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This question is from John B. Conway's "Functions of One Complex Variable I". It is stated:

Let $G$ be a region and suppose that $f:G\to\mathbb{C}$ is analytic such that $f(G)$ is a subset of a circle. Show that $f$ is constant.

I'm confused as to how to even start approaching this problem. The chapter has discussed Mobius transforms, the Cauchy-Riemann equations, and branches of the logarithm. The most logical choice here (I think) is to use the Cauchy-Riemann equations somehow. Any hints/suggestions would be helpful. Thanks in advance!

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    $\begingroup$ Do you know about the open mapping theorem? $\endgroup$
    – user61527
    Feb 13 '14 at 18:02
  • $\begingroup$ See also: math.stackexchange.com/questions/675187/… $\endgroup$
    – froggie
    Feb 13 '14 at 18:03
  • $\begingroup$ @T.Bongers The open mapping theorem is in the next chapter (so we haven't covered it yet in class). Reading that theorem, I can definitely see how to apply it, but in lieu of that specific theorem, is there any other suggestion? Thanks. $\endgroup$ Feb 13 '14 at 19:54
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For simplicity let's say that the circle is centered at the origin. (Think about why this is okay, and what you need to add when the circle is centered at some $a$ in the complex plane.)

Here is the simplest method I can think of. See if you can follow along before looking at the spoiler.

What does it mean for $f(G)$ to be a subset of a circle?

It means that $|f(z)| = r$ for all $z$ in $G$. If you square both sides you have:

$$f(z) \overline{f(z)} =|f(z)|^2 = r^2$$

Now, we can take the derivative of both sides to obtain

$$f'(z) \overline{f(z)} = 0. $$

And since $f$ is non-zero (why?), you can divide to conclude that

$f'(z) = 0 $ for all $z$ in $G$.

This allows you to conclude that $f$ is constant on $G$.

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  • $\begingroup$ I guess if you want to be really exact, you should include the case where $f$ maps into a circle of radius $0$... $\endgroup$
    – Braindead
    Feb 13 '14 at 23:06
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    $\begingroup$ How did you get the derivative as $f'(z) \overline{f(z)} = 0. $ ? What am I missing ? $\endgroup$
    – Rusty
    Feb 21 '16 at 9:02
  • $\begingroup$ @SMath $|f(z)|^2=f(z)\bar{f(z)}=c$. Take the derivative with respect to $z$ using the Leibniz rule. $\endgroup$
    – Braindead
    Feb 27 '16 at 4:24
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    $\begingroup$ How can you know that $\overline{f}$ is differentiable? In fact, it shouldn't be unless $f$ is constant, right? $\endgroup$
    – user223391
    Apr 2 '17 at 21:59

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