If z - x - iy and w -1- iz - z - i- then - w -1 implies that- on the complex planeA- z lies on the real axisB- None of the aboveC- z lies on the imaginary axisD- z lies on the unit circle

The correct option is A z lies on the real axisSince, |w|=1 ⇒1 iz/z 1=1 ⇒ |z i|=|1 iz| ⇒ |z i|=|z+i| [∵ |1 iz|=| i||z+i|=|z+i|] ∵ It is a perpendicular ...

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Byju's Answer

Standard XI

Mathematics

Modulus of a Complex Number

If z = x + iy...

Question

If $z = x + i y a n d w = \frac{1 - i z}{z - i},$ then $| w | = 1$ implies that, on the complex plane

z lies on the imaginary axis

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z lies on the real axis

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z lies on the unit circle

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None of the above

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Solution

The correct option is B z lies on the real axis
Since, $| w | = 1 \Rightarrow ∣ ∣ \begin{matrix} \frac{1 - i z}{z - 1} \end{matrix} ∣ ∣ = 1 \Rightarrow | z - i | = | 1 - i z | \Rightarrow | z - i | = | z + i | [∵ | 1 - i z | = | - i | | z + i | = | z + i |]$

$∵$ It is a perpendicular bisector of (0, 1) and (0, -1) i.e. x-axis. Thus, z lies on the real axis.

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Similar questions

If $z = x + i y$ and $ω = \frac{1 - i z}{z - i}$ , then $| ω | = 1$ implies that, in the complex plane $z$ lies on the real axis.

Q. State true or false:

z = x + i y

and

ω = \frac{1 - i z}{z - i}

, then

| ω | = 1

implies that in the complex plane

z

lies on the real axis.

Q. If

z = x + i y a n d w = \frac{1 - i z}{z - i},

then

| w | = 1

implies that, on the complex plane

Q. Let

z = x + i y

and

ω = \frac{1 - i z}{z - i}

. If

| ω | = 1

, then

z

lies on

Q. If the complex number

z = x + i y

satisfies the condition

|z + 1| = 1

, then z lies on

(a) x−axis
(b) circle with centre (−1, 0) and radius 1
(c) y−axis
(d) none of these

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If z=x+iy and w=1−izz−i, then |w|=1 implies that, on the complex plane

The correct option is B z lies on the real axisSince, |w|=1⇒∣∣1−izz−1∣∣=1⇒|z−i|=|1−iz|⇒|z−i|=|z+i| [∵|1−iz|=|−i||z+i|=|z+i|] ∵ It is a perpendicular bisector of (0, 1) and (0, -1) i.e. x-axis. Thus, z lies on the real axis.

If $z = x + i y a n d w = \frac{1 - i z}{z - i},$ then $| w | = 1$ implies that, on the complex plane