Answer
$$\beta=\frac{(1+z)^{2}-1}{(1+z)^{2}+1}=\frac{z^{2}+2 z}{z^{2}+2 z+2} $$
Work Step by Step
$\text { From } f=c / \lambda \text { and } \mathrm{Eq} .37-31,$ we get
$$\lambda_{0}=\lambda \sqrt{\frac{1-\beta}{1+\beta}}=\left(\lambda_{0}+\Delta \lambda\right) \sqrt{\frac{1-\beta}{1+\beta}} $$
Dividing both sides by $\lambda_{0}$ leads to
$$1=(1+z) \sqrt{\frac{1-\beta}{1+\beta}} $$
$\text { where } z=\Delta \lambda / \lambda_{0} . \text { We solve for } \beta: $
$$\beta=\frac{(1+z)^{2}-1}{(1+z)^{2}+1}=\frac{z^{2}+2 z}{z^{2}+2 z+2} $$
