How does $\frac{1}{z}-\frac{z}{\sqrt{L^2+z^2}\left(L+\sqrt{L^2+z^2}\right)}$ become $\frac{L}{z\sqrt{L^2+z^2}}$?

Question

I have the following expression involving lengths $z$ and $L$

$$\dfrac{1}{z}-\dfrac{z}{\sqrt{L^2+z^2}\left(L+\sqrt{L^2+z^2}\right)}$$

I 'need' to simplify this to simplify this down to $$\dfrac{L}{z\sqrt{L^2+z^2}}$$

to get the answer in the back of the book (Griffiths' Introduction to Electrodynamics), but I'm getting no where. Would anyone mind pointing me in the right direction?

Answer 1

Start with LCM to make both fractions have the same denominator.

\begin{align*} \dfrac{1}{z}-\dfrac{z}{\sqrt{L^2+z^2}\big(L+\sqrt{L^2+z^2}\big)} \\ =\dfrac{\sqrt{L^2+z^2}\big(L+\sqrt{L^2+z^2}\big)}{z\sqrt{L^2+z^2}\big(L+\sqrt{L^2+z^2}\big)} -\dfrac{z^2} {z\sqrt{zL^2+z^2}\big(L+\sqrt{L^2+z^2}\big)}\\ \\ =\dfrac{\sqrt{L^2+z^2}\big(L+\sqrt{L^2+z^2}\big)-z^2}{z\sqrt{L^2+z^2}\big(L+\sqrt{L^2+z^2}\big)} =\dfrac{L \sqrt{L^2 + z^2} + L^2} {z \sqrt{L^2 + z^2} \big(\sqrt{L^2 + z^2} + L\big)}\\ \\ =\dfrac{\qquad L\quad\space \big(\sqrt{L^2 + z^2} + L\big)} {z \sqrt{L^2 + z^2} \big(\sqrt{L^2 + z^2} + L\big)} = \dfrac{L}{z \sqrt{L^2 + z^2}} \end{align*}