TITLE: Complex proof - Not sure where to go from here. (homework)
QUESTION [0 upvotes]: Knowing $2\pi r  =\dfrac{h}{m \left(\sqrt{\frac{e^2}{mr}}\right)}$,
How do I prove $r = \dfrac{h^2}{((2\pi)^2m e^2)}$?
I started by dividing both sides by $2\pi$ to get
$r = \dfrac{h}{m\left(\sqrt{\frac{e^2}{mr}}\right) 2\pi } $
then I squared both sides to get
$r^2 = \dfrac{h^2}{m^2 (\frac{e^2}{mr})(2\pi)^2}$
But I am not sure where to go from here to yield my expected result.
All help is appreciated.

REPLY [1 votes]: $$r^2 = \frac{h^2}{m^2 (\frac{e^2}{mr})(2\pi)^2}$$ 
$$ r^2 = \frac{h^2}{m (\frac{e^2}{r})(2\pi)^2}$$
$$ r^2 = \frac{r h^2}{m ({e^2})(2\pi)^2}$$
Multiply by $\displaystyle \frac{1}{r}$ both sides
$$ r = \frac{h^2}{m ({e^2})(2\pi)^2}$$

REPLY [0 votes]: $ r^2 = \frac{h^2}{m^2 \left(\frac{e^2}{mr}\right)(2\pi)^2}$
$r^2 = \frac{h^2}{\left(\frac{me^2}{r}\right)(2\pi)^2}$
$4\pi^2r^2 = \frac{h^2}{\left(\frac{me^2}{r}\right)}$
$4\pi^2r^2 = \frac{rh^2}{me^2}$
$4\pi^2r^2me^2 = rh^2$
$4\pi^2r^2me^2 - rh^2=0$
$r(4\pi^2rme^2 - h^2)=0$
$r=0, r=\frac{h^2}{4\pi^2me^2} = \frac{h^2}{((2\pi)^2me^2)}$
$r>0$ probably (in your context) so you can reject the solution $r=0$
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