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

$$\mathrm{If}\mid \mathrm{z}\mid =1,\mathrm{prove}\mathrm{that}\frac{\mathrm{z}-1}{\stackrel{-}{\mathrm{z}}-1}(\mathrm{z}\ne 1)\mathrm{is}\mathrm{a}\mathrm{pure}\mathrm{imaginary}$$

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

$$z=rcos\theta +irsin\theta =r(cos\theta +isin\theta )$$$$\mid z\mid =r=1$$$$\frac{z-1}{\stackrel{-}{z}-1}=\frac{cos\theta +isin\theta -1}{cos\theta -isin\theta -1}=\frac{-2{sin}^2\frac{\theta }{2}+i2sin\frac{\theta }{2}cos\frac{\theta }{2}}{-2{sin}^2\frac{\theta }{2}-i2sin\frac{\theta }{2}cos\frac{\theta }{2}}$$$$=\frac{-2sin\frac{\theta }{2}(sin\frac{\theta }{2}-icos\frac{\theta }{2})}{-2sin\frac{\theta }{2}(sin\frac{\theta }{2}+icos\frac{\theta }{2})}$$$$=\frac{{(sin\frac{\theta }{2}-icos\frac{\theta }{2})}^2}{{\left(sin\frac{\theta }{2}\right)}^2-i^2{\left(cos\frac{\theta }{2}\right)}^2}$$$$=\frac{{sin}^2\frac{\theta }{2}-2isin\frac{\theta }{2}cos\frac{\theta }{2}+i^2{cos}^2\frac{\theta }{2}}{{sin}^2\frac{\theta }{2}+{cos}^2\frac{\theta }{2}}$$$$=-({cos}^2\frac{\theta }{2}-{sin}^2\frac{\theta }{2}+i\times 2sin\frac{\theta }{2}cos\frac{\theta }{2})$$$$=-(cos\theta +isin\theta )$$$$=-z$$$$plscheckthequestion...$$$$$$

Commented by Tawa1 last updated on 29/Dec/18

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

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

$$\Rightarrow \frac{z+1}{z-1}ithink$$

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

$$\frac{z+1}{z-1}$$$$z=x+iy$$$$\stackrel{-}{z}=x-iy$$$$\frac{x+iy-1}{x-iy-1}$$$$\frac{(x+1)+iy}{(x-1)-iy}$$$$\left[\frac{(x+1)+iy}{(x-1)-iy}\right]\times \left[\frac{(x-1)+iy}{(x-1)+iy}\right]$$$$\frac{x^2+y^2-1}{x^2+y^2-2x}+\frac{2xi}{x^2+y^2-2x}$$$$pureimaginary$$$$\frac{x^2+y^2-1}{x^2+y^2-2x}=0$$$$x^2+y^2-1=0$$$$x^2+y^2=1$$$$\mid x+iy\mid =1$$$$\mid z\mid =1$$$$$$

Commented by Tawa1 last updated on 29/Dec/18

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

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