How to prove that an analytic function is conservative field

Question

A function $f(z)=u(x,y)+iv(x,y)$ is called conservative field if $v_x=u_y$ everywhere. if $f(z)$ is analytic then it is conservative field if and only if $f(z)=z+z_o$, where $z_o$ is a constant.

Can any one help how to prove this!

Answer 1

Hint: for the forward direction, note that all analytic functions satisfy the Cauchy-Riemann equations, one of which is $-v_x = u_y$. What can you conclude?

I'll let you work out the specifics, but you essentially get a function in which $v$ is independent of $x$ and $u$ is independent of $y$. The only function that satisfies this property is $f(z) = z + z_0$ (alternatively, you could solve the integral $\int 0 dx$, which basically boils down to a line integral, if you want to be rigorous).

For the other direction, try breaking up the function into its real and imaginary parts. Remember that $z = x + iy$, so the function $f(z) = z$, for instance, can be written in the form $u + iv = x + iy$, where $u = x$ and $v = y$. Then find the partial derivatives as you would in the usual sense.