The locus of the point z-x-iy satisfying the equation - z-1 z-1 - -1 is given by-

The correct option is B x=0 | z 1/z+1 |=1 ∴ | z 1/z+1 |=1 ∴|z 1| = |z+1| i.e. lie on the ⊥ nbsp; bisector the line joining (1,0) ...

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Standard XII

Mathematics

Optimization Problems

The locus of ...

Question

The locus of the point $z = x + i y$ satisfying the equation $∣ ∣ ∣ \frac{z - 1}{z + 1} ∣ ∣ ∣ = 1$ is given by?

x = 0

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y = 0

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x = y

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x + y = 0

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Solution

The correct option is B $x = 0$
$∣ ∣ \frac{z - 1}{z + 1} ∣ ∣ = 1$
$∴ ∣ ∣ \frac{z - 1}{z + 1} ∣ ∣ = 1$
$∴ | z - 1 | = | z + 1 |$
i.e. lie on the $⊥$ bisector the line joining (1,0) & (-1,0)
i.e the y axis
$∴$ 2 lies on $x = 0$ .

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The locus of the point z=x+iy satisfying the equation ∣∣∣z−1z+1∣∣∣=1 is given by?

The correct option is B x=0∣∣z−1z+1∣∣=1∴∣∣z−1z+1∣∣=1∴|z−1|=|z+1|i.e. lie on the ⊥ bisector the line joining (1,0) & (-1,0)i.e the y axis∴ 2 lies on x=0.

The locus of the point $z = x + i y$ satisfying the equation $∣ ∣ ∣ \frac{z - 1}{z + 1} ∣ ∣ ∣ = 1$ is given by?

The correct option is B $x = 0$
$∣ ∣ \frac{z - 1}{z + 1} ∣ ∣ = 1$
$∴ ∣ ∣ \frac{z - 1}{z + 1} ∣ ∣ = 1$
$∴ | z - 1 | = | z + 1 |$
i.e. lie on the $⊥$ bisector the line joining (1,0) & (-1,0)
i.e the y axis
$∴$ 2 lies on $x = 0$ .