In a previous part, I solved for $B$ in terms of $A$ and resulted in this:
$$B = \frac{i\beta}{(1-i\beta)}A$$
Now I need to find $R=\frac{|B|^2}{|A|^2}$. To do so, I did this:
$$\frac{B}{A}=\frac{i\beta}{(1-i\beta)}$$ $$\frac{|B|^2}{|A|^2}=\frac{(i\beta)^2}{((1-i\beta))^2}$$ $$\frac{|B|^2}{|A|^2}=\frac{-\beta^2}{1-2i\beta-\beta ^2}$$ $$\frac{|B|^2}{|A|^2}=\frac{\beta^2}{\beta ^2+2i\beta - 1}$$ $$\frac{|B|^2}{|A|^2}=\frac{\beta^2}{\beta + i}$$
...but the book continues the example to say that it's actually $$R=\frac{\beta^2}{1+\beta^2}$$
Where did I make a mistake? This is from Introduction to Quantum Mechanics by David Griffiths (1st edition I think), pg. 57.
This part is wrong
$$\frac{|B|^2}{|A|^2}=\frac{(i\beta)^2}{((1-i\beta))^2}$$
You need to take absolute value of the right hand side and then square, not just square it. (Read about complex numbers, and what is the absolute value of a complex number. You need to find the real part and the imaginary part of the expression on the right hand side. )