Key Concepts and Formulas

• Modulus of a Complex Number: For $z = x + iy$, $|z| = \sqrt{x^2 + y^2}$. Geometrically, $|z|$ represents the distance of the point $(x, y)$ from the origin.
• Modulus of a Quotient: For complex numbers $z_1$ and $z_2$ (where $z_2 \neq 0$), $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$.
• Geometric Interpretation of $|z - z_0|$: The expression $|z - z_0|$ represents the distance between the complex number $z$ and the fixed complex number $z_0$ in the Argand plane. If $|z - z_1| = |z - z_2|$, then $z$ lies on the perpendicular bisector of the line segment joining $z_1$ and $z_2$.

Step-by-Step Solution

Step 1: Applying the Modulus Property to the Given Equation

We are given $\omega = \frac{z}{z - \frac{1}{3}i}$ and $|\omega| = 1$. Our goal is to find the locus of $z$. We will start by applying the modulus property of quotients to the expression for $\omega$.

$|\omega| = \left| \frac{z}{z - \frac{1}{3}i} \right| = \frac{|z|}{\left|z - \frac{1}{3}i\right|}$

Since $|\omega| = 1$, we have

$\frac{|z|}{\left|z - \frac{1}{3}i\right|} = 1$

Step 2: Equating the Moduli and Geometric Interpretation

Multiplying both sides by $\left|z - \frac{1}{3}i\right|$ (assuming $z \neq \frac{1}{3}i$), we get

$|z| = \left|z - \frac{1}{3}i\right|$

This equation states that the distance of $z$ from the origin $(0, 0)$ is equal to its distance from the point $\left(0, \frac{1}{3}\right)$ in the Argand plane. Geometrically, this means that $z$ lies on the perpendicular bisector of the line segment joining the origin and the point $\left(0, \frac{1}{3}\right)$. Therefore, we expect $z$ to lie on a straight line.

Step 3: Deriving the Equation of the Locus (Algebraic Method)

Let $z = x + iy$, where $x$ and $y$ are real numbers. Substituting this into the equation $|z| = \left|z - \frac{1}{3}i\right|$, we get

$|x + iy| = \left|x + iy - \frac{1}{3}i\right|$ $|x + iy| = \left|x + i\left(y - \frac{1}{3}\right)\right|$

Using the definition of the modulus, we have

$\sqrt{x^2 + y^2} = \sqrt{x^2 + \left(y - \frac{1}{3}\right)^2}$

Squaring both sides, we get

$x^2 + y^2 = x^2 + \left(y - \frac{1}{3}\right)^2$

Subtracting $x^2$ from both sides, we have

$y^2 = \left(y - \frac{1}{3}\right)^2$

Expanding the right side, we get

$y^2 = y^2 - \frac{2}{3}y + \frac{1}{9}$

Subtracting $y^2$ from both sides, we have

$0 = -\frac{2}{3}y + \frac{1}{9}$

Solving for $y$, we get

$\frac{2}{3}y = \frac{1}{9}$ $y = \frac{1}{9} \cdot \frac{3}{2} = \frac{1}{6}$

Therefore, $y = \frac{1}{6}$.

Step 4: Identifying the Locus

The equation $y = \frac{1}{6}$ represents a straight line parallel to the x-axis. This confirms our geometric intuition from Step 2.

Common Mistakes & Tips

• Remember Modulus Properties: Correctly apply properties such as $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$.
• Geometric Interpretation: Visualize complex number operations on the Argand plane to gain intuition.
• Algebraic Precision: Be careful with algebraic manipulations, especially when squaring and simplifying.

Summary

By using the properties of the complex modulus and converting the given equation into a distance relationship, we found that the locus of $z$ is a straight line. Specifically, $z$ lies on the line $y = \frac{1}{6}$, which is the perpendicular bisector of the line segment joining the origin and the point $\left(0, \frac{1}{3}\right)$.

The final answer is \boxed{a straight line}, which corresponds to option (C).