IIT JEE 1983 Maths - MCQ Question 12

Algebra Level 4

There exists a complex number z z is such that ω = 1 − i z z − i \omega=\dfrac{1-iz}{z-i} and ∣ ω ∣ = 1 |\omega|=1 .

Which is true about z z in the complex plane?

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z z lies on unit circle None of the others z z lies on real axis z z lies on imaginary axis

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Shubhamkar Ayare by Shubhamkar Ayare , 22, India

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1 solution

Chew-Seong Cheong
Dec 31, 2016

∣ ω ∣ = 1 ∣ 1 − i z z − i ∣ = 1 ∣ 1 − i z ∣ = ∣ z − i ∣ Let z = x + i y ∣ 1 − i ( x + i y ) ∣ = ∣ x + i y − i ∣ ∣ 1 + y − i x ∣ = ∣ x + i ( y − 1 ) ∣ ( y + 1 ) 2 + x 2 = x 2 + ( y − 1 ) 2 y 2 + 2 y + 1 + x 2 = x 2 + y 2 − 2 y + 1 2 y = − 2 y ⟹ y = 0 ⟹ z = x real \begin{aligned} |\omega| & = 1 \\ \left| \frac {1-iz}{z-i} \right| & = 1 \\ |1-i{\color{#3D99F6}z}| & = |{\color{#3D99F6}z}-i| & \small \color{#3D99F6} \text{Let }z = x+iy \\ |1-i({\color{#3D99F6}x+iy})| & = |{\color{#3D99F6}x+iy}-i| \\ |1+y-ix| & = |x+i(y-1)| \\ (y+1)^2 + x^2 & = x^2 +(y-1)^2 \\ y^2+2y+1 + x^2 & = x^2+y^2-2y+1 \\ 2y & = - 2y \\ \implies y & = 0 \\ \implies z & = x \quad \color{#3D99F6} \text{real} \end{aligned}

⟹ z lies on the real axis. \implies \boxed{z \text{ lies on the real axis.}}

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