If `Re ((1)/(z)) = 3,` then z lies on

Options

A. circle with centre on Y-axis
B. circle with centre on X-axis not passing through origin
C. circle with centre on X-axis passing through origin
D. None of the above

Answer

To solve the problem, we start with the condition given in the question:

**Given:**
\[ \text{Re}\left(\frac{1}{z}\right) = 3 \]

Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. We want to find the condition on \( z \).

### Step 1: Express \( \frac{1}{z} \)

We can express \( \frac{1}{z} \) as follows:
\[ \frac{1}{z} = \frac{1}{x + iy} \]

### Step 2: Multiply by the conjugate

To simplify \( \frac{1}{z} \), we multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{1}{z} = \frac{1}{x + iy} \cdot \frac{x - iy}{x - iy} = \frac{x - iy}{x^2 + y^2} \]

### Step 3: Identify the real part

Now, we can separate the real and imaginary parts:
\[ \frac{1}{z} = \frac{x}{x^2 + y^2} - i \frac{y}{x^2 + y^2} \]
Thus, the real part of \( \frac{1}{z} \) is:
\[ \text{Re}\left(\frac{1}{z}\right) = \frac{x}{x^2 + y^2} \]

### Step 4: Set the real part equal to 3

According to the problem statement, we set the real part equal to 3:
\[ \frac{x}{x^2 + y^2} = 3 \]

### Step 5: Rearrange the equation

Rearranging gives us:
\[ x = 3(x^2 + y^2) \]
This can be rewritten as:
\[ 3x^2 + 3y^2 - x = 0 \]

### Step 6: Rearranging into standard form

Rearranging this equation, we have:
\[ 3x^2 + 3y^2 = x \]
\[ 3x^2 + 3y^2 - x = 0 \]

### Step 7: Completing the square

To express this in the standard form of a circle, we can complete the square:
\[ 3\left(x^2 + y^2\right) = x \]
Dividing through by 3:
\[ x^2 + y^2 = \frac{x}{3} \]
Rearranging gives:
\[ x^2 - \frac{x}{3} + y^2 = 0 \]
Completing the square for \( x \):
\[ \left(x - \frac{1}{6}\right)^2 + y^2 = \frac{1}{36} \]

### Step 8: Identify the center and radius

From the equation \( \left(x - \frac{1}{6}\right)^2 + y^2 = \left(\frac{1}{6}\right)^2 \), we can see that the center of the circle is \( \left(\frac{1}{6}, 0\right) \) and the radius is \( \frac{1}{6} \).

### Conclusion

Thus, the circle is centered at \( \left(\frac{1}{6}, 0\right) \) and has a radius of \( \frac{1}{6} \), which means it does not pass through the origin. Therefore, the correct answer is:

**Answer:** Circle with center on the x-axis and not passing through the origin.

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