it is differentiable and where it is analytic in $\mathbb{C}$

For a function to be differentiable at a point it should fulfil C-R equations, we have

$u(x,y)=x^2$ and $v(x,y)=y^2$

So $u_x=2x=2y=v_y$ and $u_y=0=0-v_x$

So the function is differentiable iff $2x=2y\iff x=y$

So the function is differentiable on the straight line $y=x$

So $f(z)$ is not analytic? as it is only differentiable on a straight line and not in areas are a point? on the other hand the function seems to have no problematic points (singularities) so it is analytic?

Answer 1

That depends on the definition of analytic function that you use. But if it is the usual one (which implies that if $f$ is analytic at a point $z$ then it is analytic in all points of a neighborhood of $z$) than $f$ is analytic nowhere.

Answer 2

If you want differentiable as a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ ($f(x,y)=(x^2,y^2)$). This function is differentiable at all points, since its components have continuous partial derivatives everywhere.

The Cauchy-Riemann equations give you where it is complex-differentiable. You have found that complex derivative exist at every point of the line $y=x$.

To be analytic at a point it has to have complex derivative at a neighborhood of the point. This is because being analytic means that the function is equal to a convergent power series with a radius of convergence larger than $0$. In the interior of the disc of convergence, the sum of a power series has complex derivatives. Take into account that while this is a simple result, it is something that still requires a proof. As such, it is good to make it explicit in the proof since what you get from Cauchy-Riemann is existence of complex derivatives, a related, but different condition.

This function doesn't have complex derivative on any open set since the line $y=x$ has an empty interior. Therefore, it is not analytic anywhere.