`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

Options

A. `|w_(1)| lt 1`
B. `|w_(1)| = 1`
C. `|w_(2)| lt 1`
D. `|w_(2)| = 1`

Answer

B, D

Explanation

`(b,d)` `a+barc ne 0`
Now, `(a+barc)w_(1)^(2)+(b+barb)w_(1)+(bara+c)=0`
`implies (bara+c)w_(1)^(2)+(b+barb)barw_(1)+(a+barc)=0` ,brgt Since, `barw_(1) ne 0`, we get `(a+barc)(1)/(barw_(1)^(2))+(b+barb)(1)/(barw_(1))+(bara+c)=0`
Hence, `(1)/(barw_(1))` is also a root, `w_(2) ne (1)/(barw_(1))`, so `w_(1)=(1)/(barw_(1))`
`implies |w_(1)|=1`
Similarly `|w_(2)|=1`