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Take $A_{1}=\{ z: 1<|z|<R_1 \}$ and $A_{2}=\{ z: 1<|z|<R_2 \}$, wirh $R_1, R_2 > 1$. Then $A_1$ and $A_2$ are biholomorphic if and only if $R_1= R_2$.

My work: If $R_1= R_2$ then $A_1$ is biholomorphic to $A_2$ is clear. But for the converse,

I asserted that if $f$ is a biholomorhism between $A_1$ and $A_2$ then $f$ maps the inner circle of radius $1$ of $A_1$ to the inner circle to the inner circle of $A_2$, I used open mapping theorem to prove this. But I cannot see anything after this.

epsilon_delta
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  • You can also use the Schwarz reflection principle, see for example https://math.stackexchange.com/q/2714517/42969. – Martin R Jul 15 '20 at 14:47
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    actually your statement is not true about inner circle mapping to inner circle as you have an inversion $z \to R_1/z$ which works for $R_1=R_2$; a proof of the statement based on harmonic functions is in this answer (the added part) https://math.stackexchange.com/questions/3638871/conformal-mapping-between-two-disks-with-holes/3639060#3639060 – Conrad Jul 15 '20 at 14:59

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