Show that holomorphic function $f: \mathbb{C} \rightarrow \mathbb{C}$ is constant

Question

Let's $f: \mathbb{C} \rightarrow \mathbb{C}$ be a holomorphic function such that values $f$ are on line $y=ax+b$. Show that $f$ is constant.

I think I should use Cauchy-Riemann equations but I don't know what does mean that values $f$ are on line $y=ax+b$. Can you explain me that?

Thanks in advance.

Ivo Terek · Accepted Answer · 2015-05-06T19:23:23.263

5

Your hypothesis reads: $$f(x+iy) = u(x,y) + i(a u(x,y) + b),$$ that is, the point $f(x+iy)$ as a point in $\Bbb R^2$ belongs to the said line. The Cauchy-Riemann equations read, abbreviating the notation: $$u_x = a u_y, \quad u_y = -au_x.$$ So $u_x = -a^2u_x \implies (1+a^2)u_x = 0$ and... oh, you can conclude now! (if you need one more push, please tell me)

edited May 06 '15 at 19:23

answered May 06 '15 at 19:21

Ivo Terek

77,665

Is it sure that $a,b \in \mathbb R$? – Alex M. May 06 '15 at 19:23
1

Pretty much sure, although the exercise should have said it. Otherwise the image of $ax+b$ with $a,b \in \Bbb C$ is the entire plane, no? – Ivo Terek May 06 '15 at 19:24
I understand it differently: the range of $f$ is a subset of the line (i.e. 1-dimensional affine real space) $t \mapsto (t, at+b)$, with $a,b \in \mathbb C$. – Alex M. May 06 '15 at 19:30
1

Also makes sense. Well, the result remains true. The argument that comes to mind is that $f$ holomorphic and non-constant implies $f(\Bbb C)$ dense. Since the line isn't dense in $\Bbb C$, $f$ must be constant.. – Ivo Terek May 06 '15 at 19:31
Much better, this second argument. – Alex M. May 06 '15 at 19:38
It's a neat little bomb. OP can see the proof here. – Ivo Terek May 06 '15 at 19:40
@IvoTerek, in your answer, where do you use that $a,b\in \Bbb R$? – Silent May 06 '19 at 02:37
1

@Silent to ensure $1+a^2\neq 0$ and so $u_x=0$. If $a=\pm i$, what would happen? – Ivo Terek May 06 '19 at 02:41
@IvoTerek, thank you very much – Silent May 06 '19 at 02:42

score 2 · Answer 2 · answered May 06 '15 at 23:01

2

Herewith my second answer:

Holomorphic functions are conformal except where their derivatives are zero. It can't be conformal if its image is a straight line.

answered May 06 '15 at 23:01

Michael Hardy

1

score 1 · Answer 3 · answered May 06 '15 at 19:51

I'll assume you meant $x$ and $y$ are the real and imaginary parts of $f(z)$ and $a$ and $b$ are real.

Suppose $f'(z_0)\ne 0$. Choose $\varepsilon$ so close to $0$ that when $|\Delta z|=\varepsilon$ then $f(z_0+\Delta z)$ differs from $f(z_0)+f'(z_0)\,\Delta z$ by less than $\Delta z/1000000$. Then as $\Delta z$ goes around the circle $\{z:|z-z_0|=\varepsilon\}$, then $f(z_0)+f'(z_0)\,\Delta z$ goes around the circle $\{w:|w-f(z_0)| = |f'(z_0)|\varepsilon\}$. That circle is not on a straight line, and $f(z)$ stays close to $f(z_0)+f'(z_0)\,\Delta z$.

score 1 · Answer 4 · answered May 06 '15 at 19:59

1

This also follows immediately from the open mapping theorem.

answered May 06 '15 at 19:59

zhw.

105,693

score 0 · Answer 5 · answered May 06 '15 at 20:03

How about this: you can rotate/translate your line such that your new line matches with real axis (or with imaginary axis, your chose). Now, that new function, say $g=g(z)$, has $\operatorname{Im}(g)=0$, so you can use C-R equations to prove that $g=\operatorname{const}.$, and from that you can find that $f$ is also constant (because $g(z)=(f(z)-b)e^{-i\alpha}$, where $\alpha = \arctan a$, something like that).

Show that holomorphic function $f: \mathbb{C} \rightarrow \mathbb{C}$ is constant

5 Answers5