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In Real & Complex Analysis, exercise 7 chapter 14, we are asked to find $a$ so that $$f_a(z) = \frac{z}{1+az^2}$$ is one-to-one on the unit disc $U$ and then describe $f_a(U)$.

I easily solved the first part to get $|a| \le 1$. For $a = 0$ $f(U) = U$. But I'm stuck on the second part. I'm still developing intuition here and would appreciate help.

My thoughts so far:

Call $D_r = \{z:|z|<r\}$, $T_r = \{z:|z|=r\}$. Since $f_a$ is continuous and one-to-one, $f_a(T_r)$ is a simple curve and $f_a(D_r)$ is the open set bounded by $f_a(T_r)$. As $r\to1$ we get $f(U)$. If $|a| < 1$, $f(T)$ is well defined but doesn't look like a familiar curve. Things get more complicated for $|a| = 1$ since there are singularities on $T$.

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