The j-function of a point on the boundary of the fundamental domain is real valued. I am trying to prove the following statement:
Let $z \in \mathbb{D} $ (the standard fundamental domain for $SL_{2}(\mathbb{Z})$). Prove that if $z$ lies on the boundary of $\mathbb{D}$, or if $Re(z)=0$, then $j(z) \in \mathbb{R}$.
So far I have shown that $j(i)=1728$ and $j(e^{(2\pi i)/3})=0$ but I am not sure how to proceed...

Hint: Show that if $f(\tau)=g(\exp(\alpha\mathrm{i}\tau))$ for some Laurent series $g$ with real coefficients and real $\alpha>0$, then $f(-\bar{\tau}) = \bar{f}(\tau)$. Then use the symmetries of $j$ to show that on the boundary of the fundamental domain, as well as for $\operatorname{Re}\tau=0$, you get $j(\tau) = j(-\bar{\tau}) = \bar{j}(\tau)$.
