If |z−i|=1 and arg (z)=θ where θ∈(0,π2), then
cotθ−2z
i
|z−i|=1 represents a circle with centre at (0,1) and radius 1. As 0<θ<π2,θ lies the semi-circle in the first quadrant.
As ∠O=π2 and ∠POA=π2−θ, we get
cos(π2−θ)=OPOA⇒sinθ=|z|2
Also, since ∠POA=π2−θ
2iz=2i¯z|z|2=2i|z||z|2(cosθ−isinθ)=2|z|[cos(π2−θ)+isin(π2−θ)]=1sinθ[sinθ+icosθ]=1+icotθ⇒−2z=i−cotθ⇒cotθ−2z=i