Closed curve for an analityc function











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If I have an analityc function $f(z)$ on a domain $B subset mathbb{C}$ and a simple and closed curve $C$ that encloses $B$, and if $|f|$ is constant over C, the $f$ is constant on $B$?



I haven’t found a counterexample but i dont know how to apply the maximum principle or apply some analityc continuation. Please some help with this.










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    $f(z) = z$ on $|z| = 1$ ? Otherwise $f(z) =c frac{z+b}{overline{b}z+1}$ with $|b| < 1$ on $|z|=1$
    – reuns
    Nov 18 at 4:24

















up vote
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down vote

favorite












If I have an analityc function $f(z)$ on a domain $B subset mathbb{C}$ and a simple and closed curve $C$ that encloses $B$, and if $|f|$ is constant over C, the $f$ is constant on $B$?



I haven’t found a counterexample but i dont know how to apply the maximum principle or apply some analityc continuation. Please some help with this.










share|cite|improve this question


















  • 2




    $f(z) = z$ on $|z| = 1$ ? Otherwise $f(z) =c frac{z+b}{overline{b}z+1}$ with $|b| < 1$ on $|z|=1$
    – reuns
    Nov 18 at 4:24















up vote
0
down vote

favorite









up vote
0
down vote

favorite











If I have an analityc function $f(z)$ on a domain $B subset mathbb{C}$ and a simple and closed curve $C$ that encloses $B$, and if $|f|$ is constant over C, the $f$ is constant on $B$?



I haven’t found a counterexample but i dont know how to apply the maximum principle or apply some analityc continuation. Please some help with this.










share|cite|improve this question













If I have an analityc function $f(z)$ on a domain $B subset mathbb{C}$ and a simple and closed curve $C$ that encloses $B$, and if $|f|$ is constant over C, the $f$ is constant on $B$?



I haven’t found a counterexample but i dont know how to apply the maximum principle or apply some analityc continuation. Please some help with this.







complex-analysis analytic-functions maximum-principle analytic-continuation






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asked Nov 18 at 4:17









J.Rodriguez

15310




15310








  • 2




    $f(z) = z$ on $|z| = 1$ ? Otherwise $f(z) =c frac{z+b}{overline{b}z+1}$ with $|b| < 1$ on $|z|=1$
    – reuns
    Nov 18 at 4:24
















  • 2




    $f(z) = z$ on $|z| = 1$ ? Otherwise $f(z) =c frac{z+b}{overline{b}z+1}$ with $|b| < 1$ on $|z|=1$
    – reuns
    Nov 18 at 4:24










2




2




$f(z) = z$ on $|z| = 1$ ? Otherwise $f(z) =c frac{z+b}{overline{b}z+1}$ with $|b| < 1$ on $|z|=1$
– reuns
Nov 18 at 4:24






$f(z) = z$ on $|z| = 1$ ? Otherwise $f(z) =c frac{z+b}{overline{b}z+1}$ with $|b| < 1$ on $|z|=1$
– reuns
Nov 18 at 4:24

















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