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I am trying to construct a non analytic , non constant, continuous function $f $ such that $f$ :$ \mathbb C\setminus\{0\}\to \mathbb C$ with $f(z)= f\left(\frac{z}{|z|}\right)$. I got stuck how to start this problem

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2 Answers 2

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How about $f(z)=z/|z|$? This is continuous on $\mathbb{C}-\{0\}$, not analytic since $|z|$ is involved, non constant and we have $f(z/|z|)=(z/|z|)/|z/|z||=(z/|z|)/1=z/|z|=f(z)$

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  • $\begingroup$ i want to know the points where f is not analytic except zero $\endgroup$
    – maths
    Mar 28, 2020 at 8:35
  • $\begingroup$ @PrashantDattatrey Well, check for yourself. for $h\in\mathbb{R}^+$, compute $[f(z+h)-f(z)]/h$ and take the limit as $h\to0^+$. Compute also $[f(z+ih)-f(z)]/(ih)$ and take the limit as$h\to0^+$. Are these two equal? $\endgroup$ Mar 28, 2020 at 8:43
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Note that the condition of non-constant is trivial for any non-analytic $f$.

Using polar coordinates $z=r.e^{i\theta}$

$f(r.e^{i\theta})=f(\frac{r.e^{i\theta}}{|r.e^{i\theta}|})$

$f(r.e^{i\theta})=f(e^{i\theta})$

A convenient choice will be $f(r.e^{i\theta})=\theta^2\implies f(z)=(Arg(z))^2$ which is defined on $\mathbb C/{0}$.

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  • $\begingroup$ $ f(z) = arg (z)$ is not continuous on $\Bbb C\setminus0$ unless you delete a ray (originating from origin) from complex plane $\endgroup$
    – maths
    Mar 28, 2020 at 17:09
  • $\begingroup$ You are right. I fixed that now. $\endgroup$ Mar 28, 2020 at 17:23
  • $\begingroup$ This Arg (Z) should be principal argument $\endgroup$
    – maths
    Mar 28, 2020 at 17:32
  • $\begingroup$ yes it is. There is big A. $\endgroup$ Mar 28, 2020 at 17:33

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