# Analytic Functions and Equality of their Complex Derivatives

1. May 26, 2014

### NihilTico

1. The problem statement, all variables and given/known data

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 $D$

2. Relevant equations

Cauchy-Riemann Equations

Possibly Mean Value Theorem

3. 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.

2. May 26, 2014

### 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...?

3. May 26, 2014

### NihilTico

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?

4. May 26, 2014

### CAF123

Yeah, looks fine.