Question

The complex number \(z\) which satisfy the equation \(\left|\frac{1+z}{1-z}\right| = 1\) lies on

• A. a circle \(x^2+y^2=1\)
• B. the x-axis
• C. the y-axis
• D. the line \(x+y=1\)

Hint:

\(|z-a| = |z-b|\) $\Rightarrow$ perpendicular bisector of points \(a\) and \(b\).

Answer 1

The Correct Option is B

Concept: \[ \left|\frac{a}{b}\right| = 1 \Rightarrow |a| = |b| \]

Step 1: Apply condition.
\[ |1+z| = |1-z| \]

Step 2: Let \(z = x+iy\).
\[ |1+z|^2 = (1+x)^2 + y^2 \] \[ |1-z|^2 = (1-x)^2 + y^2 \]

Step 3: Equate.
\[ (1+x)^2 + y^2 = (1-x)^2 + y^2 \] \[ (1+x)^2 = (1-x)^2 \Rightarrow 4x = 0 \Rightarrow x=0 \]

Step 4: Conclusion.
\[ z = iy \Rightarrow \text{point lies on y-axis} \] Hence real part = 0 $\Rightarrow$ corresponds to option (B).