1. ## Analytic Function

I was asked to find f'(z) where f = (x + iy)/(x^2 + y^2) and state where f is analytic.

Definition of analytic is differentiable on an open set.
I say that f(z) = x/(x^2 + y^2) + iy/(x^2 + y^2), and then in order to show that f is differentiable on open set, I show that the partial derivatives exist and are continuous.

In other words, limits as approach z from positive x direction and positive y direction should be the same. Let z = u(x, y) + iv(x, y)

Then, by some manipulation of limits and such, you get f'(z) = du/dx + idv/dx from x direction and f'(z) = dv/dy - idu/dy. When I computed these, I got du/dx + idv/dx = (x^2 - y^2)/(x^2 + y^2)^2 + i2xy/(x^2 + y^2)^2

and dv/dy - idu/dy = (y^2 - x^2)/(x^2 + y^2)^2 - i2xy/(x^2+y^2)^2

Thus, my statement is that f is differentiable only when x^2 - y^2 = 0, meaning x = y, x = -y, where x and y are nonzero (if x = y = 0, then there would be 0/0 expression which is not really defined).

However, x^2 - y^2 = 0 at discrete points, so f is not differentiable on any open set containing one of the points that satisfies that equality: thus, I conclude that z is analytic NOWHERE. Is my reasoning correct?

2. Originally Posted by amoeba
I was asked to find f'(z) where f = (x + iy)/(x^2 + y^2) and state where f is analytic.

Definition of analytic is differentiable on an open set.
I say that f(z) = x/(x^2 + y^2) + iy/(x^2 + y^2), and then in order to show that f is differentiable on open set, I show that the partial derivatives exist and are continuous.

In other words, limits as approach z from positive x direction and positive y direction should be the same. Let z = u(x, y) + iv(x, y)

Then, by some manipulation of limits and such, you get f'(z) = du/dx + idv/dx from x direction and f'(z) = dv/dy - idu/dy. When I computed these, I got du/dx + idv/dx = (x^2 - y^2)/(x^2 + y^2)^2 + i2xy/(x^2 + y^2)^2

and dv/dy - idu/dy = (y^2 - x^2)/(x^2 + y^2)^2 - i2xy/(x^2+y^2)^2

Thus, my statement is that f is differentiable only when x^2 - y^2 = 0, meaning x = y, x = -y, where x and y are nonzero (if x = y = 0, then there would be 0/0 expression which is not really defined).

However, x^2 - y^2 = 0 at discrete points, so f is not differentiable on any open set containing one of the points that satisfies that equality: thus, I conclude that z is analytic NOWHERE. Is my reasoning correct?

I think it is , but haven't you studied the Cauchy-Riemann equations for analycity? It's way simpler than what you did: as $\displaystyle f(x)=\frac{x}{x^2+y^2}+\frac{y}{x^2+y^2}i=u(x,y)+i v(x,y)$, the C-R condition tells us f is analytic iff C-R equations are true, i.e. iff $\displaystyle u_x=v_y\,,\,\,u_y=-v_x$ , and you can check this doesn't happen.

Tonio