Find the locus of z if `omega= (z)/(z- (1)/(3)i), |omega| =1`

Answer

To find the locus of \( z \) given that \( \omega = \frac{z}{z - \frac{1}{3}i} \) and \( |\omega| = 1 \), we can follow these steps:

### Step 1: Set up the equation
We start with the given equation:
\[ \omega = \frac{z}{z - \frac{1}{3}i} \]
Given that \( |\omega| = 1 \), we can express this as:
\[ |\omega| = 1 \implies \left| \frac{z}{z - \frac{1}{3}i} \right| = 1 \]

### Step 2: Apply the modulus property
Using the property of modulus, we can rewrite the equation:
\[ |z| = |z - \frac{1}{3}i| \]

### Step 3: Substitute \( z \) in terms of \( x \) and \( y \)
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then we have:
\[ |z| = \sqrt{x^2 + y^2} \]
and
\[ |z - \frac{1}{3}i| = |x + i(y - \frac{1}{3})| = \sqrt{x^2 + \left(y - \frac{1}{3}\right)^2} \]

### Step 4: Set up the equation with the moduli
Now, we can equate the two moduli:
\[ \sqrt{x^2 + y^2} = \sqrt{x^2 + \left(y - \frac{1}{3}\right)^2} \]

### Step 5: Square both sides
Squaring both sides to eliminate the square roots gives us:
\[ x^2 + y^2 = x^2 + \left(y - \frac{1}{3}\right)^2 \]

### Step 6: Expand and simplify
Expanding the right side:
\[ x^2 + y^2 = x^2 + \left(y^2 - \frac{2y}{3} + \frac{1}{9}\right) \]
This simplifies to:
\[ x^2 + y^2 = x^2 + y^2 - \frac{2y}{3} + \frac{1}{9} \]

### Step 7: Cancel terms and rearrange
Cancel \( x^2 + y^2 \) from both sides:
\[ 0 = -\frac{2y}{3} + \frac{1}{9} \]
Rearranging gives:
\[ \frac{2y}{3} = \frac{1}{9} \]

### Step 8: Solve for \( y \)
Multiplying both sides by 3:
\[ 2y = \frac{1}{3} \]
Dividing by 2:
\[ y = \frac{1}{6} \]

### Step 9: Conclusion
The locus of \( z \) is the line:
\[ y = -\frac{1}{6} \]