Answer (A, B) z2 + 2λz + 1 = 0 ∴ z = -λ ± √λ2−1 ∴ z = -λ ± i √1−λ2 if - 1 < λ < 1 or z = -λ ± Ki , where √1−λ2 = K > 0 ∴ z + λ = Ki or
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z+λ
= - Ki ∴ (z + λ) (
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z+λ
) = K2 |z + λ| = K ∴ z lies on a circle with centre –λ and radius K If λ > 1 , we have z = -λ + √1−λ2 or -λ - √1−λ2 ∴ z = - λ + K1 or z = -λ - K1 where K12 = λ2 - 1 ⇒ z = - λ ± K1 if z1 and z2 are roots then |z1| = |−λ+K1| and |z2| = |λ + K1| |z1||z2| = |K12−λ2| = 1 ⇒ |z1| < 1 then |z2| > 1 ∴ If one root inside circle then other is outside.