If $f:\mathbb{C}\setminus \{0\}\to \mathbb{C}$ is a function such that $f(z)=f(\frac z{|z|})$ and its restriction to unit circle is continous,then
$(1)\lim _{z\to 0} f(z)$ exist.
$(2)f$ is analytic but not necessarily a constant function
$(3)f$ is continous but not necessarily analytic.
$(4)f$ is a constant function.
The title can be justified in the following way:-
Let $z=re^{i\theta}$, Then $f(re^{i\theta})=f(e^{i\theta})$
Hence, I can imagine rays of equal values radiating out from zero in every direction .
Now ,if $z=re^{i\theta}$, then as $r\to 0$, then $z\to 0$
$\lim_{z\to 0}f(z)= \lim_{r\to 0}f(e^{i\theta})=f(e^{i\theta})$, which depends on $\theta$ and hence the limit does not exist.
I am quite confident that there exist no such non-constant analytic function but it is getting difficult for me to write an analytical proof. (I tried to arrive at a contradiction assuming $f$ is a analytic on a bounded domain and using maximum modulus theorem but I failed.)
Since the restriction of $f$ to the unit circle is continous, that means for any fixed $\theta_0$, $\lim_{\theta\to \theta_0}f(e^{i\theta})=f(e^{i\theta_0})$
Now let $z_0=r_0e^{i\theta_0}, z=re^{i\theta}$ and $\epsilon \gt 0$ be given, then $|f(z_0)-f(z)| =|f(e^{i\theta_0})-f(e^{i\theta})|\lt \epsilon $, if $| \theta-\theta_0|\lt \delta$, where the existence of $\delta$ for $\epsilon$ is garaunteed by continuity of $f$ on the unit circle.
So $f$ is continous.
$f$ may not be constant as suggested by above proof.
Please go through my work and help me complete it.
Thanks for your valuable time.
