If `A=|z in C: z=x+ix-1` for all `x in R}` and `|z| le |omega|` for all z, `omega in A`, then z is equal to

Options

A. `1/2(1+i)`
B. `-1/2(1-i)`
C. `-1/2(1+i)`
D. `1/3(1-2i)`

Answer

To solve the problem, we need to analyze the given conditions step by step.

### Step 1: Understand the set A
The set \( A \) is defined as:
\[ A = \{ z \in \mathbb{C} : z = x + i(x - 1) \text{ for all } x \in \mathbb{R} \} \]
This means that any complex number \( z \) in \( A \) can be expressed in terms of a real number \( x \).

### Step 2: Express \( z \) in terms of \( x \)
From the definition of \( z \):
\[ z = x + i(x - 1) = x + ix - i \]
We can rewrite this as:
\[ z = x + ix - i = x + i(x - 1) \]

### Step 3: Identify the modulus condition
We are given the condition:
\[ |z| \leq |\omega| \quad \text{for all } z, \omega \in A \]
This means that the modulus of \( z \) must be less than or equal to the modulus of any \( \omega \) in the set \( A \).

### Step 4: Calculate the modulus of \( z \)
The modulus of \( z \) can be calculated as follows:
\[ |z| = |x + i(x - 1)| = \sqrt{x^2 + (x - 1)^2} \]
Calculating \( (x - 1)^2 \):
\[ (x - 1)^2 = x^2 - 2x + 1 \]
Thus,
\[ |z| = \sqrt{x^2 + (x^2 - 2x + 1)} = \sqrt{2x^2 - 2x + 1} \]

### Step 5: Find the minimum modulus
To find the minimum modulus, we need to minimize \( |z|^2 = 2x^2 - 2x + 1 \). This is a quadratic function in \( x \).

The vertex of a quadratic \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \):
\[ x = -\frac{-2}{2 \cdot 2} = \frac{1}{2} \]

### Step 6: Calculate the minimum modulus at the vertex
Substituting \( x = \frac{1}{2} \) into the expression for \( |z|^2 \):
\[ |z|^2 = 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) + 1 = 2 \cdot \frac{1}{4} - 1 + 1 = \frac{1}{2} \]
Thus,
\[ |z| = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

### Step 7: Find the corresponding \( y \) value
Now substituting \( x = \frac{1}{2} \) back into the expression for \( z \):
\[ z = \frac{1}{2} + i\left(\frac{1}{2} - 1\right) = \frac{1}{2} + i\left(-\frac{1}{2}\right) = \frac{1}{2} - \frac{1}{2}i \]

### Step 8: Final expression for \( z \)
Thus, we can express \( z \) as:
\[ z = \frac{1}{2}(1 - i) \]

### Conclusion
The final answer is:
\[ z = \frac{1}{2}(1 - i) \]