The $ā$-function satisfies $ā'^2=4(ā-e_1)(ā-e_2)(ā-e_3)$. Supposing that none of $e_1,e_2,e_3$ is zero, show that the equation $ā(z)=0$ has two distinct solutions $z= Ā±a$. For any two distinct points $b,cāā/Ī$ write down a meromorphic function whose only poles are simple poles at $b$ and $c$.
Solution.
Since $ā'(a)^2=-4e_1e_2e_3ā 0$, $a$ is not a ramification point, so $ā(z)=0$ has two distinct solutions, let one of them be $a$, then the other is $-a$, since $ā$ is even.
The linear map $\frac{2a}{b-c}(z-b)+a$ maps $b,c$ to $a,-a$.
So $\frac1{ā(\frac{2a}{b-c}(z-b)+a)}$ is a meromorphic function whose only poles are simple poles at $b$ and $c$.