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w(1), w(2) be roots of (a+barc)z^(2)+(b+...

`w_(1)`, `w_(2)` be roots of `(a+barc)z^(2)+(b+barb)z+(bara+c)=0`. If `|z_(1)| lt 1`, `|z_(2)| lt 1`, then

A

`|w_(1)| lt 1`

B

`|w_(1)| = 1`

C

`|w_(2)| lt 1`

D

`|w_(2)| = 1`

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the given quadratic equation and its roots. Let's break down the steps systematically. ### Step-by-Step Solution: 1. **Identify the given quadratic equation**: The equation is given as: \[ (a + \bar{c})z^2 + (b + \bar{b})z + (\bar{a} + c) = 0 \] where \( \bar{c} \) and \( \bar{b} \) are the complex conjugates of \( c \) and \( b \), respectively. 2. **Roots of the equation**: Let \( w_1 \) and \( w_2 \) be the roots of the quadratic equation. By Vieta's formulas, we know: \[ w_1 + w_2 = -\frac{(b + \bar{b})}{(a + \bar{c})} \] \[ w_1 w_2 = \frac{(\bar{a} + c)}{(a + \bar{c})} \] 3. **Conditions on the roots**: We are given that \( |w_1| < 1 \) and \( |w_2| < 1 \). This implies that both roots lie inside the unit circle in the complex plane. 4. **Reciprocal roots**: If \( |w_1| < 1 \), then \( \frac{1}{w_1} \) will lie outside the unit circle, and similarly for \( w_2 \). This suggests that the roots \( w_1 \) and \( w_2 \) are related to their reciprocals. 5. **Transforming the equation**: We can transform the equation by substituting \( z = \frac{1}{w} \). The new equation becomes: \[ (a + \bar{c})\left(\frac{1}{w}\right)^2 + (b + \bar{b})\left(\frac{1}{w}\right) + (\bar{a} + c) = 0 \] Multiplying through by \( w^2 \) gives: \[ (a + \bar{c}) + (b + \bar{b})w + (\bar{a} + c)w^2 = 0 \] 6. **Roots of the transformed equation**: The roots of this new equation will be \( \frac{1}{w_1} \) and \( \frac{1}{w_2} \). Since \( |w_1| < 1 \) and \( |w_2| < 1 \), it follows that \( \left|\frac{1}{w_1}\right| > 1 \) and \( \left|\frac{1}{w_2}\right| > 1 \). 7. **Conclusion**: The roots \( w_1 \) and \( w_2 \) are such that their magnitudes are less than 1, indicating that they lie inside the unit circle in the complex plane. The reciprocal roots lie outside the unit circle. ### Final Answer: The correct options based on the conditions provided are those that reflect the relationship between the roots and their magnitudes.

To solve the problem, we need to analyze the given quadratic equation and its roots. Let's break down the steps systematically. ### Step-by-Step Solution: 1. **Identify the given quadratic equation**: The equation is given as: \[ (a + \bar{c})z^2 + (b + \bar{b})z + (\bar{a} + c) = 0 ...
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