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Question Number 54543 by Tawa1 last updated on 05/Feb/19

Prove that: ((z^2 − 1)/(z^2 + 1)) = i tan(θ) where z = cos(θ) + i sin(θ)

$$\mathrm{Prove}\:\mathrm{that}:\:\:\:\frac{\mathrm{z}^{\mathrm{2}} \:−\:\mathrm{1}}{\mathrm{z}^{\mathrm{2}} \:+\:\mathrm{1}}\:\:=\:\:\mathrm{i}\:\mathrm{tan}\left(\theta\right) \\ $$$$\mathrm{where}\:\:\:\:\:\mathrm{z}\:\:=\:\:\mathrm{cos}\left(\theta\right)\:+\:\mathrm{i}\:\mathrm{sin}\left(\theta\right) \\ $$

Commented by maxmathsup by imad last updated on 06/Feb/19

we have z =e^(iθ) ⇒((z^2 −1)/(z^2 +1)) =((e^(2iθ) −1)/(e^(2iθ) +1)) =((e^(iθ) −e^(−iθ) )/(e^(iθ) +e^(−iθ) )) =i (((e^(iθ) −e^(−iθ) )/(2i))/((e^(iθ) +e^(−iθ) )/2)) =i ((sinθ)/(cosθ)) =itanθ

$${we}\:{have}\:{z}\:={e}^{{i}\theta} \:\Rightarrow\frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}\:=\frac{{e}^{\mathrm{2}{i}\theta} −\mathrm{1}}{{e}^{\mathrm{2}{i}\theta} \:+\mathrm{1}}\:=\frac{{e}^{{i}\theta} \:−{e}^{−{i}\theta} }{{e}^{{i}\theta} \:+{e}^{−{i}\theta} }\:={i}\:\frac{\frac{{e}^{{i}\theta} \:−{e}^{−{i}\theta} }{\mathrm{2}{i}}}{\frac{{e}^{{i}\theta} \:+{e}^{−{i}\theta} }{\mathrm{2}}}\:={i}\:\frac{{sin}\theta}{{cos}\theta}\:={itan}\theta \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

((z^2 −1)/(z^2 +1)) =((z−(1/z))/(z+(1/z))) z=cosθ+isinθ=e^(iθ) (1/z)=(1/e^(iθ) )=e^(−iθ) =cos(−θ)+isin(−θ)=cosθ−isinθ so ((z−(1/z))/(z+(1/z)))=((2isinθ)/(2cosθ))=itanθ

$$\frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\frac{{z}−\frac{\mathrm{1}}{{z}}}{{z}+\frac{\mathrm{1}}{{z}}} \\ $$$${z}={cos}\theta+{isin}\theta={e}^{{i}\theta} \\ $$$$\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{{e}^{{i}\theta} }={e}^{−{i}\theta} ={cos}\left(−\theta\right)+{isin}\left(−\theta\right)={cos}\theta−{isin}\theta \\ $$$${so}\:\:\:\frac{{z}−\frac{\mathrm{1}}{{z}}}{{z}+\frac{\mathrm{1}}{{z}}}=\frac{\mathrm{2}{isin}\theta}{\mathrm{2}{cos}\theta}={itan}\theta \\ $$

Commented by Tawa1 last updated on 05/Feb/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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