If `z = x + yi, omega = (2-iz)/(2z-i) and |omega|=1`, find the locus of z in the complex plane

Answer

To solve the problem, we need to find the locus of the complex number \( z = x + yi \) given that \( \omega = \frac{2 - iz}{2z - i} \) and \( |\omega| = 1 \).

### Step-by-Step Solution:

1. **Substituting \( z \)**:
We start by substituting \( z = x + yi \) into the expression for \( \omega \):
\[ \omega = \frac{2 - i(x + yi)}{2(x + yi) - i} \]
Simplifying the numerator:
\[ 2 - ix - y = 2 - ix - y \]
So the numerator becomes:
\[ 2 - ix - y = 2 - y - ix \]

Now simplifying the denominator:
\[ 2(x + yi) - i = 2x + 2yi - i = 2x + (2y - 1)i \]

Thus, we have:
\[ \omega = \frac{2 - y - ix}{2x + (2y - 1)i} \]

2. **Finding the modulus of \( \omega \)**:
The modulus of \( \omega \) can be expressed as:
\[ |\omega| = \frac{|2 - y - ix|}{|2x + (2y - 1)i|} \]

Calculating the modulus of the numerator:
\[ |2 - y - ix| = \sqrt{(2 - y)^2 + x^2} \]

Calculating the modulus of the denominator:
\[ |2x + (2y - 1)i| = \sqrt{(2x)^2 + (2y - 1)^2} \]

Therefore, we have:
\[ |\omega| = \frac{\sqrt{(2 - y)^2 + x^2}}{\sqrt{4x^2 + (2y - 1)^2}} \]

3. **Setting the modulus equal to 1**:
Since \( |\omega| = 1 \), we set up the equation:
\[ \frac{\sqrt{(2 - y)^2 + x^2}}{\sqrt{4x^2 + (2y - 1)^2}} = 1 \]

Squaring both sides gives:
\[ (2 - y)^2 + x^2 = 4x^2 + (2y - 1)^2 \]

4. **Expanding both sides**:
Expanding the left-hand side:
\[ (2 - y)^2 + x^2 = 4 - 4y + y^2 + x^2 \]

Expanding the right-hand side:
\[ 4x^2 + (2y - 1)^2 = 4x^2 + (4y^2 - 4y + 1) = 4x^2 + 4y^2 - 4y + 1 \]

5. **Setting the equation**:
Now we have:
\[ 4 - 4y + y^2 + x^2 = 4x^2 + 4y^2 - 4y + 1 \]

Rearranging gives:
\[ 4 - 1 + y^2 + x^2 - 4y^2 = 4x^2 \]
Simplifying this leads to:
\[ 3 - 3y^2 + x^2 - 4x^2 = 0 \]
or:
\[ 3 - 3y^2 - 3x^2 = 0 \]

6. **Final expression**:
Dividing through by 3 gives:
\[ x^2 + y^2 = 1 \]

### Conclusion:
The locus of \( z \) in the complex plane is a circle centered at the origin with a radius of 1.