If `z= x + yi and omega = ((1- zi))/(z-i)`, then `|omega|=1` implies that in the complex plane

Options

A. z lies on the imaginary axis
B. z lies on the real axis
C. z lies on the unit circle
D. None of these

Answer

To solve the problem, we start with the given complex numbers:

Let \( z = x + yi \) and \( \omega = \frac{1 - zi}{z - i} \).

We are tasked with finding the conditions under which \( |\omega| = 1 \).

### Step 1: Set up the equation for modulus
Since \( |\omega| = 1 \), we can write:
\[ |\omega| = \left| \frac{1 - zi}{z - i} \right| = 1 \]
This implies:
\[ |1 - zi| = |z - i| \]

### Step 2: Calculate the modulus of the numerator and denominator
Now, we need to calculate the modulus of both sides:
1. For \( |1 - zi| \):
\[ |1 - zi| = |1 - (x + yi)i| = |1 - (xi + y)| = |1 - yi - x| = \sqrt{(1 - x)^2 + y^2} \]
2. For \( |z - i| \):
\[ |z - i| = |(x + yi) - i| = |x + (y - 1)i| = \sqrt{x^2 + (y - 1)^2} \]

### Step 3: Set the moduli equal
From the previous step, we have:
\[ \sqrt{(1 - x)^2 + y^2} = \sqrt{x^2 + (y - 1)^2} \]

### Step 4: Square both sides
Squaring both sides to eliminate the square roots gives:
\[ (1 - x)^2 + y^2 = x^2 + (y - 1)^2 \]

### Step 5: Expand both sides
Expanding both sides results in:
\[ (1 - 2x + x^2) + y^2 = x^2 + (y^2 - 2y + 1) \]
This simplifies to:
\[ 1 - 2x + x^2 + y^2 = x^2 + y^2 - 2y + 1 \]

### Step 6: Cancel common terms
We can cancel \( x^2 \) and \( y^2 \) from both sides:
\[ 1 - 2x = -2y + 1 \]

### Step 7: Simplify the equation
This simplifies to:
\[ -2x = -2y \]
or
\[ x = y \]

### Step 8: Interpret the result
The equation \( x = y \) represents a line in the complex plane where the real part is equal to the imaginary part. This line is the line \( y = x \), which is the 45-degree line through the origin.

### Conclusion
Thus, the condition \( |\omega| = 1 \) implies that \( z \) lies on the line \( y = x \) in the complex plane.

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