Question

Let \( w = \frac{1-iz}{z-i} \). If \( |w| = 1 \), which of the following must be true?

• A. \(z\) lies inside the unit circle
• B. \(z\) lies on real axis
• C. \(z\) lies on imaginary axis
• D. \(z\) lies outside the unit circle
• E. \( \text{Re } z < 0 \)

Hint:

Convert modulus equations into algebra using \(z=x+iy\).

Answer 1

The Correct Option is B

Concept: If \( |w|=1 \), then numerator and denominator magnitudes are equal.

Step 1: Use modulus property. \[ |w| = \left|\frac{1-iz}{z-i}\right| = 1 \Rightarrow |1-iz| = |z-i| \]

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

Step 3: Equate. \[ (1+y)^2 + x^2 = x^2 + (y-1)^2 \] Cancel \(x^2\): \[ (1+y)^2 = (y-1)^2 \] Expand: \[ 1 + 2y + y^2 = y^2 - 2y + 1 \] \[ 2y = -2y \Rightarrow 4y=0 \Rightarrow y=0 \] Thus \(z=x\) is real.