Let z- x- iy and - - 1- izz-i- If - -1- then z lies on

Given: ω = 1 izz i, |ω| =1 ⇒ | 1 izz i |=1 ⇒ |1 iz|= |z i| ⇒ | i(z+i)| = |z i| ⇒ |z+i|= |z i| Therefore, z lies on the perpendicular bisector of line segmen ...

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Byju's Answer

Standard XII

Mathematics

Inequality of 2 Complex Numbers

Let z= x+ iy ...

Question

Let $z = x + i y$ and $ω = \frac{1 - i z}{z - i}$ . If $| ω | = 1$ , then $z$ lies on

the imaginary axis

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on the real axis

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a unit circle

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a parabola

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Solution

The correct option is B on the real axis
$| ω | = 1 \Rightarrow ∣ ∣ ∣ \frac{1 - i z}{z - i} ∣ ∣ ∣ = 1 \Rightarrow | 1 - i z | = | z - i | \Rightarrow | - i (z + i) | = | z - i | \Rightarrow | z + i | = | z - i |$

Therefore, $z$ lies on the perpendicular bisector of line segment joining $(0, - 1)$ and $(0, 1) .$
Hence, the locus of $z$ is $y = 0$ , so $z$ lies on the real axis.

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Let z=x+iy and ω=1−izz−i. If |ω|=1, then z lies on

The correct option is B on the real axis|ω|=1⇒∣∣∣1−izz−i∣∣∣=1⇒|1−iz|=|z−i|⇒|−i(z+i)|=|z−i|⇒|z+i|=|z−i| Therefore, z lies on the perpendicular bisector of line segment joining (0,−1) and (0,1). Hence, the locus of z is y=0, so z lies on the real axis.

Let $z = x + i y$ and $ω = \frac{1 - i z}{z - i}$ . If $| ω | = 1$ , then $z$ lies on