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

Correct option is 2z lies on the real axis

Questions

If $z = x + i y$ and $w = \frac{1 - i z}{z - i},$ then $| w | = 1$ implies, that, in the complex plane

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z lies on the imaginary axis

z lies on the real axis

z lies on the unit circle

none of these

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detailed solution

Correct option is B

|w|=1 ⇒ 1−izz−i=1⇒ −i2−iz=|z−i|⇒ |z+i|=|z−i| ⇒ |(−i)(z+i)|=|z−i|⇒z lies on the perpendicular bisector of the segment join-ing i and -i⇒z lies on the real axis.

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

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Ready to Test Your Skills?

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If $z = x + i y$ and $w = \frac{1 - i z}{z - i},$ then $| w | = 1$ implies, that, in the complex plane