Z=1−2icosθ1+icosθZ=−Zˉ⇒1−2icosθ1+icosθ=−(1−2icosθ1+icosθ)(1+icosθ)(1−2icosθ)=−(1−2icosθ)(1+icosθ)(1+icosθ)(1+2icosθ)=−(1−2icosθ)(1−icosθ)1+3icosθ−2cos2θ=−(1−3icosθ−2cos2θ)2−4cos2θ=0⇒cos2θ=21⇒θ=−4π,−43π,4π,43π,45π,47π sum =3π
The sum of all possible values of θ∈[−π,2π], for which 1−2icosθ1+icosθ is purely imaginary, is equal
Held on 8 Apr 2024 · Verified 6 Jul 2026.
3π
2π
5π
4π
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