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Question

Let z1 be a complex root of the equation anzn+an1zn1++a1z+a0=3 (|z|<1), where |ai|<2 for i=0,1,2,,n. Then

A
13<|z1|<1
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B
|z1|<13
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C
12<|z1|<1
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D
|z1|<12
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Solution

The correct option is A 13<|z1|<1
anzn+an1zn1++a1z+a0=3
Putting z1 in the equation,
|anzn1+an1zn11++a1z1+a0|=|3|
|an||z1|n+|an1||z1|n1++|a1||z1|+|a0|
|anzn1+an1zn11++a1z1+a0|
=3
|an||z1|n+|an1||z1|n1++|a1||z1|+|a0|3
2(|z1|n+|z1|n1++|z1|+1)>3 [|ai|<2]
|z1|n+|z1|n1++|z1|+1>32
1+|z1|++|z1|n1+|z1|n>32

Now, 1+|z1|++|z1|n1+|z1|n=1|z1|n+11|z1|
So, 1|z1|n+11|z1|>32
22|z1|n+1>33|z1| (|z1|<11|z1|>0)
0<2|z1|n+1<3|z1|1
3|z1|1>0
|z1|>13

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