<s>[INST] 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. 
 [/INST] 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 $c = x - \frac{1}{x}$.
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