[JEE Advanced 1983] If $z=x+iy$ and $\omega=\dfrac{1-iz}{z-i}$, then $|\omega|=1$ implies that
If $z=x+iy$ and $\omega=\dfrac{1-iz}{z-i}$, then $|\omega|=1$ implies that
(a) $z$ lies on the imaginary axis
(b) $z$ lies on the real axis
(c) $z$ lies on the unit circle
(d) none of these
1 Answer
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✓ Vidaara Team
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· 1mo ago
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Correct answer: (b) $z$ lies on the real axis
$1-iz=-i(z+i)$, so $|\omega|=\frac{|z+i|}{|z-i|}=1\Rightarrow|z+i|=|z-i|\Rightarrow z$ is on the real axis.
JEE Advanced 1983 · Complex Numbers — verified solution by the Vidaara Team.
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