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If z =(x+iy) and w =( 1 - iz) / (z - 1) such that | w | = 1 then show that z is purely real .

$given w= \frac{1 - i z}{z - i} Squaring on both sides w^{2} = {(\frac{1 - i z}{z - i})}^{2} \Rightarrow 1 = \frac{1 + i^{2} z^{2} - 2 i z}{z^{2} + i^{2} - 2 i z} (∵ | w | = 1 \Rightarrow w^{2} = 1) \Rightarrow z^{2} - 1 - 2 i z = 1 - z^{2} - 2 i z \Rightarrow 2 z^{2} = 2 \Rightarrow z^{2} = 1 \Rightarrow (x + i y)^{2} = 1 \Rightarrow x^{2} + i^{2} y^{2} + 2 i x y = 1 \Rightarrow x^{2} - y^{2} + 2 i x y = 1 Imaginary part is zero, so Z is purely real$

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