If `omega = z//[z-(1//3)i] and |omega| = 1`, then find the locus of z.

Options

A. a straight line
B. a parabola
C. an ellipse
D. a circle

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: Write down the given equation
We have:
\[ \omega = \frac{z}{z - \frac{1}{3}i} \]
and it is given that \( |\omega| = 1 \).

### Step 2: Apply the modulus condition
Since \( |\omega| = 1 \), we can write:
\[ |\omega| = \left| \frac{z}{z - \frac{1}{3}i} \right| = 1 \]
This implies:
\[ |z| = |z - \frac{1}{3}i| \]

### Step 3: Express \( z \) in terms of its real and imaginary parts
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 from the modulus condition
From the equality \( |z| = |z - \frac{1}{3}i| \), we have:
\[ \sqrt{x^2 + y^2} = \sqrt{x^2 + \left(y - \frac{1}{3}\right)^2} \]

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

### Step 6: Simplify the equation
Cancel \( x^2 \) from both sides:
\[ y^2 = \left(y - \frac{1}{3}\right)^2 \]
Expanding the right-hand side:
\[ y^2 = y^2 - \frac{2y}{3} + \frac{1}{9} \]

### Step 7: Rearrange the equation
Subtract \( y^2 \) from both sides:
\[ 0 = -\frac{2y}{3} + \frac{1}{9} \]
Multiplying through by 9 to eliminate the fraction:
\[ 0 = -6y + 1 \]
Thus:
\[ 6y = 1 \quad \Rightarrow \quad y = \frac{1}{6} \]

### Step 8: Conclusion about the locus of \( z \)
Since \( y = \frac{1}{6} \) is constant, the locus of \( z \) is a horizontal line in the complex plane where the imaginary part is \( \frac{1}{6} \). Thus, we can express the locus as:
\[ \text{Locus of } z: \quad y = \frac{1}{6} \]