Doublet derivation

Question

I'm just following these notes. I am not sure how they went from the third line to the fourth line where it is expanded. Any chance someone could point me in the right direction? Thanks :-)

Answer 1

Examining the log term we have

$\large\log\left(\frac{1+\epsilon/z}{1-\epsilon/z}\right)=\log\left((1+\epsilon/z)\frac{1}{1-\epsilon/z}\right)$

Now the fraction $\frac{1}{1-\epsilon/z}$ has the following Taylor series expansion

$\large \frac{1}{1-\epsilon/z}=1+\sum_{n=1}^{\infty}(\epsilon/z)^n\sim 1+(\epsilon/z)+O\left((\epsilon^2/z^2\right))$

Interestingly, as per the OP's comment, the left hand side of the above equation can be regarded as a sum to infinity of a geometric progression of the terms on the right.

Thus we have

$\large\log\left(\frac{1+\epsilon/z}{1-\epsilon/z}\right)=\log\left\{(1+\epsilon/z)\left[1+(\epsilon/z)+O\left((\epsilon^2/z^2\right)\right]\right\}$