= Solution
If such an $f$ existed, boundedness would make its isolated singularity at zero removable. Let $F$ be the holomorphic extension to the disc. Continuity gives $1\leq|F(0)|\leq2$. Equality at either endpoint contradicts the minimum-modulus principle applied to $F$ or the <maximum modulus principle>, since $F$ is nonconstant. Hence $F(0)\in A$. Surjectivity of $f$ supplies some nonzero $z$ with $f(z)=F(0)$, contradicting injectivity of the extension. Therefore
$$
\boxed{\text{no such biholomorphic map exists}.}
$$
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