If z1 and z2 are two complex numbers and if arg (z1+z2/z1-z2)=(π/2) but |z1+z2| ≠|z1-z2| then the figure formed by the points represented by 0, z1, z2 and z1+z2 is

Question

If z1 and z2 are two complex numbers and if arg (z1+z2/z1-z2)=(π/2) but |z1+z2| ≠|z1-z2| then the figure formed by the points represented by 0, z1, z2 and z1+z2 is (A) a parallelogram but not a rectangle or a rhombus (B) a rectangle but not a square (C) a rhombus but not a square (D) a square

Tardigrade Admin · Accepted Answer

We have vertices A(0), B(z1), C(z1+z2) and D(z2). arg (z1+z2/z1-z2)=(π/2) i.e., diagonals A C and B D are perpendicular. Also, |z1+z2| ≠|z1-z2| i.e., diagonals A C and B D have different length. Therefore, A B C D is rhombus but not a square. Hence, it is a rhombus.

Q. If z1​ and z2​ are two complex numbers and if arg z1​−z2​z1​+z2​​=2π​ but ∣z1​+z2​∣=∣z1​−z2​∣ then the figure formed by the points represented by 0,z1​,z2​ and z1​+z2​ is

a parallelogram but not a rectangle or a rhombus

a rectangle but not a square

a rhombus but not a square

a square

Q. If $z_{1}$ and $z_{2}$ are two complex numbers and if arg $\frac{z _{1} + z _{2}}{z _{1} - z _{2}} = \frac{π}{2}$ but $∣ z_{1} + z_{2} ∣ \neq = ∣ z_{1} - z_{2} ∣$ then the figure formed by the points represented by $0, z_{1}, z_{2}$ and $z_{1} + z_{2}$ is