Proving Analytic Function Derivatives with CR Equations

[g1990]

Homework Statement

Let g(z) be an analytic function. I have to show that g'(z) is also analytic, using only the CR eqns. I am given that the 2nd partials are continuous

Homework Equations

let f(x,y)=u(x,y)+iv(x,y)
CR: du/dx=dv/dy and du/dy=-dv/dx
continuous 2nd partials: d/dx(du/dy)=d/dy(du/dx) and d/dx(dv/dy)=d/dy(dv/dx)

The Attempt at a Solution

I am confused as to how to express the derivative df/dz. would f'=du/dx+du/dy+idv/dx+idv/dy?

 

[Office_Shredder]

I don't think you want df/dz. If a function satisfies the CR equations and has continuous partials, what can you say about it?

 

[g1990]

I have to solve this problem using only calculus. I can't use any theorems like using Taylor series. Maybe I can say that it is harmonic?

 

[g1990]

nvm- got it! f'(z)=du/dx+idv/dx. Taking these and the real and imaginary parts, we can get the CR eqns for f' using the CR eqns for f and the fact that the mixed partials are equal.