Analytic Functions and Equality of their Complex Derivatives

[NihilTico]

Homework Statement

If f and g are both analytic in a domain D and if f'(z)=g'(z) for every z\in{D}, show that f and g differ by a constant in D

Homework Equations

Cauchy-Riemann Equations

Possibly Mean Value Theorem

The Attempt at a Solution

I'm pretty sure I'm making a mountain out of a mole hill here, this question has stumped me for a while, and I'm certain that I'm missing some small insight that would make the answer evident.

The central premise of the question as I understand it is, for f(x,y)=u(x,y)+iv(x,y) and g(x,y)=a(x,y)+ib(x,y), we should have:

u_x(x,y)=v_y(x,y)=c_x(x,y)=d_y(x,y)
and
u_y(x,y)=-v_x(x,y)=c_y(x,y)=-d_x(x,y).

I think that I am able to integrate these functions as one normally would, for example:

u(x,y)=\int{u_{x}(x,y)dx}+A(y)+B
u(x,y)=\int{u_{y}(x,y)dy}+C(x)+D

where I've taken the liberty of adding the function and constant of integration outside of the integral. My impression from this is that I should be able to use this idea and the Cauchy-Riemann equations to show that all of the functions that appear from the integral vanish (or at least become constant) and that I am just left with the sum of two constants. Unfortunately I haven't actually gotten anywhere with that.


Another approach I've tried was to consider an open disk contained D and centered at a+ib with an arbitrary point given by z=x+iy and to apply the Mean Value Theorem by writing:

u_{x}(x,y)-u_{x}(a,y)+u_{x}(a,y)-u_{x}(a,b)=(x-a)u_{xx}(X)+(y-b)u_{xy}(Y)

and doing this for each of u_x(x,y)=v_y(x,y)=c_x(x,y)=d_y(x,y) and u_y(x,y)=-v_x(x,y)=c_y(x,y)=-d_x(x,y), however, I haven't seen any path to take with this either that would be relevant to the question at hand.

This has been bugging me, and I'm convinced there is a straightforward way of going about this.

 

[CAF123]

f'(z) = g'(z) on D and the derivatives exist everywhere within D. Then this means f'(z) - g'(z) vanishes everywhere within D. So...?

 

[NihilTico]

f'(z) = g'(z) on D and the derivatives exist everywhere within D. Then this means f'(z) - g'(z) vanishes everywhere within D. So...?

Hahaha! I knew it was going to be something I would want to smack myself about!

So let me see if I got this then:

f'(z)-g'(z)=0, hence \int{f'(z)-g'(z)}dz=\int{(0)}dz\rightarrow{f(z)-g(z)}=C for some complex number C, right?

 

[CAF123]

Yeah, looks fine.