Krdytkyu

The proposition that I am required to prove is the following

Let $D subset mathbb{C}$ be a domain and suppose $f(x, y)=u(x,
y)+iv(x, y)$ is holomorphic on $D$. Let $a, b, cin mathbb{R}$ such
that $a^2+b^2>0$ and $$au(x, y)+bv(x, y)=c quad forall z=(x, y)in D$$

Prove $f$ is constant

I will write $u(x, y)$ as $u$ in short, and similarly for $v(x, y)$.
My attempt is as follows:
Since f is holomorphic we have:
$$dfrac{partial u}{partial x}=dfrac{partial v}{partial y}qquad dfrac{partial u}{partial y}=-dfrac{partial v}{partial x}$$
If $a=0neq b$, then $bv=c implies v=Kinmathbb{R}$

$$implies dfrac{partial u}{partial x}=dfrac{partial v}{partial y}=dfrac{partial u}{partial y}=-dfrac{partial v}{partial x}=0$$
$$implies f'(z)=0 quad forall zin D$$
Similarly for $aneq 0=b$
Now the remaining case to check is if $aneq 0neq b$

Taking partial derivatives both sides of the given relation we get:

$$adfrac{partial u}{partial x}+bdfrac{partial v}{partial x}=0$$
$$implies dfrac{partial u}{partial x}=dfrac{b}{a}bigg(-dfrac{partial v}{partial x}bigg)=dfrac{b}{a}dfrac{partial u}{partial y}$$
Similarly taking partial derivatives w.r.t $y$ we get:
$$dfrac{partial u}{partial x}=-dfrac{a}{b}dfrac{partial u}{partial y}$$
$$implies dfrac{a}{b}=dfrac{-b}{a}$$
$$implies a^2+b^2=0$$
This is a contradiction, so it is not possible that $aneq 0neq b$. Therefore for all possible values of $a$ and $b$, the function $f$ is constant.

Yes, it is correct.

Note that, after having proved it for the case in which $a=0$ or $b=0$, you could use that to prove it in the general case, since$$au+bv=operatorname{Re}bigl((a-bi)(u+vibigr).$$

In the case $a ne 0 ne b$ you obtained the identities
$$
frac{partial u}{partial x} = frac ba frac{partial u}{partial y} \
frac{partial u}{partial x} = -frac ab frac{partial u}{partial y}
$$
At that point you should conclude that
$ frac{a}{b}=frac{-b}{a}$ or $frac{partial u}{partial x} = frac{partial v}{partial x} = 0$. The first case leads to a contradiction, therefore $f'(z)=0$.

You don't have to consider the cases $a=0$ or $b=0$ separately if you proceed
as follows: Differentiating with respect to $x$ gives

$$
afrac{partial u}{partial x}+bfrac{partial v}{partial x}=0
$$

Differentiating with respect to $y$ and applying the Cauchy-Riemann
equations gives
$$
bfrac{partial u}{partial x}-afrac{partial v}{partial x}=0
$$

That is a linear equation system, and its coefficient determinant is $-a^2-b^2 ne 0$. It follows that it only has the trivial solution
$$
frac{partial u}{partial x} = frac{partial v}{partial x} = 0
$$