(complex variables)Let $D \subset C$ be an open connected...

Question

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.

Answer 1

(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 $D$ .

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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 $D$.

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$b)$ 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 $D$

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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.

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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.

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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.

Answer 2

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.