Let z be a complex number satisfying `z^(2) + 2zlambda + 1=0` , where `lambda ` is a parameter which can take any real value.
The roots of this equation lie on a certain circle if

Options

A. `-1 lt lambda lt 1`
B. `lambda gt 1`
C. `lambda lt 1`
D. none of these

Explanation

To solve the problem, we need to analyze the quadratic equation given by:

\[ z^2 + 2z\lambda + 1 = 0 \]

where \( \lambda \) is a real parameter. We want to find the conditions under which the roots of this equation lie on a certain circle.

### Step 1: Identify the coefficients
The given quadratic equation can be identified as:
- \( a = 1 \)
- \( b = 2\lambda \)
- \( c = 1 \)

### Step 2: Use the quadratic formula
The roots of a quadratic equation \( az^2 + bz + c = 0 \) can be found using the quadratic formula:

\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substituting the values of \( a \), \( b \), and \( c \):

\[ z = \frac{-2\lambda \pm \sqrt{(2\lambda)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]

### Step 3: Simplify the expression
Calculating the discriminant:

\[ (2\lambda)^2 - 4 \cdot 1 \cdot 1 = 4\lambda^2 - 4 \]

Thus, the roots become:

\[ z = \frac{-2\lambda \pm \sqrt{4(\lambda^2 - 1)}}{2} \]

This simplifies to:

\[ z = -\lambda \pm \sqrt{\lambda^2 - 1} \]

### Step 4: Express the roots in terms of real and imaginary parts
Now, we can analyze the roots further. The expression \( \sqrt{\lambda^2 - 1} \) can be real or imaginary depending on the value of \( \lambda \).

1. If \( \lambda^2 - 1 \geq 0 \) (i.e., \( \lambda \leq -1 \) or \( \lambda \geq 1 \)), then the roots are real.
2. If \( \lambda^2 - 1 < 0 \) (i.e., \( -1 < \lambda < 1 \)), then the roots are complex.

For the case where \( -1 < \lambda < 1 \):

\[ z = -\lambda \pm i\sqrt{1 - \lambda^2} \]

### Step 5: Identify the circle
Now, let’s express \( z \) in the standard form \( a + ib \):

- Real part \( a = -\lambda \)
- Imaginary part \( b = \sqrt{1 - \lambda^2} \)

To find the condition for the roots to lie on a circle, we need to check if:

\[ a^2 + b^2 = 1 \]

Calculating \( a^2 + b^2 \):

\[ (-\lambda)^2 + \left(\sqrt{1 - \lambda^2}\right)^2 = \lambda^2 + (1 - \lambda^2) = 1 \]

This shows that the roots lie on a circle of radius 1 centered at the origin.

### Conclusion
The roots of the equation \( z^2 + 2z\lambda + 1 = 0 \) lie on a circle of radius 1 centered at the origin if \( \lambda \) lies in the interval \( (-1, 1) \).