Prove that if c ϵ C and $x = \frac{(c+\sqrt{c^2+4})}{2}$ so $x-\frac{1}{x} $ ϵ C. Prove that if c ϵ C and $x = \frac{(c+\sqrt{c^2+4})}{2}$ so  $x-\frac{1}{x} $ ϵ C.  I have no idea how do this. Please help me. 

We have
$x-\frac{1}{x}$ = $\frac{x^2 - 1}{x}$ = $\frac{(x-1)(x+1)}{x}$ 
x-1 = $\frac{c + \sqrt{ c ^ {2}+4 }}{2}$ -1 = $\frac{c + \sqrt{ c ^ {2}+4 }-2}{2}$
x+1 = $\frac{c + \sqrt{ c ^ {2}+4 }+2}{2}$
So : $x-\frac{1}{x}$ = $\frac{\frac{c + \sqrt{ c ^ {2}+4 }-2}{2}*\frac{c + \sqrt{ c ^ {2}+4 }+2}2}{\frac{c + \sqrt{ c ^ {2}+4 }}{2}}$ = $\frac{(c + \sqrt{ c ^ {2}+4 })^2 - 4 }{4}$*$\frac{2}{c + \sqrt{ c ^ {2}+4 }}$ = $\frac{2c^2 + 2c \sqrt{ c ^ {2}+4 }}{2(c + \sqrt{ c ^ {2}+4} )}$=c
We have proved $x-\frac{1}{x}$ ϵ C
