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Question Number 51643 by Tawa1 last updated on 29/Dec/18

If ∣z∣ = 1, prove that ((z − 1)/(z^− − 1)) (z ≠ 1) is a pure imaginary

$$\mathrm{If}\:\:\:\mid\mathrm{z}\mid\:=\:\mathrm{1},\:\:\:\:\mathrm{prove}\:\mathrm{that}\:\:\:\:\:\frac{\mathrm{z}\:−\:\mathrm{1}}{\overset{−} {\mathrm{z}}\:−\:\mathrm{1}}\:\:\:\:\:\:\left(\mathrm{z}\:\neq\:\mathrm{1}\right)\:\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{pure}\:\mathrm{imaginary} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18

z=rcosθ+irsinθ=r(cosθ+isinθ) ∣z∣=r=1 ((z−1)/(z^− −1))=((cosθ+isinθ−1)/(cosθ−isinθ−1))=((−2sin^2 (θ/2)+i2sin(θ/2)cos(θ/2))/(−2sin^2 (θ/2)−i2sin(θ/2)cos(θ/2))) =((−2sin(θ/2)(sin(θ/2)−icos(θ/2)))/(−2sin(θ/2)(sin(θ/2)+icos(θ/2)))) =(((sin(θ/2)−icos(θ/2))^2 )/((sin(θ/2))^2 −i^2 (cos(θ/2))^2 )) =((sin^2 (θ/2)−2isin(θ/2)cos(θ/2)+i^2 cos^2 (θ/2))/(sin^2 (θ/2)+cos^2 (θ/2))) =−(cos^2 (θ/2)−sin^2 (θ/2)+i×2sin(θ/2)cos(θ/2)) =−(cosθ+isinθ) =−z pls check the question...

$${z}={rcos}\theta+{irsin}\theta={r}\left({cos}\theta+{isin}\theta\right) \\ $$$$\mid{z}\mid={r}=\mathrm{1} \\ $$$$\frac{{z}−\mathrm{1}}{\overset{−} {{z}}−\mathrm{1}}=\frac{{cos}\theta+{isin}\theta−\mathrm{1}}{{cos}\theta−{isin}\theta−\mathrm{1}}=\frac{−\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}+{i}\mathrm{2}{sin}\frac{\theta}{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}}{−\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}−{i}\mathrm{2}{sin}\frac{\theta}{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}} \\ $$$$=\frac{−\mathrm{2}{sin}\frac{\theta}{\mathrm{2}}\left({sin}\frac{\theta}{\mathrm{2}}−{icos}\frac{\theta}{\mathrm{2}}\right)}{−\mathrm{2}{sin}\frac{\theta}{\mathrm{2}}\left({sin}\frac{\theta}{\mathrm{2}}+{icos}\frac{\theta}{\mathrm{2}}\right)} \\ $$$$=\frac{\left({sin}\frac{\theta}{\mathrm{2}}−{icos}\frac{\theta}{\mathrm{2}}\right)^{\mathrm{2}} }{\left({sin}\frac{\theta}{\mathrm{2}}\right)^{\mathrm{2}} −{i}^{\mathrm{2}} \left({cos}\frac{\theta}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\frac{{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}−\mathrm{2}{isin}\frac{\theta}{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}+{i}^{\mathrm{2}} {cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}+{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}} \\ $$$$=−\left({cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}−{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}+{i}×\mathrm{2}{sin}\frac{\theta}{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}\right) \\ $$$$=−\left({cos}\theta+{isin}\theta\right) \\ $$$$=−{z} \\ $$$${pls}\:{check}\:{the}\:{question}... \\ $$$$ \\ $$

Commented by Tawa1 last updated on 29/Dec/18

God bless you sir. The question is not true sir ?

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{The}\:\mathrm{question}\:\mathrm{is}\:\mathrm{not}\:\mathrm{true}\:\mathrm{sir}\:? \\ $$

Commented by peter frank last updated on 29/Dec/18

⇒ ((z+1)/(z−1)) i think

$$\Rightarrow\:\frac{{z}+\mathrm{1}}{{z}−\mathrm{1}}\:\:\:\:{i}\:{think} \\ $$

Answered by peter frank last updated on 29/Dec/18

((z+1)/(z−1)) z=x+iy z^− =x−iy ((x+iy−1)/(x−iy−1)) (((x+1)+iy)/((x−1)−iy)) [ (((x+1)+iy)/((x−1)−iy))]×[(((x−1)+iy)/((x−1)+iy ))] ((x^2 +y^2 −1)/(x^2 +y^2 −2x))+((2xi)/(x^2 +y^2 −2x)) pure imaginary ((x^2 +y^2 −1)/(x^2 +y^2 −2x))=0 x^2 +y^2 −1=0 x^2 +y^2 =1 ∣x+iy∣=1 ∣z∣=1

$$\frac{{z}+\mathrm{1}}{{z}−\mathrm{1}} \\ $$$${z}={x}+{iy} \\ $$$$\overset{−} {{z}}={x}−{iy} \\ $$$$\frac{{x}+{iy}−\mathrm{1}}{{x}−{iy}−\mathrm{1}} \\ $$$$\:\frac{\left({x}+\mathrm{1}\right)+{iy}}{\left({x}−\mathrm{1}\right)−{iy}} \\ $$$$\left[\:\frac{\left({x}+\mathrm{1}\right)+{iy}}{\left({x}−\mathrm{1}\right)−{iy}}\right]×\left[\frac{\left({x}−\mathrm{1}\right)+{iy}}{\left({x}−\mathrm{1}\right)+{iy}\:}\right] \\ $$$$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{x}}+\frac{\mathrm{2}{xi}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{x}} \\ $$$${pure}\:{imaginary} \\ $$$$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{x}}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mid{x}+{iy}\mid=\mathrm{1} \\ $$$$\mid{z}\mid=\mathrm{1} \\ $$$$ \\ $$

Commented by Tawa1 last updated on 29/Dec/18

God bless you sir

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

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