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Question

Let z1,z2 and z3 represent the vertices A, B and C of the triangle ABC in the Argand plane, such that |z1|=|z2|=|z3|=5, then the value of z1sin 2A+z2sin 2 B+z3sin 2C  is 


Solution


|z1|=|z2|=|z3|=5
|z|=5 is the circumcircle of triangle ABC 
AOB=2C,BOC=2A and COA=2B
We have,
z2z1=OBOAei2C=ei2C
Similarly,
z3z1=OCOAei2B=ei2B
 Now,
z1sin 2A+z2sin 2B+z3sin 2C
=z1(sin 2A+z2z1sin 2B+z3z1sin 2C)
=z1(sin 2A+ei2Csin 2B+e2iBsin 2C)
=z1(sin 2A + sin 2B cos 2C + i sin 2B sin 2C + sin 2C cos 2B - i sin 2C sin 2B)
=z1(sin 2A + sin (2B + 2C))
=z1(sin 2A + sin(2π2A))=0

Mathematics

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