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Question Number 126552 by mathocean1 last updated on 21/Dec/20
u ; v∈ C such that ∣u∣=∣v∣=1 and uv≠−1. show that ((u+v)/(1+uv)) ∈ R.
$${u}\:;\:{v}\in\:\mathbb{C}\:{such}\:{that}\:\mid{u}\mid=\mid{v}\mid=\mathrm{1} \\ $$$${and}\:{uv}\neq−\mathrm{1}. \\ $$$${show}\:{that}\:\frac{{u}+{v}}{\mathrm{1}+{uv}}\:\in\:\mathbb{R}. \\ $$
Answered by Olaf last updated on 21/Dec/20
u = e^(iα) , v = e^(iβ) ((u+v)/(1+uv)) = ((e^(iα) +e^(iβ) )/(1+e^(iα) e^(iβ) )) = ((e^(i((α+β)/2)) (e^(i((α−β)/2)) +e^(−i((α−β)/2)) ))/(e^(i((α+β)/2)) (e^(i((α+β)/2)) +e^(−i((α+β)/2)) ))) = ((cos(((α−β)/2)))/(cos(((α+β)/2)))) ∈R
$${u}\:=\:{e}^{{i}\alpha} ,\:{v}\:=\:{e}^{{i}\beta} \\ $$$$\frac{{u}+{v}}{\mathrm{1}+{uv}}\:=\:\frac{{e}^{{i}\alpha} +{e}^{{i}\beta} }{\mathrm{1}+{e}^{{i}\alpha} {e}^{{i}\beta} }\:=\:\frac{{e}^{{i}\frac{\alpha+\beta}{\mathrm{2}}} \left({e}^{{i}\frac{\alpha−\beta}{\mathrm{2}}} +{e}^{−{i}\frac{\alpha−\beta}{\mathrm{2}}} \right)}{{e}^{{i}\frac{\alpha+\beta}{\mathrm{2}}} \left({e}^{{i}\frac{\alpha+\beta}{\mathrm{2}}} +{e}^{−{i}\frac{\alpha+\beta}{\mathrm{2}}} \right)} \\ $$$$=\:\frac{\mathrm{cos}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)}{\mathrm{cos}\left(\frac{\alpha+\beta}{\mathrm{2}}\right)}\:\in\mathbb{R} \\ $$

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