Given π<α,β<2π
1+i(2sinα)1−isinα is purely imaginary
⇒1+4sin2α(1−isinα)(1−i(2sinα)) is purely imaginary
⇒1+4sin2α1−2sin2α=0
⇒sin2α=21
⇒α=45π,47π
Also 1+i(−2cosβ)1+icosβ is purely real
⇒1+4cos2β(1+icosβ)(1+2icosβ) is purely real
⇒3cosβ=0
⇒β=23π
Now Zαβ=sin2α+icos2β
⇒Zαβ=sin25π+icos3π=1−i
or Zαβ=sin27π+icos3π=−1−i
(α,β)∈S∑(iZαβ+iZˉαβ1)=[i(1−i)+i(1+i)1]+[i(−1−i)+i(−1+i)1]
=[i+1−2i(1−i)]+[−i+1+2i(1+i)]
=2−1=1