Let z1 be a complex root of the equation anzn+an−1zn−1+⋯+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 A13<|z1|<1 anzn+an−1zn−1+⋯+a1z+a0=3 Putting z1 in the equation, |anzn1+an−1zn−11+⋯+a1z1+a0|=|3| ⇒|an||z1|n+|an−1||z1|n−1+⋯+|a1||z1|+|a0| ≥|anzn1+an−1zn−11+⋯+a1z1+a0| =3 ⇒|an||z1|n+|an−1||z1|n−1+⋯+|a1||z1|+|a0|≥3 ⇒2(|z1|n+|z1|n−1+⋯+|z1|+1)>3[∵|ai|<2] ⇒|z1|n+|z1|n−1+⋯+|z1|+1>32 ⇒1+|z1|+⋯+|z1|n−1+|z1|n>32
Now, 1+|z1|+⋯+|z1|n−1+|z1|n=1−|z1|n+11−|z1| So, 1−|z1|n+11−|z1|>32 ⇒2−2|z1|n+1>3−3|z1|(∵|z1|<1⇒1−|z1|>0) ⇒0<2|z1|n+1<3|z1|−1 ⇒3|z1|−1>0 ⇒|z1|>13