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It is well known that if $f$ is holomorphic and $|f|$ is a constant function on a domain $D,$ then $f$ is constant in $D.$ In the proof of it, we use the differentiability of $f$. Thus, in this case, it can be easily shown that $f^2$ is constant in $D$ implies $f$ is constant in $D.$

Is there any continuous complex function $f$, not holomorphic, such that either $|f|$ or $f^2$ is continuous on $D$?

I would be grateful if you give any comment for my question. Thanks in advance.

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Take $f(z)=\frac{z}{|z|}$ and you'll get $|f|\equiv1$.

If there exists a function $g$ continuous such that $g^2$ constantly equal to $c$, then roughly speaking it is clear that $g=\pm\sqrt c$.

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  • $\begingroup$ Thanks for your answer. The first example is good, but I don’t understand the second one. Is there a continuous function $g$satisfying $g^2$ constant? $\endgroup$
    – 04170706
    Apr 9, 2019 at 1:29
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    $\begingroup$ No because if $g^2=c \neq 0$, $g=\pm {\sqrt c}$ for some fixed determination of the square root and the usual arguments show that the sets on which you get plus and minus respectively are both open and closed in the domain - closed trivial by continuity, open same $\endgroup$
    – Conrad
    Apr 9, 2019 at 1:59
  • $\begingroup$ Ok! I understand it. If $f^2=\alpha$ on a domain, then $f=\sqrt{\alpha}$ on $D$ or $f=-\sqrt{\alpha}$ on $D$. $\endgroup$
    – 04170706
    Apr 9, 2019 at 3:24
  • $\begingroup$ You wrote "on a domain". When I read "domain" I mean an open connected subset of $\Bbb C$. In this case it is true that you cannot find a non constant function on $D$ such that $f$ is nonconstant and its square does. But if you drop this assumption and you have, say $D=D_1\sqcup D_2$, then you can define $g:=1$ on $D_1$ and $g:=-1$ on $D_2$ which is nonconstant, but $g^2\equiv1$. $\endgroup$
    – Joe
    Apr 9, 2019 at 11:22

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