Let z be a complex number satisfying `z^(2) + 2zlambda + 1=0` , where `lambda ` is a parameter which can take any real value.
For every large value of `lambda` the roots are approximately.

Options

A. `-2lambda, 1//lambda`
B. `-lambda,-1//lambda`
C. `-2lambda, -(1)/(2lambda)`
D. none of these

Explanation

To solve the equation \( z^2 + 2z\lambda + 1 = 0 \) for large values of the parameter \( \lambda \), we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation can be identified in the standard form \( az^2 + bz + c = 0 \), where:
- \( a = 1 \)
- \( b = 2\lambda \)
- \( c = 1 \)

### Step 2: Apply the quadratic formula
The roots of a quadratic equation are given by the 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 \( b^2 - 4ac \):
\[ (2\lambda)^2 - 4 \cdot 1 \cdot 1 = 4\lambda^2 - 4 \]
Now substituting this back into the formula for \( z \):
\[ z = \frac{-2\lambda \pm \sqrt{4\lambda^2 - 4}}{2} \]
Factoring out the 4 from the square root:
\[ z = \frac{-2\lambda \pm 2\sqrt{\lambda^2 - 1}}{2} \]
This simplifies to:
\[ z = -\lambda \pm \sqrt{\lambda^2 - 1} \]

### Step 4: Analyze for large values of \( \lambda \)
For large values of \( \lambda \), we can approximate \( \sqrt{\lambda^2 - 1} \):
\[ \sqrt{\lambda^2 - 1} \approx \sqrt{\lambda^2} = \lambda \]
Thus, the roots become:
\[ z \approx -\lambda \pm \lambda \]

### Step 5: Calculate the approximate roots
Calculating the two cases:
1. \( z \approx -\lambda + \lambda = 0 \)
2. \( z \approx -\lambda - \lambda = -2\lambda \)

### Conclusion
For large values of \( \lambda \), the roots of the equation \( z^2 + 2z\lambda + 1 = 0 \) are approximately:
\[ z \approx 0 \quad \text{and} \quad z \approx -2\lambda \]