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

To determine the locus of the complex numbers \(z\) that satisfy the given condition \(\left|\dfrac{i+z}{i-z}\right| = 1\), we will follow a logical and mathematical approach:

1. We start by letting \(z = x + yi\), where \(x\) and \(y\) are real numbers. Thus, we have: \(\dfrac{i+z}{i-z} = \dfrac{i + x + yi}{i - x - yi}\).
2. We then multiply the numerator and denominator by the conjugate of the denominator: \(\left(i + x + yi\right)\left(i + x - yi\right) = (i + x)^2 + y^2\).
3. The magnitude condition given is: \(\left|\dfrac{i+z}{i-z}\right| = 1\), which implies: \(\left|i + x + yi\right| = \left|i - x - yi\right|\).
4. Calculating the moduli, we have: \(\sqrt{x^2 + (y+1)^2} = \sqrt{x^2 + (y-1)^2}\).
5. By squaring both sides: \(x^2 + (y+1)^2 = x^2 + (y-1)^2\).
6. Simplifying gives: \(y^2 + 2y + 1 = y^2 - 2y + 1\).
7. On further simplification: \(4y = 0 \Rightarrow y = 0\).

Hence, the complex numbers \(z\) satisfying the condition \(\left|\dfrac{i+z}{i-z}\right| = 1\) lie on the x-axis.

Therefore, the correct answer is: the x-axis.