If z1 and z2 are two complex numbers such that |z1−z2z1+z2|=1, then

Clearly both z1 and z2 cannot be 0. Supposez2≠0. We can write |z1−z2z1+z2|=1, as |z1/z2−1z1/z2+1|=1 or as |z1z2−1|=|z1z2+1|.This shows that z1/z2 lies on the perpendicular bisector of the segment joining A(−1 + i0) and B(1 + i0) Thus, z1/z2 lies on the imaginary axis.∴ z1/z2=ia for some a ∈ R.⇒ z2/z1=1/ia=−i/a⇒ z2=ikz1 for some k ∈ RAlternative Solution: O(0), A(z1), B(z1+z2) and c(z2) are the vertices of a parallelogram.Its diagonals OB(z1+z2) and CA(z1−z2) will be equal if and only if OABC is a rectangle, that is, z2/z1 is purely imaginary.

If z1 and z2 are two complex numbers such that |z1−z2z1+z2|=1, then

Clearly both z1 and z2 cannot be 0. Supposez2≠0. We can write |z1−z2z1+z2|=1, as |z1/z2−1z1/z2+1|=1 or as |z1z2−1|=|z1z2+1|.This shows that z1/z2 lies on the perpendicular bisector of the segment joining A(−1 + i0) and B(1 + i0) Thus, z1/z2 lies on the imaginary axis.∴ z1/z2=ia for some a ∈ R.⇒ z2/z1=1/ia=−i/a⇒ z2=ikz1 for some k ∈ RAlternative Solution: O(0), A(z1), B(z1+z2) and c(z2) are the vertices of a parallelogram.Its diagonals OB(z1+z2) and CA(z1−z2) will be equal if and only if OABC is a rectangle, that is, z2/z1 is purely imaginary.

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If $z_{1}$ and $z_{2}$ are two complex numbers such that $| \frac{z_{1} - z_{2}}{z_{1} + z_{2}} | = 1,$ then

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$z_{2} = k z_{1}, k \in$ R

$z_{2} = i k z_{1}, k \in$ R

$z_{1} = z_{2}$

none of these

answer is B.

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Detailed Solution

Clearly both $z_{1}$ and $z_{2}$ cannot be $0.$ Suppose

$z_{2} \neq 0.$ We can write $| \frac{z_{1} - z_{2}}{z_{1} + z_{2}} | = 1,$ as $| \frac{z_{1} / z_{2} - 1}{z_{1} / z_{2} + 1} | = 1$ or as $| \frac{z_{1}}{z_{2}} - 1 | = | \frac{z_{1}}{z_{2}} + 1 | .$

This shows that $z_{1} / z_{2}$ lies on the perpendicular bisector of the segment joining $A (- 1 + i 0)$ and $B (1 + i 0)$

Thus, $z_{1} / z_{2}$ lies on the imaginary axis.

$∴$ $z_{1} / z_{2} = i a$ for some $a \in R .$

$\Rightarrow$ $z_{2} / z_{1} = 1 / i a = - i / a$

$\Rightarrow$ $z_{2} = i k z_{1}$ for some $k \in R$

Alternative Solution:

$O (0), A (z_{1}), B (z_{1} + z_{2})$ and $c (z_{2})$ are the vertices of a parallelogram.

Its diagonals $O B (z_{1} + z_{2})$ and $C A (z_{1} - z_{2})$ will be equal if and only if $O A B C$ is a rectangle, that is, $z_{2} / z_{1}$ is purely imaginary.