<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] 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 $x_0+\frac{1}{x_0} = c \in C$
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