Question

If $z_{1}$ and $z_{2}$ are two complex numbers and if arg $\frac{z_{1}+z_{2}}{z_{1}-z_{2}}=\frac{\pi}{2}$ but $\left|z_{1}+z_{2}\right| \neq\left|z_{1}-z_{2}\right|$ then the figure formed by the points represented by $0, z_{1}, z_{2}$ and $z_{1}+z_{2}$ 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

Answer 1

Answer: (C)

We have vertices $A(0), B\left(z_{1}\right), C\left(z_{1}+z_{2}\right)$ and $D\left(z_{2}\right)$. $\arg \frac{z_{1}+z_{2}}{z_{1}-z_{2}}=\frac{\pi}{2}$
i.e., diagonals $A C$ and $B D$ are perpendicular.
Also, $\left|z_{1}+z_{2}\right| \neq\left|z_{1}-z_{2}\right|$
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.