I'm currently in an introductory complex analysis class. I'm working through the following problem:
Let $\Bbb{H}$ be the upper half-plane and $B$ be left half of the fundamental region:
$$ B = \{\tau \in \Bbb{H}: |\tau|>1, -\frac{1}{2}<\textit{Re}(\tau)<0 \} $$
Show that the modular invariant $j$ is a conformal equivalence between $B$ and $\Bbb{H}$.
I have only learned a bit about conformal equivalence and modular invariance; the instructor hasn't been too helpful. Can anyone guide me on how to solve this exercise?
$j$ is biholomorphic from $U=|\tau|>1,|\Re(\tau)|\in (0,1/2)$ to $\Bbb{C}-L$ where $L$ is the curve which is the image of the boundary of $U$.
$L$ is the real axis. This is because $J$ has real coefficients thus it is real on $\Re(\tau)=0$ and $\Re(\tau)=\pm 1/2$. On $|\tau|=1$ it is real too because we have the following correspondence by some modular transformations
Next, let $B= \{|\tau|>1,\Re(\tau)\in (-1/2,0)\}$.
$\Im(j(-1/4+i))>0$ gives that $JjB)$ is an open subset of $\Im(z)>0$.
And $j(a+ib)=\overline{j(-a+ib)}$ gives that $\Bbb{C-R} = j(B)\cup\overline{j(B)}$.
Whence $j(B)=\Bbb{H}$.