AITS complex nos

z^2z_1+zz_2+z_3 = 0 \Rightarrow \overline{z^2z_1+zz_2+z_3}= \overline{z^2}\overline{z_1} + \overline{z}\overline{z_2} + \overline{z_3}=0

Now |z| =1 \Rightarrow \overline{z} = \frac{1}{z} and |z_1| = \sqrt 3 \Rightarrow \overline{z_1} = \frac{3}{z_1} etc.

So we have after dividing by 3 throughout, \frac{1}{z^2z_1} + \frac{1}{zz_2}+\frac{1}{ z_3}=0

So if p=z^2z_1; q=zz_2; r=z_3 we have

p+q+r=0; \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0 \Rightarrow pq+qr+rp=0

pq+qr+rp=0 \Rightarrow q(p+r) + rp = 0 \Rightarrow q^2=pr.

This means (zz_2)^2 = z^2z_1z_3 or z_2^2=z_1z_3

Hence z_1, z_2, z_3 are in GP

Find the mistake
z²z₁+ z₂z + z₃=0 is quadratic equation in z.Let its roots be 1,μ
μ+1=-(z₂z₁) and μ=z₃/z₁
Substituting value of μ ,z₃z₁+1=-z₂z₁ giving z₁+z₂+z₃=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