Given that $R^2 = z^2 + \rho^2$, I wish to show that,
$$ \ln \Bigg [ \frac{\sqrt{(L - z)^2 + \rho^2} + L - z}{\sqrt{(L + z)^2 + \rho^2} - L - z} \Bigg ], \tag{3.30} $$
is approximately equal to:
$$ \ln \Bigg [ \frac{1 - z/R + L/R}{1 - z/R - L/R} \Bigg ] \tag{3.35} . $$
if $z \ggg L$. But I am unable to get this result. I get:
$$ \ln \Bigg [ \frac{1 - 2z/L + R^2/2L^2}{R^2/2L^2} \Bigg ], $$
by the usual trick: take $L^2$ common with the square. Use the Taylor Exapansion. Take $L$ common in both the numerator and denominator and cancel it out.
What am I missing?
Edit:
Here's the snippet from the textbook. I have underlined the relevant portion for your reference:
Please note that I dropped all the constants in my post. I hope the snippet is readable. If not, access the book at: http://shrek.unideb.hu/~adamkardos/temp/boox/Modern%20electrodynamics.pdf
Please correct if I am making any mistake.
I think I understand what they did...these physicists really are...gees!
The gist is that $\;z>>L\;$, so $\;\sqrt{(L\pm z)^2+\rho^2}\approx\sqrt{z^2+\rho^2}=R\;$ (becasue this is going to be the argument of a logarithm), and then they just did divide by $\;R\;$ ...!:
$$\frac{\sqrt{(L-z)^2+\rho^2}+L-z}{\sqrt{(L-z)+\rho^2}-L-z}\approx\frac{R+L-z}{R-L+z}\stackrel{:R}=\frac{1+\frac LR-\frac zR}{1-\frac LR+\frac zR}$$
...and that's all!