<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] 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
$$x-\dfrac1x = \dfrac{c+\sqrt{c^2+4}}2 - \dfrac{\sqrt{c^2+4}-c}2 = c$$
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