If `1,alpha,alpha^(2),………..,alpha^(n-1)` are the n, `n^(th)` roots of unity and `z_(1)` and `z_(2)` are any two complex numbers such that `sum_(r=0)^(n-1)|z_(1)+alpha^(R ) z_(2)|^(2)=lambda(|z_(1)|^(2)+|z_(2)|^(2))`, then `lambda=`

Options

A. n
B. `(n-1)`
C. `(n+1)`
D. 2n

Answer

To solve the problem, we need to analyze the given expression and simplify it step by step.

### Step-by-Step Solution:

1. **Understanding the Roots of Unity**:
The numbers \(1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\) are the \(n\)th roots of unity, where \(\alpha = e^{2\pi i / n}\). This means that \(\alpha^n = 1\) and the sum of all \(n\)th roots of unity is zero.

2. **Setting Up the Expression**:
We need to evaluate the expression:
\[ \sum_{r=0}^{n-1} |z_1 + \alpha^r z_2|^2 \]
Using the property of modulus, we can expand this as:
\[ |z_1 + \alpha^r z_2|^2 = (z_1 + \alpha^r z_2)(\overline{z_1 + \alpha^r z_2}) = |z_1|^2 + |z_2|^2 + z_1 \overline{\alpha^r z_2} + \overline{z_1} \alpha^r z_2 \]

3. **Summing Over \(r\)**:
Now we sum this expression from \(r = 0\) to \(n-1\):
\[ \sum_{r=0}^{n-1} |z_1 + \alpha^r z_2|^2 = \sum_{r=0}^{n-1} \left( |z_1|^2 + |z_2|^2 + z_1 \overline{\alpha^r z_2} + \overline{z_1} \alpha^r z_2 \right) \]
This can be simplified to:
\[ n |z_1|^2 + n |z_2|^2 + \sum_{r=0}^{n-1} \left( z_1 \overline{\alpha^r z_2} + \overline{z_1} \alpha^r z_2 \right) \]

4. **Evaluating the Sum of Roots**:
The sum \(\sum_{r=0}^{n-1} \alpha^r\) is equal to zero because it represents the sum of all \(n\)th roots of unity. Therefore:
\[ \sum_{r=0}^{n-1} z_1 \overline{\alpha^r z_2} = z_1 \overline{z_2} \sum_{r=0}^{n-1} \alpha^r = 0 \]
and similarly,
\[ \sum_{r=0}^{n-1} \overline{z_1} \alpha^r z_2 = 0 \]

5. **Final Expression**:
Thus, we have:
\[ \sum_{r=0}^{n-1} |z_1 + \alpha^r z_2|^2 = n |z_1|^2 + n |z_2|^2 \]
This can be factored as:
\[ n (|z_1|^2 + |z_2|^2) \]

6. **Setting the Equation**:
According to the problem statement:
\[ \sum_{r=0}^{n-1} |z_1 + \alpha^r z_2|^2 = \lambda (|z_1|^2 + |z_2|^2) \]
Therefore, we can equate:
\[ n (|z_1|^2 + |z_2|^2) = \lambda (|z_1|^2 + |z_2|^2) \]

7. **Solving for \(\lambda\)**:
Assuming \(|z_1|^2 + |z_2|^2 \neq 0\), we can divide both sides by \(|z_1|^2 + |z_2|^2\):
\[ \lambda = n \]

### Conclusion:
Thus, the value of \(\lambda\) is \(n\).