The function $f$ given by $z$ if $|z|\leq 1/2$ and $|z|e^{2i(1-|z|)\arg z}$ if $1/2\leq |z|\leq 1$ and $0\leq \arg z<2\pi$ is a joint discontinuous extension of the identity on $|z|\leq 1/2$ and the constant function $1$ on $|z|=1$. Is there an explicit smooth extension of this piecewise defined function which has no zeros on the annulus $1/2\leq |z|\leq 1$? For its mere existence (but possibly with zeros) see e.g
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The function $f(z)= z+(1-z)e^{2} e^{ \frac{1}{\frac{1}{2}-|z|}}$ for $1/2\leq |z|\leq 1$ is a smooth extension but has a unique zero in $]-1,-1/2[$. – ray Aug 11 '21 at 07:18
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There is not even a continuous extension of the function $q(z)=z$ on $|z|\leq 1/2$ and $q(z)=1$ on $|z|=1$ to the annulus $A:=\{1/2\leq |z|\leq 1\}$ that is zero-free on $A$. In fact, as is well known (Brouwer) the identity function on $|z|=1$ has no zero-free continuous extension to the unit disk $|z|\leq 1$. In particular, the function $g(z)=1$ on $|z|\leq 1/2$ and $g(z)=z$ on $|z|=1$ has no zero-free continuous extension to $A$. Now multiplication by $\overline z$ shows that $k(z):=\overline z g(z)$ has no zero-free continuous extension to $A$ either. As $q=\overline k$, we are done.
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