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. 
 A: We have $\dfrac1x = \dfrac2{c+\sqrt{c^2+4}} = \dfrac2{c+\sqrt{c^2+4}}\cdot \dfrac{c-\sqrt{c^2+4}}{c-\sqrt{c^2+4}} = \dfrac{\sqrt{c^2+4}-c}2$
Hence, we have
$$x-\dfrac1x = \dfrac{c+\sqrt{c^2+4}}2 - \dfrac{\sqrt{c^2+4}-c}2 = c$$
A: 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
A: Hint: Notice that your condition implies
$$x-\frac{1}{x}=c$$
that is equivalent to $x^2-cx-1=0$.
A: Notice that $a x^2 + b x + c = 0$ has solutions $\frac{-b \pm \sqrt{b^2 - 4 a c}}{2a}$ by the quadratic formula. Using this result and working backwards, we see that $x = \frac{(c+\sqrt{c^2+4})}{2}$ implies that $ x^2 -c x -1 = 0$. So $x^2 -1 = cx$ then dividing both sides by x, we get $\frac{x^2 - 1}{x}=c$ So $c = x - \frac{1}{x}$.
A: In general the root to the equation $ax^2+bx +c = 0$ has roots $x =\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. 
We notice that the expresion of $x$ in your question is similar to this form, too. Thus by comparing the coefficient, we get $x_0 = \frac{c\pm\sqrt{c^2+4}}{2}$ is a root of the equation $x^2 -cx -1 = 0$. 
Take $x = \frac{1}{x_0}$, we get $\frac{1}{x_0^2}-c\frac{1}{x_0}-1=0$. Thus by Vieta's formulas get $x_0+\frac{1}{x_0} = c \in C$
