Let f(z) is of the form `alpha z +beta` , where `alpha, beta,z` are complex numbers such that `|alpha| ne |beta|`.f(z) satisfies following properties :
(i)If imaginary part of z is non zero, then `f(z)+bar(f(z)) = f(barz)+ bar(f(z))`
(ii)If real part of of z is zero , then `f(z)+bar(f(z)) =0`
(iii)If z is real , then `bar(f(z)) f(z) gt (z+1)^2 AA z in R`
`(4x^2)/((f(1)-f(-1))^2)+y^2/((f(0))^2)=1 , x , y in R`, in (x,y) plane will represent :

Options

A. hyperbola
B. circle
C. ellipse
D. pair of line

Answer

To solve the problem step by step, we will analyze the properties of the function \( f(z) = \alpha z + \beta \) where \( \alpha, \beta, z \) are complex numbers, and derive the required equation in the \( (x,y) \) plane.

### Step 1: Analyze the properties of \( f(z) \)

Given the properties of \( f(z) \):

1. **Property (i)**: If the imaginary part of \( z \) is non-zero,
\[ f(z) + \overline{f(z)} = f(\overline{z}) + \overline{f(z)} \]

2. **Property (ii)**: If the real part of \( z \) is zero,
\[ f(z) + \overline{f(z)} = 0 \]

3. **Property (iii)**: If \( z \) is real,
\[ \overline{f(z)} f(z) > (z + 1)^2 \quad \forall z \in \mathbb{R} \]

### Step 2: Substitute and simplify the first property

Substituting \( f(z) = \alpha z + \beta \) and \( \overline{f(z)} = \overline{\alpha} \overline{z} + \overline{\beta} \):

From property (i):
\[ \alpha z + \beta + \overline{\alpha} \overline{z} + \overline{\beta} = \alpha \overline{z} + \beta + \overline{\alpha} z + \overline{\beta} \]

Cancelling \( \beta \) and \( \overline{\beta} \):
\[ \alpha z + \overline{\alpha} \overline{z} = \alpha \overline{z} + \overline{\alpha} z \]

Rearranging gives:
\[ (\alpha - \overline{\alpha}) z = (\overline{\alpha} - \alpha) \overline{z} \]

This implies:
\[ (\alpha - \overline{\alpha})(z - \overline{z}) = 0 \]

Since the imaginary part of \( z \) is non-zero, we conclude:
\[ \text{Im}(\alpha) = 0 \quad \Rightarrow \quad \alpha \text{ is real.} \]

### Step 3: Analyze the second property

From property (ii):
\[ f(z) + \overline{f(z)} = 0 \quad \text{when } \text{Re}(z) = 0 \]

Substituting gives:
\[ \alpha z + \beta + \overline{\alpha} \overline{z} + \overline{\beta} = 0 \]

Since \( \alpha \) is real, \( \overline{\alpha} = \alpha \):
\[ \alpha z + \alpha \overline{z} + \beta + \overline{\beta} = 0 \]

This implies:
\[ \beta + \overline{\beta} = 0 \quad \Rightarrow \quad \text{Re}(\beta) = 0 \quad \Rightarrow \quad \beta \text{ is purely imaginary.} \]

### Step 4: Write \( \alpha \) and \( \beta \)

Let:
\[ \alpha = a \quad (a \in \mathbb{R}), \quad \beta = bi \quad (b \in \mathbb{R}) \]

### Step 5: Evaluate \( f(1), f(-1), f(0) \)

Calculating:
- \( f(1) = a + bi \)
- \( f(-1) = -a + bi \)
- \( f(0) = bi \)

### Step 6: Substitute into the final equation

The final equation is:
\[ \frac{4x^2}{(f(1) - f(-1))^2} + \frac{y^2}{(f(0))^2} = 1 \]

Calculating \( f(1) - f(-1) \):
\[ f(1) - f(-1) = (a + bi) - (-a + bi) = 2a \]

Calculating \( (f(0))^2 \):
\[ (f(0))^2 = (bi)^2 = -b^2 \]

Substituting these into the equation gives:
\[ \frac{4x^2}{(2a)^2} + \frac{y^2}{-b^2} = 1 \]

This simplifies to:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Conclusion

This is the equation of a hyperbola.