Then prove that the points A(z1),B(z2),C(3) and D(2) (taken in clockwise sense) are concyclic.
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Solution
(3−z12−z1)(2−z23−z2)=k ∴arg(3−z12−z1⋅2−z23−z2)=arg(k) Now if k>0, arg(k)=0 ∴arg(3−z12−z1)+arg(2−z23−z2)=0 or arg(3−z12−z1)=arg(3−z22−z2) Hence, chord DC subtends same angle at A and B. Therefore, point A, B, C, D are concyclic. Ans: 1