For $w: \mathrm{C}^+ \to D(0,1), w(z) = \dfrac {az+b}{cz+d}, a,b,c,d \in \mathrm {C}$, if $z \in \mathrm {R}$ then $|w| = |a/c| = 1$

Question

For context, my textbook is defining a transformation from $\mathrm {C}^+$ to $D(0,1)$ (D stands for disk). It finds out that $(-b/a)=\overline{(-d/c)}= \beta$. Then $w = \dfrac{a}{c} \dfrac{z - \beta}{z - \overline{\beta}}$. Now the book is stating that for $z \in \mathrm {R}$ $|w| =^{(1)} |a/c| =^{(2)} 1$. I understand $(1)$. Yet, I don't get $(2)$. How is $|a/c| = 1$? I assume this comes from the fact that $w(z) \in D(0,1)$, but I can't find a connection. Any hints? Thanks in advance.