Q.
Let z1 and z2 be two complex numbers such that ∣z2∣=1 and z1+1z1−1=(z2+1z2−1)2, then find the value of ∣z1−1∣+∣z1+1∣.
518
107
Complex Numbers and Quadratic Equations
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Answer: 2
Solution:
∵∣z2∣=1⇒z2=eiθ=cosθ+isinθ ∴z2+1z2−1=cosθ+1+isinθcosθ−1+isinθ=2cos22θ+i⋅2sin2θcos2θ−2sin22θ+i⋅2sin2θcos2θ=cos2θisin2θ=itan2θ ∴z1+1z1−1=−tan22θ⇒z1−1=−z1tan22θ−tan22θ ⇒z1=1+tan22θ1−tan22θ=cosθ
which is purely real and lies in [−1,1] ∴∣z1−1∣+∣z1+1∣=2