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.
Best Answer
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.