Question

The complex numbers $z$ satisfying $\left|\dfrac{i+z}{i-z}\right| = 1$ lie on

• A. the circle $x^{2} + y^{2} = 1$
• B. the $x$-axis
• C. the $y$-axis
• D. the line $x + y = 1$

Hint:

$|z - a| = |z - b|$ is the perpendicular bisector of the segment joining $a$ and $b$ in the complex plane. Here $a = -i$ and $b = i$, so bisector is the real axis ($x$-axis).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
$|z_1/z_2| = 1 \Rightarrow |z_1| = |z_2|$. Use this to find the locus.
Step 2: Detailed Explanation:
$|i+z| = |i-z|$. Let $z = x+iy$: $|x+i(y+1)| = |x+i(y-1)|$.
$x^2+(y+1)^2 = x^2+(y-1)^2 \Rightarrow 4y = 0 \Rightarrow y = 0$.
So $z$ lies on the $x$-axis.
Step 3: Final Answer:
The locus is the $x$-axis.