TargetIIT

plz solve

A(z[ss]1[/ss]) , B(z[ss]2[/ss]) and C(z[ss]3[/ss]) are 3 points  lying on the curve
|z|=[sqrt]3[/sqrt]

1. If z[p]2[/p]z[ss]1[/ss]+ z[ss]2[/ss]z + z[ss]3[/ss]=0 has one root of modulus unity.
z[ss]1[/ss],z[ss]2[/ss],z[ss]3[/ss]are in:
a)AP
b)Gp
c)HP
D)all of the above

2.If z[p]2[/p]z[ss]1[/ss]+ z[ss]2[/ss]z + z[ss]3[/ss]=0 and
z[p]2[/p]z[ss]2[/ss]+ z[ss]3[/ss]z + z[ss]1[/ss]=0 each have a common root unity  then
triangle ABC is:
a)isosceles
b)equilateral
c)right
d)scalene

[hide]ans
1.b
2.b[/hide]

[im]http://codecogs.izyba.com/gif.latex?z%5E2z_1+zz_2+z_3%20%3D%200%20%5CRightarrow%20%5Coverline%7Bz%5E2z_1+zz_2+z_3%7D%3D%20%5Coverline%7Bz%5E2%7D%5Coverline%7Bz_1%7D%20+%20%5Coverline%7Bz%7D%5Coverline%7Bz_2%7D%20+%20%5Coverline%7Bz_3%7D%3D0[/im]

Now [im]http://codecogs.izyba.com/gif.latex?%7Cz%7C%20%3D1%20%5CRightarrow%20%5Coverline%7Bz%7D%20%3D%20%5Cfrac%7B1%7D%7Bz%7D[/im] and [im]http://codecogs.izyba.com/gif.latex?%7Cz_1%7C%20%3D%20%5Csqrt%203%20%5CRightarrow%20%5Coverline%7Bz_1%7D%20%3D%20%5Cfrac%7B3%7D%7Bz_1%7D[/im] etc.

So we have after dividing by 3 throughout, [im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7B1%7D%7Bz%5E2z_1%7D%20+%20%5Cfrac%7B1%7D%7Bzz_2%7D+%5Cfrac%7B1%7D%7B%20z_3%7D%3D0[/im]

So if [im]http://codecogs.izyba.com/gif.latex?p%3Dz%5E2z_1%3B%20q%3Dzz_2%3B%20r%3Dz_3[/im] we have

[im]http://codecogs.izyba.com/gif.latex?p+q+r%3D0%3B%20%5Cfrac%7B1%7D%7Bp%7D%20+%20%5Cfrac%7B1%7D%7Bq%7D%20+%20%5Cfrac%7B1%7D%7Br%7D%20%3D%200%20%5CRightarrow%20pq+qr+rp%3D0[/im]

[im]http://codecogs.izyba.com/gif.latex?pq+qr+rp%3D0%20%5CRightarrow%20q%28p+r%29%20+%20rp%20%3D%200%20%5CRightarrow%20q%5E2%3Dpr.[/im]

This means [im]http://codecogs.izyba.com/gif.latex?%28zz_2%29%5E2%20%3D%20z%5E2z_1z_3[/im] or [im]http://codecogs.izyba.com/gif.latex?z_2%5E2%3Dz_1z_3[/im]

Hence [im]http://codecogs.izyba.com/gif.latex?z_1%2C%20z_2%2C%20z_3[/im] are in GP

Find the mistake
z[p]2[/p]z[ss]1[/ss]+ z[ss]2[/ss]z + z[ss]3[/ss]=0  is quadratic equation in z.Let its roots be 1,μ
μ+1=-([frac]z[ss]2[/ss][/]z[ss]1[/ss][/frac])    and  μ=z[ss]3[/ss]/z[ss]1[/ss]
Substituting value of μ ,[frac]z[ss]3[/ss][/]z[ss]1[/ss][/frac]+1=-[frac]z[ss]2[/ss][/]z[ss]1[/ss][/frac]    giving  z[ss]1[/ss]+z[ss]2[/ss]+z[ss]3[/ss]=0

@prophet sir plz help, I can't understand

"p+q+r=0   therefore

1/p  + 1/q + 1/r =0  .............."

how??

@cipher jus see wat are p ,q and r

oops[4][4]

thanks to everyone who answered. I got it now