Consider the region S of complex numbers a such that `|z^(2) - az + 1|=1`, where `|z|=1` . Then area of S in the Argand plane is

Options

A. `pi+8`
B. `pi + 4`
C. `2pi+ 4`
D. `pi + 6`

Explanation

To solve the problem, we need to find the area of the region \( S \) defined by the equation \( |z^2 - az + 1| = 1 \) for \( |z| = 1 \).

### Step-by-Step Solution:

1. **Understanding the Equation**:
We start with the equation \( |z^2 - az + 1| = 1 \). Since \( |z| = 1 \), we can express \( z \) in terms of the exponential form:
\[ z = e^{i\theta} \]
where \( \theta \) is a real number.

2. **Substituting \( z \)**:
Substitute \( z = e^{i\theta} \) into the equation:
\[ |(e^{i\theta})^2 - a e^{i\theta} + 1| = 1 \]
This simplifies to:
\[ |e^{2i\theta} - ae^{i\theta} + 1| = 1 \]

3. **Rearranging the Equation**:
We can rewrite the expression inside the modulus:
\[ |e^{2i\theta} - ae^{i\theta} + 1| = |(e^{i\theta} - \frac{a}{2})^2 + (1 - \frac{a^2}{4})| \]
This is a transformation into a form that resembles a circle.

4. **Identifying the Circle**:
The equation \( |z^2 - az + 1| = 1 \) can be interpreted geometrically. The center of the circle is at \( \frac{a}{2} \) and the radius is \( 1 \).

5. **Finding the Range of \( a \)**:
Since \( |z| = 1 \), the term \( 2\cos\theta \) varies from \(-2\) to \(2\). Thus, the center of the circle \( \frac{a}{2} \) must lie within this range for the circle to be valid.

6. **Calculating Area**:
The area of the circle can be calculated using the formula for the area of a circle:
\[ \text{Area} = \pi r^2 \]
where \( r \) is the radius. Here, \( r = 1 \), so the area of the circle is:
\[ \text{Area} = \pi \cdot 1^2 = \pi \]

7. **Combining Areas**:
The total area \( S \) consists of the area of two semicircles and a rectangle. The rectangle has a width of \( 4 \) and a height of \( 2 \), giving it an area of \( 8 \). Thus, the total area is:
\[ \text{Total Area} = 2 \cdot \frac{\pi}{2} + 8 = \pi + 8 \]

### Final Answer:
The area of the region \( S \) in the Argand plane is:
\[ \boxed{\pi + 8} \]