Q.

Let  A(ω2) , B(2iω)  and C(-4) be three points lying on the Argand plane. Now a point P is taken on the circumcircle of the triangle ABC such that PA. BC = PC. AB, (where P, A, B, C are in order). If z is the complex number associated with the mid – point of PB, then the value of |Z|2  is, ( ω is a non – real cube rot of unity) 
 

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Detailed Solution

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Applying Rotation formula at P and B we get
p+4pω2×2iωω22iω+4=1p(2iωω2+2iω+4)=ω2(2iω+4)4(2iωω2)p=2i8iω+8ω24iωω2+4=2(i4iω+4ω2)(4ω2i+4iω)×iω=2iω

So,   z=p+b2=0|Z|2=0
 

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