If the imaginery part of `(z-3)/(e^(itheta))+(e^(itheta))/(z-3)` is zero, then `z` can lie on

Options

A. a circle with unit radius
B. a circle with radius `3` units
C. a straight line through the point `(3,0)`
D. a parabola with the vertex `(3,0)`

Explanation

To solve the problem, we need to analyze the expression given and find the conditions under which the imaginary part is zero. Let's go through the solution step by step.

### Step 1: Write the given expression
We are given the expression:

\[ \frac{z - 3}{e^{i\theta}} + \frac{e^{i\theta}}{z - 3} \]

### Step 2: Set the imaginary part to zero
We need to find when the imaginary part of this expression is zero.

### Step 3: Substitute \( z - 3 \)
Let \( z - 3 = r e^{i\phi} \), where \( r \) is the modulus and \( \phi \) is the argument of the complex number \( z - 3 \).

### Step 4: Rewrite the expression
Substituting \( z - 3 \) into the expression gives:

\[ \frac{r e^{i\phi}}{e^{i\theta}} + \frac{e^{i\theta}}{r e^{i\phi}} = r e^{i(\phi - \theta)} + \frac{1}{r} e^{i(\theta - \phi)} \]

### Step 5: Simplify the expression
This can be simplified to:

\[ r e^{i(\phi - \theta)} + \frac{1}{r} e^{-i(\phi - \theta)} \]

### Step 6: Find the imaginary part
The imaginary part of this expression is given by:

\[ \text{Im}\left(r e^{i(\phi - \theta)}\right) + \text{Im}\left(\frac{1}{r} e^{-i(\phi - \theta)}\right) \]

Using the property \(\text{Im}(e^{i\theta}) = \sin(\theta)\), we have:

\[ r \sin(\phi - \theta) + \frac{1}{r} \sin(\theta - \phi) = r \sin(\phi - \theta) - \frac{1}{r} \sin(\phi - \theta) \]

### Step 7: Set the imaginary part to zero
Setting the imaginary part to zero gives:

\[ (r - \frac{1}{r}) \sin(\phi - \theta) = 0 \]

### Step 8: Solve the equation
This equation can be satisfied in two cases:

1. \( r - \frac{1}{r} = 0 \)
2. \( \sin(\phi - \theta) = 0 \)

### Step 9: Analyze the first case
From \( r - \frac{1}{r} = 0 \):

\[ r^2 = 1 \implies r = 1 \]

This means that:

\[ |z - 3| = 1 \]

This indicates that \( z \) lies on a circle of radius 1 centered at \( 3 \) in the complex plane.

### Step 10: Analyze the second case
From \( \sin(\phi - \theta) = 0 \):

\[ \phi - \theta = n\pi \quad (n \in \mathbb{Z}) \]

This means that \( \phi = \theta + n\pi \), which indicates that the argument of \( z - 3 \) is aligned with the angle \( \theta \).

### Conclusion
From the analysis, we conclude that:

1. \( z \) lies on a circle of radius 1 centered at \( 3 \).
2. The argument of \( z - 3 \) can be any angle that is a multiple of \( \pi \) away from \( \theta \).

Thus, the final conclusion is that \( z \) can lie on a circle with radius 1 centered at \( 3 \) and also represents a straight line through the point \( 3 \) in the complex plane.