Let `z_1, z_2a n dz_3` represent the vertices `A ,B ,a n dC` of the triangle `A B C ,` respectively, in the Argand plane, such that `|z_1|=|z_2|=5.` Prove that `z_1sin2A+z_2sin2B+z_3sin2C=0.`

Explanation

`|z_(1)|=|z_(2)|=|z_(3)|=5`
`implies|z|=5` is the circumcircle of triangle ABC
`impliesangleAOB=2C,angleBOC=2A," and "angleCOA=2B`

We have,
`(z_(2))/(z_(1))=(OB)/(OA)e^(i2C)=e^(i2C`
Similarly, `(z_(3))/(z_(1))=(OC)/(OA)e^(-i2B)=e^(-i2B`
Now, `z_(1)sin2A+z_(2)sin2B+z_(3)sin2C`
`=z_(1)(sin2A+(z_(2))/(z_(1))sin2B+(z_(3))/(z_(1))sin2C)`
`=z_(1)(sin2A+sin2Be^(i2C)+sin2Ce^(-2iB))`
`=z_(1)(sin2A+sin2Bcos2C+isin2Bsin2C+sin2Ccos2B-isin2Csin2B)`
`=z_(1)(sin2A+sin(2B+2C))`
`=z_(1)(sin2A+sin(2pi-2A))`
=0