(complex variables)Let $D \subset C$ be an open connected... Let $D \subset \mathbb{C}$ be an open connected.
a) Use the Cauchy-Riemann equations to prove that if $F: D \to \mathbb{R}$ is holomorphic, then $F$ is constant.
b)Let $f, g: D \to \mathbb{C}$ be holomorphic functions. Prove that if $\bar f g$ is a holomorphic function, then $f$ is constant or $g \equiv 0$.
I don't know if I got it right but here goes.
a) $F(z) = u+iv$.If $F$ is holomorphic then the function is differentiable at all points and $u_x=v_y, u_y =-v_x$.
For constant $F$ then we have that $F'(z) = u+iv = 0 \to au_x + bv_y = 0 = au_y + bv_y$ using Cauchy-Riemann equations $(a-b)u_x = -(a+b)v_x$.
Case $1$
$a=0 \to -bu_x = -bv_x \to au_x + bv_y = 0 \to v_y = 0 \to u_x=-u_y=v_x=v_y=0 \to F$ is constant.
Case $2$
$a \ne 0 \to au_x + bv_x = 0 \to u_x + \frac{b}{a}v_x = 0\tag1$
$-av_x + bu_x \to av_x = bu_x \to v_x = \frac{b}{a} u_x\tag2$
$(2) \to (1)$
$ u_x +\frac{b}{a}v_x = 0 \to  u_x +\frac{b}{a} \frac{b}{a} u_x = 0 \to u_x(1 + \frac{b^2}{a^2}) = (a^2+b^2)u_x=0 \to u_x=0 \to u_x=-u_y=v_x=v_y=0$.
So $F$ is constant.
b)
I thought that if I prove that $f$ is constant in a neighborhood of each point where g doesn't vanish, it would work. But I couldn't execute the plan.
How am I going? too bad?
Thanks.
 A: Your answer to a) does not make sense. Why are you starting with $F$ being  a constant  and  what are $a$ and $b$? For  a correct proof note that $v\equiv 0$ by hypothesis so $v_x=v_y=0$. By C-R equations we get $u_x=u_y=0$ which forces $u$ to be a constant (by connectedness of $D$).  Hence, $F$ is  a constant too.
Answer for b): Consider $E=\{z \in D: g(z) \neq 0\}$. Suppose $E \neq \emptyset$. On this open set $\overline f =\frac 1 g (\overline f g)$ is analytic and so is $f$. In any open disk contained in $E$ we can apply a) to $f+\overline f$ and $i(f-\overline f)$ to see that $f$ is a constant. But $D$ is connected, so $f$ being a constant in some open disk contained in $D$ implies that it is constant everywhere.
A: 
(a) Use the Cauchy-Riemann equations to prove that if $F:D\to \Bbb{R}$ is
holomorphic, then $F$ is constant.

$F:D\to \Bbb{R}$ holomorphic.
Let $F(z)=u(z)+iv(z)=a$ $\quad$ where $a\in \Bbb{R}$
Then $v=0$ on $D\subset \Bbb{R^2}$ implies $v_x=0=v_y$
Hence by C-R equation we have $u_x=v_y=0$ and $u_y=-v_x=0$
Then $F'=u_x+iv_x=0$ on $D$ .
Since $D$ is open connected and $F'(z) =0$ on $D$ implies $F$ is constant on $D$ .

Alternative: (Using definition of derivative)
$F'(z) =\lim_{h\to 0} \frac{f(z+h) -f(z) }{h}$
$h\to 0$ along the real axis,
$F'(z) =\lim_{h\to 0} \frac{F(z+h) -F(z) }{h}$
$F'(z) $ is purely real
$h\to 0$ along the imaginary axis,
$F'(z) =\lim_{h\to 0} \frac{f(z+ih) -f(z) }{ih}$
Here $F'(z) $ is purely imaginary
Hence either $F$ is not differentiable or $F'(z) =0$  ( as $0$ is the only complex number which is purely real as well as purely imaginary).
But $F$ is holomorphic is given, hence $F'(z) =0$ .
Hence $F$ is holomorphic on a open connected set with derivative $0$ everywhere on $D$ implies $F$ is constant on $D$.


$b)$ Let $f,g:D\to \Bbb{C }$ be holomorphic functions. Prove that if $f\overline{g}$ is a holomorphic function, then $f$ is constant or
$g≡0$.

If you already know that $f$ is holomorphic on $D$ iff $ \begin{align}
\frac{\partial}{\partial \bar z} f = 0.
\end{align}$
Where
$\frac{\partial f}{\partial\bar{z}}:=\frac{1}{2}\left(\frac{\partial f}{\partial x}-\frac{1}{i}\frac{\partial f}{\partial y} \right) .$
Then $f\overline{g}$ is holomorphic implies
$ \begin{align}
\frac{\partial}{\partial \bar z} f\overline{g} = 0.
\end{align}$
Implies $ \begin{align}
\overline{\frac{\partial}{\partial \ z} (\overline{f}} g) = 0.
\end{align}$
Where $ \begin{align}
\frac{\partial}{\partial \bar z} (\overline{f}g )={g}\frac{\partial\overline{f}}{\partial \bar z}  + \overline{f} \frac{\partial g}{\partial \bar z} \end{align}$
Since $g$ is holomorphic,
$ \begin{align}
\frac{\partial}{\partial \bar z} g = 0.
\end{align}$
Hence $  \begin{align}0=\overline{\frac{\partial}{\partial \bar z} (\overline{f}g )}
=g\overline{\frac{\partial f}{\partial z}} 
\end{align}$
Implies $ g{\frac{\partial f}{\partial z}} =0$
Hence $g=0$ or ${\frac{\partial f}{\partial z}} =0$ i.e $f$ is constant on $D$

Alternative:
Lemma: If $f$ is holomorphic on a domain (open and connected) $D$ be such that $\overline{f}$ is also holomorphic then $f$ is constant.
Proof: Let $f=u+iv$ . Then $\overline{f}=u-iv$
As $f$ is holomorphic, $u_x=v_y$ and $u_y=-v_x$
Again by holomorphicity of $\overline{f}$ , we have $u_x=-v_y, u_y=v_x$
From above two set of equality,
$u_x=v_y=-u_x $ and $v_x=-u_y=-v_x$
Hence $u_x=0, v_x=0$ implies $f'(z) =0$ .
Hence $f$ is constant.

Proof of $b) $
Suppose $g\neq 0$ .
Then $\overline{f}=\frac{g\overline{f}}{g}$
As both $\overline{f}g,g$ are holomorphic and $g\neq 0$ implies $\overline{f}$ is holomorphic on $D$ .
Since $f, \overline{f}$ both are holomorphic on $D$ , by above lemma $f$ is constant.

Comments: After  reading those proof, I am sure you will find your mistake.
You have  done a huge mistake  to proof part $a) $. You have assumed the conclusion  and again derived the conclusion.Always start proof with "To show "  and go towards the conclusion through logical consequences.
